How many moles of NiCl2 can be formed in the reaction of 7.00 mol of Ni and 14.0 mol of HCl?

Answers

Answer 1

Answer:

since the concentration of limiting reactant are the same for both nickel and hydrochloric acid, they both will produce the same amount if Nickel Chloride

How Many Moles Of NiCl2 Can Be Formed In The Reaction Of 7.00 Mol Of Ni And 14.0 Mol Of HCl?
Answer 2

7 mols of [tex]NiCl_2[/tex] formed in a reaction of 7 mol of Ni and 14 mol of HCl.

The moles of Ni = 7 mol

The mols of HCl = 14 mol

It is required to calculate moles of [tex]NiCl_2[/tex]

What is a mole?

A mole corresponds to the mass of a substance that contains [tex]6.023 \times 10^{23}[/tex]particles of the substance. The mole is the SI unit for the amount of a substance. Its symbol is mol.

The reaction of Ni and HCl occurs as

[tex]Ni +2 HCl \to NiCl_2 + H_2[/tex]

Since the concentration of limiting reactant are the same for both nickel and hydrochloric acid, they both will produce the same amount of Nickel Chloride.

If we have 7 mols of Ni and 14 mols of HCl, then when the 7 moles of Ni has reacted, we will still have 7 moles of HCl unreacted.

So, the moles of [tex]NiCl_2[/tex] formed equal to the moles of Ni and HCl reacted.

Therefore, 7 mols of [tex]NiCl_2[/tex] formed in a reaction of 7 mol of Ni and 14 mol of HCl.

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Related Questions

Analyze the transition of a photon

Answers

Photons may be generated by the transition of an electron from one energy level in an atom or molecule to a lower energy level. Photons may be absorbed as they cause an electron to be raised from a lower energy level to a higher energy level (in an atom or molecule).
The photon itself does not undergo a transition of energy: it either exists (with an energy defined by its wavelength), or it doesn't exist (it was destroyed!). You could say that the emitting or absorbing atom/molecule/etc. undergoes a change, or transition, in energy. But "transition" is usually used as a name for the process of jumping in energy.
Hope it help

what is meant by density​

Answers

Answer:

The degree of compactness of a substance

The mole fraction of NaCl in an
aqueous solution is 0.132. How
many moles of NaCl are present in
1 mole of this solution?
Molar Mass
NaCl: 58.44 g/mol
H2O: 18.016 g/mol

Answers

Answer:

Moles of water are 0.868

Explanation:

Match the change to its definition.


Name of change Change

condenation gas to solid

freezing solid to liquid

melting gas to liquid

evaporation liquid to gas

sublimation solid to gas

deposition liquid to solid

Answers

Answer:I didnt understand you have already done it perfect

Consider the constitutional isomers 2-methylbut-1-ene, 2-methylbut-2-ene, and 3-methylbut-1-ene. When each of these alkenes is subjected to catalytic hydrogenation (H2, Pt), a single product results. Which of the following best describes the structural relationship among these products?

a. the product are cis-trans isomers.
b. the product are identical.
c. the product are constitutional isomers.
d. the product are enantiomers.
e. the product are diastereomers.

Answers

Answer:

Explanation:

I am almost sure that the products are identical.

Four atoms and/or ions are sketched below in accordance with their relative atomic and/or ionic radii. Which of the following sets of species are compatible with the sketch?
Explain. (a) C,Ca2+,Cl−,Br−;
(b) Sr4, Cl,Br−,Na+

(d) Al,Ra2+,Zr2+

(c) Y,K,Ca,Na+, Mg2+;

e) Fe,Rb,Co,Cs


Answers

Answer:

Hence the correct option is an option (b) Sr4, Cl,Br−,Na+.

Explanation:    

Bromine and chlorine belong to an equivalent group. As we go down the group the dimensions increases which too there's a charge on the bromine atom. therefore the size of the Br- is going to be larger in comparison to the chlorine atom.

Sr atom is within the second group, and also it's below the above-mentioned atoms.so Sr is going to be the larger one among all the atoms.

Sodium and chlorine belong to an equivalent period .size decrease from left to right. but due to the charge on sodium its size decreases and there's an opportunity that Na+ size could be adequate for Cl.      

Here we finally assume that two atoms are of an equivalent size (Na+ and Cl) which are less in size compared to the opposite two(Sr and Br-) during which one is greater (Sr)and the opposite is smaller(Br-).

A 18.0 L gas cylinder is filled with 6.20 moles of gas. The tank is stored at 33 ∘C . What is the pressure in the tank?
Express your answer to three significant figures and include the appropriate units.

Answers

Answer:

8.65 atm

Explanation:

Using ideal law equation;

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (Latm/molK)

T = temperature (K)

According to the information given in this question;

V = 18.0 L

n = 6.20 moles

R = 0.0821 Latm/molK

T = 33°C = 33 + 273 = 306K

P = ?

Using PV = nRT

P × 18 = 6.20 × 0.0821 × 306

18P = 155.76

P = 155.76/18

P = 8.65 atm

How many chromosomes do we not understand?

Answers

Answer:

we don't understand why humans have only 46 chromosomes

Answer:

46 chromosomes is what we don't understand

how is the akin of frog similar to a fish​

Answers

Answer:

Have you ever touched a fish? Most fish will feel a bit rough - due to their scales. Some, like sharks, will feel like sandpaper. Even fish with small, smoother scales will feel a bit like that. Amphibians don’t have scales, and most species will be wet to some degree - they have to keep their skin moist or they’ll die. A few groups, like toads and newts, have rougher skin, which is heavier and thicker, which allows them to retain moisture better away from water.

Functionally, the big thing about amphibian skin is that it is semi-permeable. Amphibians can breathe through their skin - all amphibians can get some oxygen through their skin, but some species of salamanders get all their oxygen that way - they have no lungs or gills. The skin can also allow water in - sort of like a paper towel. The bad thing is that other chemicals can pass through the skin, too - pollutants and other chemicals tend to affect amphibians far more than they do other groups.

Amphibians also shed their skin - fish do not. People don’t tend to see frogs shedding their skin often, though, since they eat it to regain nutrients and other resources in the skin.

Finally, since amphibian skin offers no defense against predators in the way that scales do, and limited barrier against disease the way non-amphibian skin does (shedding helps), the skin of many amphibians contain toxins, and some of them have anti-fungal properties (typically due to symbiotic bacteria). Many species have evolved chemical defenses in the skin, while others have glands that produce toxins that can be secreted outside of the skin.

The skin can withstand dessication more than the fish.

They have moist skin used as respiratory surface during deep sleep / hibernation.

They have moist skin due to secretion of mucus by glands under the skin.

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Two common methods to generate an aldehyde is by oxidation of an alcohol and through ozonolysis.

a. True
b. False

Answers

Answer:

a. True.

Explanation:

Only primary and secondary alcohols can oxidise to give an aldehyde. But a weak oxidizing agent must be used to prevent formation of a carboxylic acid or ketone.

weak oxidizing agents: Chromyl chloride, silver/oxygen/500°C

take an example of ethanol:

[tex]{ \bf{CH _{3} CH_{2}OH \: \: \frac{Ag/O_{2} }{500 \degree C} > \: \:CH _{3} CHO}}[/tex]

[tex]{ \sf{CH _{3} CHO \: \: is \: ethanal}} [/tex]

By ozonolysis:

Here, reactants are Ozone gas, Carbon tetrachloride at a temperature (<20°C), ethanoic acid, zinc and water.

take an example of propanol:

if it undergoes ozonolysis, it gives ethanal and methanal.

Answer:

A. True

Explanation:

Only primary and secondary alcohols can oxidise to give an aldehyde. But a weak oxidizing agent must be used to prevent formation of a carboxylic acid or ketone.

weak oxidizing agents: Chromyl chloride, silver/oxygen/500°C

take an example of ethanol:

By ozonolysis:

Here, reactants are Ozone gas, Carbon tetrachloride at a temperature (<20°C), ethanoic acid, zinc and water.

take an example of propanol:

if it undergoes ozonolysis, it gives ethanal and methanal.

In the graphic, 195 represents the _______.

195 Pt
78

A. Atomic Mass
B. Atomic Number
C. Neutron Number​

Answers

Answer:

ITS ANSWER IS

OPTION B. ATOMIC NUMBER

HI HAVE A NICE DAY

I performed an experiment and mixed copper nitrate and potassium iodide. When they reacted, they formed a precipitate, even though solubility rules would not lead one to predict so because potassium nitrate is obviously soluble and so should copper (II) iodide. One can deduce from the formation of a precipitate that copper is reduced. Write a proposed reaction for the oxidation reduction of copper (II) iodide. Justify the choice of the substance that reduces the copper based on experimental evidence. Also, justify the choice using the atomic structure of potassium ion and iodide ion.

Answers

Answer:

2Cu2^+ + 2I^- ----> 2Cu^+ + I2

Explanation:

The reaction performed in the experiment is;

2 Cu(NO3)2 + 4 KI → 2 CuI (s) + 4 KNO3 + I2

The iodide ions reduces Cu^2+ to Cu^+ which is insoluble in water hence the precipitate. This is so because iodine is a good oxidizing agent seeing that it requires one electron to fill its outermost shell. Potassium on the other hand is a good reducing agent since it easily looses its one electron.

The oxidation - reduction equation is as follows;

2Cu2^+ + 2e ----> 2Cu^+ reduction half equation

2I^- ----> I2 + 2e. Oxidation half equation

Balanced redox reaction equation;

2Cu2^+ + 2I^- ----> 2Cu^+ + I2

A certain atom has an energy state 3.50 eV above the ground state. When excited to this state, the atom remains for 2.0 ms, on average, before it emits a photon and returns to the ground state. (a) What are the energy and wavelength of the photon

Answers

Answer:

[tex]\lambda=3451*10^{10}m[/tex]

Explanation:

From the question we are told that:

Energy state [tex]e=3.50 eV[/tex]

Time [tex]t=2ms[/tex]

Generally the equation for energy of Photon is mathematically given by

[tex]E=e-e_0[/tex]

[tex]E=3.6*10^{-19}J[/tex]

[tex]E=5.7*10^{-19}J[/tex]

Generally the equation for Wave-length of Photon is mathematically given by

[tex]\lambda=\frac{hc}{E}[/tex]

[tex]\lambda=\frac{6.626*10^{-34}*3*10^8}{5.76*10^{-19}}[/tex]

[tex]\lambda=3451*10^{10}m[/tex]

Predict the products (if any) that will be formed by the reaction below. If no reaction occurs, write NR after the reaction arrow.

2HClO4(aq) + Co(s) -->

Answers

Answer:

The product is aqueous [tex]CO(HCl)_2[/tex] and [tex]O_2(g)[/tex].

Explanation:

Given:

⇒ [tex]2HClO_4(aq) +CO(s)[/tex]

then,

The reaction will be:

⇒ [tex]2HClO_4(aq)+CO(s) \rightarrow CO(HCl)_2 +O_2 (g)[/tex]

In the above reaction, we can see that

The products is:

aqueous [tex]CO(HCl)_2[/tex] and [tex]O_2(g)[/tex]

Thus the above is the correct answer.

Which statement describes the 3d, 4s, and 4p orbitals of Arsenic (As) based on its electronic configuration and position in the periodic table?
The 3d and 4s orbitals are completely filled, and the 4p orbital is partially filled.
The 3d orbital is completely filled, and the 4s and 4p orbitals are partially filled.
The 3d, 4s, and 4p orbitals are completely filled.
The 3d, 4s, and 4p orbitals are partially filled.

Answers

Answer:

The 3d and 4s orbitals are completely filled, and the 4p orbital is partially filled.

Explanation:

The correct answer is: The 3d and 4s orbitals are completely filled, and the 4p orbital is partially filled.

The d orbital contains 10 electrons, the s orbital takes 2 electrons and the p orbital takes six electrons.

The orbital in chemistry is defined as a region in space where there is a high probability of finding an electron. There are s, p, d, f orbitals in chemistry which correspond to sharp, principal, diffuse and fundamental.

The electronic configuration of arsenic is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3.

From this electronic configuration, we can see that the 4s and 3d orbitals are half filled while the 4p orbital is half filled.

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What is the final volume, in L. of a balloon that was initially at 173.8 mL at 17.5°C and was then heated to 78.0*C?

Answers

Answer:

0.21 L.

Explanation:

From the question given above, the following data were obtained:

Initial temperature (T₁) = 17.5°C = 17.5°C + 273 = 290.5 K

Initial volume (V₁) = 173.8 mL

Final temperature (T₂) = 78 °C = 78 °C + 273 = 351 K

Final volume (V₂) =?

V₁/T₁ = V₂/T₂

173.8 / 290.5 = V₂ / 351

Cross multiply

290.5 × V₂ = 173.8 × 351

290.5 × V₂ = 61003.8

Divide both side by 290.5

V₂ = 61003.8 / 290.5

V₂ = 210 mL

Finally, we shall convert 210 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

210 mL = 210 mL × 1 L / 1000 mL

210 mL = 0.21 L

Thus, the final volume of the balloon is 0.21 L.

Answer:

[tex]\boxed {\boxed {\sf 0.775 \ L}}[/tex]

Explanation:

1. Calculated Final Volume

We are asked to find the final volume of a balloon given a change in temperature. We will use Charles's Law, which states the volume of a gas is directly proportional to the temperature. The formula for this law is:

[tex]\frac{V_1}{T_1}= \frac{V_2}{T_2}[/tex]

The initial volume is 173.8 milliliters and the initial temperature is 17.5 degrees Celsius.

[tex]\frac {173.8 \ mL}{17.5 \textdegree C}= \frac{V_2}{T_2}[/tex]

The balloon is heated to a final temperature of 78.0 degrees Celsius, but the volume is unknown.

[tex]\frac {173.8 \ mL}{17.5 \textdegree C}= \frac{V_2}{78.0 \textdegree C}[/tex]

We are solving for the final volume, so we must isolate the variable V₂. It is being divided by 78.0 degrees Celsius. The inverse of division is multiplication, so we multiply both sides by 78.0 °C.

[tex]78.0 \textdegree C *\frac {173.8 \ mL}{17.5 \textdegree C}= \frac{V_2}{78.0 \textdegree C} * 78.0 \textdegree C[/tex]

[tex]78.0 \textdegree C *\frac {173.8 \ mL}{17.5 \textdegree C}=V_2[/tex]

The units of degrees Celsius cancel.

[tex]78.0 *\frac {173.8 \ mL}{17.5}=V_2[/tex]

[tex]78.0 *9.931428571 \ mL= V_2[/tex]

[tex]774.6514286 \ mL =V_2[/tex]

2. Convert to Liters

We are asked to give the volume in liters, so we must convert out units. Remember that 1 liter contains 1000 milliliters.

[tex]\frac { 1 \ L}{1000 \ mL}[/tex]

[tex]774.6514286 \ mL * \frac{ 1 \ L}{1000 \ mL}[/tex]

[tex]774.6514286 * \frac{ 1 \ L}{1000}[/tex]

[tex]0.7746514286 \ L[/tex]

3. Round

The original values of volume and temperature have 3 and 4 significant figures. We always round our answer to the least number of sig figs, which is 3. This is the thousandths place for the number we calculated. The 6 in the ten-thousandths place tells us to round the 4 up to a 5.

[tex]0.775 \ L[/tex]

The final volume is approximately 0.775 liters.

A reaction was performed, and the dichloromethane solvent was dried by adding magnesium sulfate drying agent. When the reaction flask was shaken, it was observed that the magnesium sulfate clumped together at the bottom of the flask. What does this observation indicate

Answers

The clumping of magnesium sulfate means that the wrong kind of drying agent have been used for the sample.

What is a drying agent?

A drying agent is also referred to as a desiccant. It is a substance that is used to remove moisture from a sample. We must recall that the drying agent to be used must not react with the sample.

Since the magnesium sulfate was found to clump together at the bottom of the flask, it means that the wrong kind of drying agent have been used for the sample.

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name a factor tht affects the value of electron affinity​

Answers

Answer:

Atomic sizeNuclear chargesymmetry of the electronic configuration
Various factors that affect electron affinity are atomic size, nuclear charge and the symmetry of the electronic configuration. Atomic size: With increase in the atomic size, the distance between the nucleus and the incoming electron also increases.

A 70.0‑g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter. The metal and water come to the same temperature at 24.6 °C. How much heat did the metal give up to the water?

Answers

Answer:

1087.84 J

Explanation:

From the question given above, the following data were obtained:

Mass of metal (Mₘ) = 70 g

Temperature of metal (Tₘ) = 80 °C

Mass of water (Mᵥᵥ) = 100 g

Temperature of water (Tᵥᵥ) = 22 °C

Equilibrium temperature (Tₑ) = 24.6 °C

Heat lost by metal (Qₘ) =?

NOTE: Specific heat capacity of water (Cᵥᵥ) = 4.184 J/gºC

Heat lost by metal (Qₘ) = Heat gained by water (Qᵥᵥ)

Qₘ = Qᵥᵥ

Thus, we shall determine the heat gained by water. This can be obtained as follow:

Qᵥᵥ = MᵥᵥCᵥᵥ(Tₑ – Tᵥᵥ)

Qᵥᵥ = 100 × 4.184 (24.6 – 22)

Qᵥᵥ = 418.4 × 2.6

Qᵥᵥ = 1087.84 J

Thus, the heat gained by water is 1087.84 J.

Heat lost by metal (Qₘ) = Heat gained by water (Qᵥᵥ)

Qₘ = Qᵥᵥ

Qᵥᵥ = 1087.84 J

Qₘ = 1087.84 J

Therefore, the heat lost by the metal is 1087.84 J

A 70.0‑g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter. After reaching a temperature of 24.6 °C, the heat given up by the metal to the water is -1.08 kJ.

What is a calorimeter?

A calorimeter is an object used for calorimetry, or the process of measuring the heat of chemical reactions or physical changes as well as heat capacity.

A 70.0‑g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter. The final temperature of the system is 24.6 °C.

Let's use the following expression to calculate the heat absorbed by the water.

Qw = c × m × ΔT

Qw = (4.184 J/g.°C) × 100 g × (24.6 °C - 22.0 °C) = 1.08 kJ

where,

Qw is the heat absorbed by the water.c is the specific heat capacity of water.m is the mass of water.ΔT is the change in the temperature for water.

According to the law of conservation of energy, the sum of the heat absorbed by the water and the heat released by the metal (Qm) is zero.

Qw + Qm = 0

Qm = -Qw = -10.8 kJ

A 70.0‑g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter. After reaching a temperature of 24.6 °C, the heat given up by the metal to the water is -1.08 kJ.

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Nitric acid and nitrogen monoxide react to form nitrogen dioxide and water, like this: At a certain temperature, a chemist finds that a 7.7 L reaction vessel containing a mixture of nitric acid, nitrogen monoxide, nitrogen dioxide, and water at equilibrium has the following composition: compound amount
HNO 16.2 g 11.0 g 18.6 g H20 236.7 g 3 NO NO
Calculate the value of the equilibrium constant K for this reaction. Round your answer to 2 significant digits.

Answers

Answer:

K = 3.3

Explanation:

Nitric acid, HNO3, reacts with nitrogen monoxide, NO, to produce nitrogen dioxide, NO2 and water H2O as follows:

2HNO3(g) + NO(g) → 3NO2(g) + H2O(g)

Where equilibrium constant, K, is:

K = [NO2]³[H2O] / [HNO3]²[NO]

[] is the molar concentration of each species at equilibrium.

To solve this question we need to find molarity of each gas and replace these in the equation as follows:

[NO2] -Molar mass NO2-46.0g/mol-

18.6g * (1mol/46.0g) = 0.404mol / 7.7L = 0.0525M

[H2O] -Molar mass:18.01g/mol-

236.7g * (1mol/18.01g) = 13.14 moles / 7.7L = 1.707M

[HNO3] -Molar mass:53.01g/mol-

16.2g * (1mol/53.01g) = 0.3056 moles / 7.7L = 0.0397M

[NO] -Molar mass: 30.0g/mol-

11.0g * (1mol/30.0g) = 0.367 moles / 7.7L = 0.0476M

Replacing:

K = [NO2]³[H2O] / [HNO3]²[NO]

K = [0.0525M]³[1.707M] / [0.0397M]²[0.0476M]

K = 3.3

The products of nuclear reaction usually have a different mass than the reactants why?

Answers

Answer:

Explanation:

The best way to explain this is to use an example

[tex]I\frac{125}{53} + e \frac{0}{-1} ====> Te\frac{125}{52}[/tex]

You have to understand what happened. A electron was shot into the nucleus of the Iodine. That electron change the entire composition of the nucleus resulting in 52 protons. The mass remained the same (125) but the nucleus was altered. The chemical became 125 52 Tellurium.  But what is important is that it takes a tremendous amount of energy to disrupt a nucleus, and a new chemical is born from that disruption.

Given the standard enthalpy changes for the following two reactions: (1) 2Fe(s) + O2(g)2FeO(s)...... ΔH° = -544.0 kJ (2) 2Zn(s) + O2(g)2ZnO(s)......ΔH° = -696.6 kJ what is the standard enthalpy change for the reaction:

Answers

Answer:

-76.3 kJ

Explanation:

Here is the complete question

Given the standard enthalpy changes for the following two reactions:

(1) 2Fe(s) + O₂(g) → 2FeO(s)......ΔH° = -544.0 kJ

(2) 2Zn(s) + O₂(g) → 2ZnO(s)......ΔH° = -696.6 kJ. What is the standard enthalpy change for the reaction:

(3) FeO(s) + Zn(s) → Fe(s) + ZnO(s)......ΔH° = ?

Solution

Since (1) 2Fe(s) + O₂(g) → 2FeO(s)......ΔH° = -544.0 kJ

reversing the reaction, we have

2FeO(s) → 2Fe(s) + O₂(g) ......ΔH° = +544.0 kJ (4)

Adding reactions (2) and (3), we have

2FeO(s) → 2Fe(s) + O₂(g) ......ΔH° = +544.0 kJ (4)

2Zn(s) + O₂(g) → 2ZnO(s)......ΔH° = -696.6 kJ  (2)

This gives

2FeO(s) + 2Zn(s) → 2Fe(s) + 2ZnO(s)......ΔH° =

The enthalpy change for this reaction is the sum of enthalpy changes for reaction (2) and (3) = ΔH° = +544.0 kJ + (-696.6 kJ)

= +544.0 kJ - 696.6 kJ)

= -152.6 kJ

Since the required reaction is (3) which is FeO(s) + Zn(s) → Fe(s) + ZnO(s)

we divide the enthalpy change for reaction (4) by 2 to obtain the enthalpy change for reaction (3).

So, ΔH° = -152.6 kJ/2 = -76.3 kJ

So, the standard enthalpy change for the reaction

FeO(s) + Zn(s) → Fe(s) + ZnO(s) is -76.3 kJ

Hydrocarbons do not dissolve in concentrated sulfuric acid, but methyl benzoate does. Explain this difference and write an equation showing the ions that are produced.

Answers

Answer:

See explanation

Explanation:

For a substance to dissolve in another, there must be some sort of interaction between the substances.

Recall that like dissolves like. That is, polar substances dissolve polar substances and non polar substances dissolve nonpolar substances.

Hydrocarbons are nonpolar hence they do not dissolve in polar sulphuric acid. Methyl benzoate is polar hence it dissolve in polar sulphuric acid.

The equation showing the ions is depicted in the image attached to this answer.

Calculate the molarity of a solution consisting of 65.5 g of K2S0 4 in 5.00 L of solution. ​

Answers

Answer:

Molarity is 0.075 M.

Explanation:

Moles:

[tex]{ \tt{ = \frac{65.5}{RFM} }}[/tex]

RFM of potassium sulphate :

[tex]{ \tt{ = (39 \times 2) + 32 + (16 \times 4)}} \\ = 174 \: g[/tex]

substitute:

[tex]{ \tt{moles = \frac{65.5}{174} = 0.376 \: moles}}[/tex]

In volume of 5.00 l:

[tex]{ \tt{5.00 \: l = 0.376 \: moles}} \\ { \tt{1 \: l = ( \frac{0.376}{5.00} ) \: moles}} \\ { \tt{molarity = 0.075 \: mol \: l {}^{ - 1} }}[/tex]

Problem 7 (Diffusion due to viscosity) If the viscosity of a solution is quadrupled, the rms-average distance of a collection of diffusing molecules from their starting point would be _________ over the same amount of time.

Answers

Answer:

1/2 the distance

Explanation:

If the viscosity of a solution is quadrupled then the distance of collection of diffusing molecules would be half over the same amount of time. The viscosity of the molecules is dependent on density of the liquid. It is independent to the volume of the liquid.

Using the molarity of vinegar, calculate the mass percent of acetic acid in the original sample. Assume the density of vinegar is 1.00 g/mL. (The formula for acetic acid is C2H4O2).

Answers

Answer:

5.37% w/w is the mass percent of vinegar assuming a molarity of 0.8935mol/L

Explanation:

Assuming the molarity of vinegar is 0.8935mol/L:

Mass percent is defined as 100 times the ratio between mass of solute (In this case, acetic acid), and the mass of the solution

To solve this question we need to find the mass of acetic acid from the moles using the molar mass and the mass of the solution from the volume in liters using the density:

Mass Acetic acid -Molar mass: 60.052g/mol-

0.8935mol * (60.052g / mol) = 53.656g Acetic Acid

Mass Solution:

1L = 1000mL * (1.00g/mL) = 1000g Solution

Mass Percent:

53.656g Acetic Acid / 1000g Solution * 100 =

5.37% w/w is the mass percent of vinegar assuming a molarity of 0.8935mol/L

The mass percent of acetic acid in the original sample of vinegar of molarity 0.8935mol/L is 5.37% w/w.

How do we calculate the mass percent?

Mass percent of any solute present in any solution will be calculated as the:

Mass % of solute = (mass of solute / mass of solution) × 100

Let the molarity of vinegar = 0.8935mol/L

Means 0.8935 moles of vinegar present in the 1 liter of the solution.

Now we calculate mass from moles as:

n = W/M, where

W = required mass

M = molar mass = 60.052g /mol

W = (0.8935mol)(60.052g/mol) = 53.656g

Mass of solution = 1L = 1000mL×(1.00g/mL) = 1000g Solution

Then the mass % of acetic acid:

Mass % = (53.656g / 1000g) × 100 = 5.37% w/w

Hence the required % mass is 5.37% w/w.

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The first step of electrophilic aromatic substitution involves the formation of the arenium ion intermediate.

a. True
b. Fasle

Answers

Answer:

True

Explanation:

Aromatic compounds undergo substitution rather than addition reactions because the aromatic structure is maintained.

Electrophilic aromatic substitution begins with attack of the electrophile on the aromatic ring to yield a delocalized intermediate called the arenium intermediate. Loss of hydrogen from this intermediate yields the final product.

The solvent for an organic reaction is prepared by mixing 70.0 mL of acetone (C3H6O) with 75.0 mL of ethyl acetate (C4H8O2). This mixture is stored at 25.0 ∘C. The vapor pressure and the densities for the two pure components at 25.0 ∘C are given in the following table. What is the vapor pressure of the stored mixture?

Answers

Answer:

The answer is "170.9 mm Hg".

Explanation:

[tex]\text{Mass of acetone = volume} \times density[/tex]

                          [tex]= 70.0 \times 0.791\\\\ = 55.37\ g\\[/tex]

[tex]\text{Moles of acetone} = \frac{mass}{molar\ mass}\\\\[/tex]

                            [tex]=\frac{55.37}{58.08}\\\\ = 0.9533\ mol[/tex]

[tex]\text{Mass of ethyl acetate = volume} \times density[/tex]

                                   [tex]= 73.0 \times 0.900\\\\ = 65.7\ g[/tex]  

[tex]\text{Moles of ethyl acetate = mass} \times\ molar\ mass[/tex]

                                    [tex]= \frac{65.7}{88.105} \\\\= 0.7457\ mol[/tex]

[tex]\text{Mole fraction of acetone x(acetone)} = \frac{0.9533}{(0.9533 + 0.7457)}\ = 0.5611\\\\[/tex] [tex]\text{Mole fraction of ethyl acetate x(ethyl acetate)} =\frac{0.7457}{(0.9533 + 0.7457) }= 0.4389[/tex]

Applying Raoult's law: [tex]\text{Vapor pressure = x(acetone)P(acetone) + x(ethyl acetate)P(ethyl acetate)}\\\\= 0.5611 \times 230.0 + 0.4389 \times 95.38\\\\ = 170.9\ mm \ Hg\\[/tex]

The solvent for an organic reaction is prepared by mixing 70.0 mL of acetone (C3H6O) with 75.0 mL of ethyl acetate (C4H8O2).

The vapor pressure of the stored mixture is: 170.03 mmHg

In the given information, there is some information that is still missing.

The parameters that we are being given include:

The volume of acetone = 70.0 mLThe volume of ethyl acetate = 75.0 mLThe standard temperature for the mixture = 25° C

The  first step we need to take is to determine the mass and number of moles  of each compound (i.e. for acetone and ethyl acetate)

For us to do that:

We need the density of acetone and ethyl acetate, which is not given:

Assuming that at a standard condition of vapour pressure:

230 mmHg of acetone has a density of 0.791 g/mL95.38 mmHg of ethyl acetate has a density of 0.900 g/mL

Then;

Using the relation:

[tex]\mathbf{Density = \dfrac{Mass}{volume}}[/tex]

Mass of acetone = Density of acetone × volume of acetone

Mass of acetone = 0.791  g/mL  × 70.0 mL

Mass of acetone = 55.37 g

Mass of ethyl acetate = Density of ethyl acetate  × volume of ethyl acetate

Mass of ethyl acetate = 0.900 g/mL  ×  75.0 mL

Mass of ethyl acetate = 67.5 g

At standard conditions;

For acetone, molar mass = 58.08 g/molFor ethyl acetate, molar mass = 88.11 g/mol

Now, using the formula for calculating the numbers of moles which can be expressed as:

[tex]\mathbf{Number \ of \ moles = \dfrac{mass}{molar \ mass}}[/tex]

For acetone:

[tex]\mathbf{Number \ of \ moles = \dfrac{55.37 \ g}{58.08 \ g/mol}}[/tex]

[tex]\mathbf{Number \ of \ moles =0.95334 \ mol}[/tex]

For ethyl acetate:

[tex]\mathbf{Number \ of \ moles = \dfrac{67.5 \ g}{88.11 \ g/mol}}[/tex]

[tex]\mathbf{Number \ of \ moles =0.76609 \ mol}[/tex]

Now, we will determine the mole fraction of each compound.

The mole fraction describes the ratio a certain constituent of a mixture to the total amount of all the constitutent in the mixture.

Using the formula:

[tex]\mathbf{mole \ fraction = \dfrac{n_A}{n_A+n_B+...n_N}}[/tex]

For Acetone:

[tex]\mathbf{mole \ fraction = \dfrac{0.95334}{0.95334+0.76609}}[/tex]

[tex]\mathbf{mole \ fraction =0.5545 }[/tex]

For ethyl acetate:

[tex]\mathbf{mole \ fraction = \dfrac{0.76609}{0.76609+0.95334}}[/tex]

[tex]\mathbf{mole \ fraction =0.4455}[/tex]

Finally, we can compute determine the vapour pressure of the stored mixture using Raoult's Law.

Raoult's Law posits that the constituent of a partial pressure in a mixture of a liquid is proportional to the mole fraction of that constituent in the mixture provided the temperature is constant.

∴ For the stored mixture = Vapor pressure of acetone + vapour pressure of ethyl acetate.

where:

Vapour pressure of the solution = (mole fraction × vapor pressure) of solvent

For acetone;

Vapor pressure = 0.5545 × 230 mmHg

Vapour pressure = 127.54 mmHg

For ethyl acetate:

Vapour pressure = 0.4455 × 95.38 mmHg

Vapour pressure ==42.49 mmHg

Thus, the vapor pressure of the stored mixture is

= (127.54 + 42.49 ) mmHg

= 170.03 mmHg

Therefore, we can conclude that the vapour pressure of the stored mixture is 170.03 mmHg

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A 25.00 gram sample of an unknown metal initially at 99.0 degrees Celcius is added to 50.00 grams of water initially at 10.55 degrees Celcius. The final temperature of the system is 20.15 degrees Celcius. Calculate the specific heat of the metal. (The specific heat of water is 4.184 J/g*C).

Answers

Answer:

1.0188 J/g*C

Explanation:

Using the formula; Q = m × c × ∆T

Q(water) = -Q(metal)

m × c × ∆T (water) = -{m × c × ∆T (metal)}

According to this question,

mass of metal = 25g

initial temp of metal = 99°C

mass of water = 50g

initial temp of water = 10.55°C

final temperature of the system = 20.15°C

c of water = 4.184 J/g*C

50 × 4.184 × (20.15 - 10.55) = 25 × c × (20.15 - 99)

209.2 × 9.6 = 25c × -78.85

2008.32 = -1971.25c

c = 2008.32 ÷ 1971.25

c of metal = 1.0188 J/g*C

What Volume of silver metal will weigh exactly 2500.0g. The density of silver

Answers

Answer:

cm3 = 2500.0 g / 10.5 g/cm3 = 238 cm3

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