Answer:
A. 1.9 × 103 g
(next one)
Which metric unit would be the best choice to report the result?
A. kg
Answer:
1. 2
2. 1.9 × 10^3 g
3. kg
Explanation:
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
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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An antacid tablet weighing 1.30g was fully neutralized at 42.00 mL(an excess amount) of 0.250MHCl. 10.00 mL of 0.100 M NaOH was then used to back titrate the excess HCl. How many moles of acid did the antacid neutralize
Answer:
0.0095 moles of acid were neutralized by the antiacid
Explanation:
The antiacid is a base that neutralize the acid in stomach. To find the moles of acid neutralized we need to find the moles of acid added initially. This acid is added in excess, then, the moles of NaOH added reacts to neutralize the moles of acid in excess. The difference between initial moles of HCl and moles of NaOH needed to titrate the excess = Moles of HCl that were neturalized by the antiacid as follows:
Moles HCl added:
42.00mL = 0.04200L * (0.250mol/L) = 0.0105 moles HCl
Moles NaOH to titrate the excess:
10.00mL = 0.01000L * (0.10mol/L) = 0.0010 moles NaOH = Moles HCl in excess.
Moles of acid that were neutralized:
0.0105 moles - 0.0010 moles =
0.0095 moles of acid were neutralized by the antiacidname a factor tht affects the value of electron affinity
Answer:
Atomic sizeNuclear chargesymmetry of the electronic configurationThe 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?
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° CThe 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/mLThen;
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/molNow, 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 solventFor 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).
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
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.
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.
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.
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
By process of incineration, a mystery substance is empirically determined to contain 40.00% carbon by weight, 6.67% hydrogen, and 53.33% oxygen. Its molecular weight ranges between 55 and 62 g/mole. a. (6 points) Determine the chemical formula of this substance
Answer:
C₂H₄O₂
Explanation:
Step 1: Divide each percentage by the atomic mass of the element
C: 40.00/12.01 = 3.331
H: 6.67/1.01 = 6.60
O: 53.33/16.00 = 3.333
Step 2: Divide all the numbers by the smallest one
C: 3.331/3.331 = 1
H: 6.60/3.331 ≈ 2
O: 3.333/3.331 ≈ 1
The empirical formula is CH₂O, with a molecular weight of 12 g/mol + 2 × 1 g/mol + 16 g/mol = 30 g/mol. The molecular weight of the compound must be a product of 30, such as 60 (between 55 and 62 g/mol). Since we have to multiply by 2 (30 to 60) to get to the molecular weight of the compound, we also have to multiply the empirical formula by 2 to get the chemical formula of the compound.
CH₂O × 2 = C₂H₄O₂
Determine whether the statement about identifying a halide is true: Regardless of any concentration of ammonium solution, the precipitates in the reaction solution of my unknown halide after 0.1M AgNO3 remain because my unknown halide solution contains Br. Select one: True False
Answer:
False
Explanation:
The statement ; Regardless of any concentration of ammonium solution the precipitate of unknown halide after 0.1M AgNO3 will remain is FALSE
This is Because the remaining concentration of AgNO3 is dependent on the solubility of Ag⁺
Determine whether each of the examples represents a colligative property or a non-colligative property. boiling point elevation Choose... color Choose... freezing point depression Choose... vapor pressure lowering Choose... density Choose...
Answer:
boiling point elevation - colligative property
color - non-colligative property
freezing point depression - colligative property
vapor pressure lowering - colligative property
density - non-colligative property
Explanation:
A colligative property is a property that depends on the number of particles present in the system.
Freezing point depression, boiling point elevation and vapour pressure lowering are all colligative properties of solutions.
Colour and density do not depend on the number of particles present hence they are not colligative properties.
The boiling point elevation, freezing point depression, and vapor pressure lowering are colligative properties. And color and density are non-colligative properties.
Explanation:
The colligative properties are the properties depending upon the number of particles of solute not on the nature of the solute.Example of colligative properties:Vapor pressure loweringElevation boiling pointDepression in freezing pointOsmotic pressureThe non-colligative properties are the properties depending upon the nature of solute and solvent.Example of non-colligative properties :ViscositySurface tensionDensitySolubilitySo, from this, we can conclude that boiling point elevation, freezing point depression, and vapor pressure lowering are colligative properties. And color and density are non-colligative properties.
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When the equation,
O2 + __C 10H 22 →
CO2 +
H2O is balanced, the coefficient of O2 is?
Please help!
how is the akin of frog similar to a fish
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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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?
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 VolumeWe 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 LitersWe 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. RoundThe 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 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
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]
What is Bose Einstein state of matter
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:
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
Analyze the transition of a photon
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
Answer:
Moles of water are 0.868
Explanation:
What is the hydrogen atoms in 39.6g of ammonium sulphate,NH4 2SO4
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).
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/LThe 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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……….is strong due to the ……………..between positive ions and negative delocalized electrons
Answer:
atom &bond
Explanation:
atom is strong due to the bond
what is meant by density
Answer:
The degree of compactness of a substance
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.
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
Give four examples illustrating each of the following terms. a. homogeneous mixture b. heterogeneous mixture c. compound e. physical change d. element f. chemical change
1. homogenous: sugar solution
2. heterogeneous: sand solution
3. compound: water
4. physical change: ice melting
5. element: hydrogen
6. chemical change: burning fire
What Volume of silver metal will weigh exactly 2500.0g. The density of silver
Answer:
cm3 = 2500.0 g / 10.5 g/cm3 = 238 cm3
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) -->
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.
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.
Answer:
Explanation:
I am almost sure that the products are identical.
20. An oxide of osmium (symbol Os) is a pale yellow solid. If 2.89 g of the compound contains 2.16 g of osmium, what is its empirical formula?
The empirical formula is OsO₄ :
Explanation:
Osmium oxide contains osmium and oxygen only.
Thus, we shall determine the mass of oxygen in osmium oxide. This can be obtained as follow:
Mass of compound = 2.89 g
Mass of Os = 2.16 g
Mass of O =?Mass of O = (Mass of compound) – (Mass of Os)
Mass of O = 2.89 – 2.16
Mass of O = 0.73 g
Finally, we shall determine the empirical formula of the compound. This can be obtained as follow:
Mass of Os = 2.16 g
Mass of O = 0.73 g
Empirical formula =..?Os = 2.16 g
O = 0.73 g
Divide by their molar mass of
Os = 2.16 / 190 = 0.011
O = 0.73 / 16 = 0.046
Divide by the smallest
Os = 0.011 / 0.011 = 1
O = 0.046 / 0.011 = 4
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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.
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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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?
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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