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.

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:

So, 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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Answer:

Answer is B) 20.0 moles

Explanation:

From the equation,

1 mole of Fe2O3 = 2 moles of Al

therefore 10.0 moles of Fe2O3 = 10×2

= 20.0 moles.

Answer:

5 × 10⁵ molecules (500,000 molecules)

Explanation:

Step 1: Convert 3 × 10⁻¹⁶ g to moles

We will use the molar mass of TCDD (321.97 g/mol).

3 × 10⁻¹⁶ g × 1 mol/321.97 g = 9 × 10⁻¹⁹ mol

Step 2: Convert 9 × 10⁻¹⁹ mol to molecules

The required conversion factor is Avogadro's number (6.02 × 10²³ molecules/mol).

9 × 10⁻¹⁹ mol × 6.02 × 10²³ molecules/1 mol = 5 × 10⁵ molecules

Answer:

7.38*10^21

Explanation:

2Na+2H20=2NaOH+H2

nNa=0.0123

number of water moles: 0.012*6*10^23=7.38*10^21

Answer:

Blue or Purple color filter

Explanation:

Given that Sodium has an emission spectrum with wavelength ≈ 590nm and for a wavelength of 590nm  the color is yellow.

Hence To filter out the color ( yellow ) to enable the measurement of the Balmer series of hydrogen spectrum, we have to use a filter that possess the complementary color of yellow ( i.e. purple(RYB color model) or blue (RGB additive color model )

therefore color filter to be used = Blue or Purple

Answer:

Low value for copper recovery

Explanation:

The percentage recovery is obtained from;

Percent recovery = amount of substance you actually collected / amount of substance you were supposed to collect × 100

Note that the fact that some of the copper nitrate solution splashed out of the beaker means that some amount copper has been lost from the system. This loss of copper leads to a lower value of copper recovered from solution.

Answer:

a)  [tex]V_1=5ul[/tex]

b)  [tex]v=20ul[/tex]

Explanation:

From the question we are told that:

initial Concentration [tex]C_1=50mg/ml[/tex]

Final Concentration [tex]C_2=10mg/ml[/tex]

Final volume needs [tex]V_2 =25ul[/tex]

Generally the equation for Volume is mathematically given by

[tex]C_1V_1=C_2V_2[/tex]

[tex]V_1=\frac{C_1V_1}{C_2}[/tex]

[tex]V_1=\frac{10*25}{50}[/tex]

[tex]V_1=5ul[/tex]

Therefore

The volume of buffer needed is

[tex]v=V_2-V_1\\\\v=25-5[/tex]

[tex]v=20ul[/tex]

Answer:

Cm = n/V

n(H2SO4) = 245/98 = 2.5 mol

Cm(H2SO4) = 2.5/1 = 2.5 M

Explanation:

Answer:

+0.151, spontaneous

Explanation:

Given that;

Co^2+(aq) + 2e ---->Co(s) -0.28 V

Pb^2+(aq) + 2e ---->Pb(s). -0.13 V

Hence Co is the anode and Pb is the cathode

E°cell = E°cathode - E°anode

So;

E°cell = -0.13 V - (-0.28 V)

E°cell = 0.15 V

The cell reaction is spontaneous since E°cell is positive.

Answer:

B

Explanation:

Ammonia is considered a base as it's pH is 11

Answer from Gauthmath

The  compound NH3 (Ammonia) can be classified as a weak Base. Below you can learn more about Ammonia.

Ammonia is a chemical compound which is derived from the combination of Nitrogen and Hydrogen. It is denoted by the chemical formula NH3.

Ammonia is a base and when it reacts with acids to gives out salts. Physically, It is a colorless gas with a distinct characteristic of a pungent smell.

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Answer:

O2 is limiting reactant

Explanation:

To find the limiting reactant we need to convert the mass of each reactant to the moles using the formula weight. And, as 1 mole of C6H12O6 reacts with 6 moles of O2, we can know wich reactant will be over first (Limiting reactant) as follows:

Moles C6H12O6:

650g * (1mol/180.16g) = 3.608 moles C6H12O6

Moles O2:

650g * (1mol/32g) = 20.31 moles O2

Now, for a complete reaction of 3.608 moles of C6H12O6 are required:

3.608 moles C6H12O6 * (6mol O2 / 1mol C6H12O6) = 21.65 moles O2

As there are just 20.31 moles of O2,

Answer:

I'll see it

Explanation:

Answer:

0.012 moles of electrons were transferred

Explanation:

We can find the number of electrons transferred from the time in seconds and the current in Amperes using the equation:

n = I*t / F

Where n are moles of electrons transferred

I is current in Amperes = 0.2053A

t is time in seconds:

94.3min*(60s/1min) = 5658s

F is faraday constant 96485A*s/mol

Replacing:

n = 0.2053A*5658s / 96485A*s/mol

n = 0.012 moles of electrons were transferred

Moles are the mass per unit molar mass of compound. The number of moles of electrons that are transferred is 0.012 moles.

An electrolytic cell is a type of electrochemical cell that uses electrical energy from external sources to conduct the chemical energy in a cell.  

The moles transferred in electrolytic cells are measured as:

[tex]\rm n = \rm \dfrac{I\times t}{F}[/tex]

Given,

Current (I) in amperes = 0.2053 A

Time (t) in seconds = 5658 sec

Faraday constant (F) = 96485 A -s/mol

Substituting values in the equation above moles (n) can be calculated as:

[tex]\begin{aligned} \rm n &= \dfrac{ 0.2053 \times 5658 }{96485}\\\\&= 0.012 \;\rm mol\end{aligned}[/tex]

Therefore, 0.012 moles are transferred in this process.

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Answer:

atom &bond

Explanation:

atom is strong due to the bond

Chemistry 11 Solutions

978Ͳ0Ͳ07Ͳ105107Ͳ1Chapter 8 Solutions and Their Properties • MHR | 85

Amount in moles, n, of the NaCl(s):

NaCl

2.5 g

m

n

M

58.44 g

2

4.2778 10 m l

ol

o

/m

u

Molar concentration, c, of the NaCl(aq):

–2 4.2778 × 10 mol

0.100

0.42778 mol/L

0.43 mol

L

/L

n

c

V

The molar concentration of the saline solution is 0.43 mol/L.

Check Your Solution

The units are correct and the answer correctly shows two significant digits. The

dilution of the original concentrated solution is correct and the change to mol/L

seems reasonable.

Section 8.4 Preparing Solutions in the Laboratory

Solutions for Practice Problems

Student Edition page 386

51. Practice Problem (page 386)

Suppose that you are given a stock solution of 1.50 mol/L ammonium sulfate,

(NH4)2SO4(aq).

What volume of the stock solution do you need to use to prepare each of the

following solutions?

a. 50.0 mL of 1.00 mol/L (NH4)2SO4(aq)

b. 2 × 102 mL of 0.800 mol/L (NH4)2SO4(aq)

c. 250 mL of 0.300 mol/L NH4

+

(aq)

What Is Required?

You need to calculate the initial volume, V1, of (NH4)2SO4(aq) stock solution

needed to prepare each given dilute solution.

The dilution gives the relationship between the molarity and the volume of the solution. The volume of stock solution with a molarity of 1.50 mol/L is 50 mL.

Dilution is said to be the addition of more volume to the concentrated solution to make it less in molar concentration. This tells about the inverse and indirect relationship between the volume and the molar concentration of the solution.

Given,

Initial volume = V₁

Initial molar concentration (M₁) = 1.50 mol/L

Final volume (V₂) = 125 mL = 0.125 L

Final molar concentration (M₂)= 0.60 mol/L

The dilution is calculated as:

M₁V₁ = M₂V₂

V₁ = M₂V₂ ÷ M₁

Substituting the values in the above formula as

V₁ = M₂V₂ ÷ M₁

V₁ = (0.60 mol/L × 0.125 L) ÷ 1.50 mol/ L

V₁ = 0.05 L

= 50 mL

Therefore, 50 mL of stock solution is needed to make a 0.60 mol/L solution.

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Explanation:

here's the answer. I just plug the expression into my calculator and find the intercept to avoid the quadratic formula

Answer:

Explanation:

Na2CO3+Ca(NO3)2=CaCO3+2NaNO3

nNa2CO3=0.02

nCa(NO3)2=0.02

mCaCO3=0.02*100=2 gram

nNaNo3=0.04

Cm=2/15

From the calculation, the mass of the product is 2 g.

A chemical reaction occurs when two more substances are mixed together. In this case, the reaction is shown by; Ca(NO3)2 + Na2CO3 ----> CaCO3(s) + 2NaNO3.

Number of moles of Na2CO3 = 100/1000 L *  0.2 mol/L = 0.02 moles

Number of moles of  Ca(NO3)2 = 200/1000 L *  0.1 mol/L = 0.02 moles

Since the reaction is equimolar, amount of the product = 0.02 moles * 100 g/mol = 2 g

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Answer:

22.1

Explanation:

Step 1: Calculate the number of formula units of lithium nitrate

A 5.08 L solution contains 2.68 × 10²¹ formula units per liter.

5.08 L × 2.68 × 10²¹ formula units/1 L = 1.36 × 10²² formula units

Step 2: Calculate the number of nitrate ions

Lithium nitrate dissociates completely in water according to the following equation.

LiNO₃(aq) ⇒ Li⁺(aq) + NO₃⁻(aq)

The molar ratio of LiNO₃ to NO₃⁻ is 1:1. The number of nitrate ions is 1/1 × 1.36 × 10²² = 1.36 × 10²².

Then,

log (1.36 × 10²²) = 22.1

The number of nitrate ions are present in the following aqueous solution is 13.6x10²¹, and log(base 10) value of this is 22.2.

Those salts which are completely dissociate into their ions in the solution form will known as strong electrolyte.

Lithium nitrate is a strong electrolyte and it shows complete dissociation as:

LiNO₃ → Li⁺ + NO₃⁻

Given that,

1 liter of solution containing = 2.68 x 10²¹ formula unit

5.08 L of solution containing = 5.08 x 2.68 x 10²¹ = 13.6x10²¹ formula unit

So, number of nitrate ion in given solution = 13.6x10²¹ formula unit

Log(13.6x10²¹) = 22.1

Hence, required values are 13.6x10²¹ and 22.1.

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Answer:

81°C

111°C

cyclohexane

Explanation:

Distillation is a process of separating two liquids based on differences in Bolling point. For two substances having different boiling points, they are collected as they are converted into vapour, condensed and move down the condenser one after the other.

Since the boiling point of cyclohexane is less than that of toluene, cyclohexane is collected first before toluene.

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⁺

1. homogenous: sugar solution

2. heterogeneous: sand solution

3. compound: water

4. physical change: ice melting

5. element: hydrogen

6. chemical change: burning fire

Answer:

NaoH= sodium hydroxide

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

Answer:

3.6487g

CuCl2 moles reacted = (0.15×250)/1000

according to balanced chemical equation

precipitated Cu(OH)2 moles = Reacted CuCl2 moles

molar mass of Cu(OH)2 = 63.5+ (17+1)×2 = 97.5

mass of precipitate = (97.5 × 0.15×250)/1000

= 3.648g

Explanation:

since hydrochloric acid is an acid and calcium hydroxide is a base, we know that this is an acid base reaction. The ions will then dissociatiate and bond with one another.

Answer:

I don't know ❤️

Answer:

The empirical formula is C4H8O

Explanation:

Step 1: Data given

Valproic acid is composed of C, H, and O

Mass of the sample = 0.165 grams

Mass of water = 0.166 grams

Mass of CO2 = 0.403 grams

Molar mass of water ( H2O) = 18.02 g/mol

Molar mass of CO2 = 44.01 g/mol

Atomic mass of C = 12.01 g/mol

Atomic mass of O = 16.0 g/mol

Atomic mass H = 1.01 g/mol (H2 = 2.02 g/mol)

Step 2: The equation

CxHyOz + O2 → CO2 + H2O

Step 3: Calculate the number of carbon in the sample

The carbon comes from CO2

Mass C = (12.01 g/mol/44.01 g/mol) * 0.403 grams

Mass C = 0.110 grams

Step 4: Calculate mass of hydrogen in the sample

The hydrogen comes from H2O

Mass H = (2.02/18.02) * 0.166 grams

MAss H = 0.0186 grams

Step 5: Calculate mass of O

The mass of O in the sample = Mass of sample - mass of H - mass of C

The mass of O = 0.165 grams - 0.110 grams - 0.0186 grams

The mass of O = 0.0364 grams

Step 6: Calculate moles

Moles C = 0.110 grams / 12.01 g/mol = 0.00916 moles

Moles H = 0.0186 / 1.01 = 0.0184 moles

Moles O = 0.0364/16.0 = 0.00228 moles

Step 7: Calculate empirical formula

We divide by the smallest amount of moles

C: 0.00916/ 0.00228 = 4

H: 0.0184/0.00228 = 8

O: 0.00228/0.00228 = 1

The empirical formula is C4H8O