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Lana71 [14]
3 years ago
6

the student determines the molar mass of the gas to be 64 g mol-1. write the expression (set-up) for calculating the percent err

or in the experimental value, assuming that the unknown gas is butane (molar mass 58 g mol-1). calculations are not required
Chemistry
1 answer:
sleet_krkn [62]3 years ago
7 0

Answer:

Percentage error = \frac{(64 - 58)}{58} x 100%

Explanation:

In order to calculate the percentage error in the student's calculation, the following formula will be used:

Percentage error = \frac{(experimental molar mass - theoretical molar mass)}{theoretical molar mass} x 100%

experimental molar mass = 64 g/mol

theoretical molar mass = 58 g/mol

Hence, the expression becomes:

Percentage error = \frac{(64 - 58)}{58} x 100%

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What is the freezing point of an aqueous solution that has 25.00 g of calcium iodide dissolved in 1250 g of water?
ozzi

Answer:

<u></u>

  • <u>- 0.380ºC</u>

Explanation:

The lowering of the freezing point of a solvent is a colligative property ruled by the formula:

  • \Delta T_f=K_f\times m\times i

Where:

  • ΔTf is the lowering of the freezing point
  • Kf is the molal freezing constant of the solvent: 1.86 °C/m
  • m is the molality of the solution
  • i is the van't Hoff factor: the number of particles (ions) per unit of ionic compound.

<u />

<u>a) molality, m</u>

  • m = number of moles of solute/ kg of solvent
  • number of moles of CaI₂ = mass in grams/ molar mass
  • number of moles of CaI₂ = 25.00g / 293.887 g/mol = 0.0850667mol
  • m = 0.0850667mol/1.25 kg = 0.068053m

<u>b) i</u>

  • Each unit of CaI₂, ideally, dissociates into 1 Ca⁺ ion and 2 I⁻ ions. Thus, i = 1 + 2 = 3

<u />

<u>c) Freezing point lowering</u>

  • ΔTf =  1.86 °C/m × 0.068053m × 3 = 0.3797ºC ≈ 0.380ºC

<h2>I have problems to upload the full answer in here, so I attach a pdf file with the whole answer.</h2>
Download pdf
6 0
3 years ago
Be sure to answer all parts. The equilibrium constant (Kp) for the reaction below is 4.40 at 2000. K. H2(g) + CO2(g) ⇌ H2O(g) +
nikklg [1K]

<u>Answer:</u>

<u>For 1:</u> The value of \Delta G for the chemical equation is -24.636 kJ/mol

<u>For 2:</u> The value of \Delta G for the chemical equation is -20.925 kJ/mol

<u>Explanation:</u>

For the given chemical equation:

H_2(g)+CO_2(g)\rightleftharpoons H_2O(g)+CO(g)

  • <u>For 1:</u>

To calculate the \Delta G for given value of equilibrium constant, we use the relation:

\Delta G=-RT\ln K_p      .....(1)

where,

\Delta G = ? kJ/mol

R = Gas constant = 8.314J/K mol

T = temperature = 2000 K

K_p = equilibrium constant in terms of partial pressure = 4.40

Putting values in above equation, we get:

\Delta G=-(8.314J/Kmol)\times 2000K\times \ln (4.40)\\\\\Delta G=-24636.12J/mol

Converting this into kilo joules, we use the conversion factor:

1 kJ = 1000 J

So, -24636.12 J/mol = -24.636 kJ/mol

Hence, the value of \Delta G for the chemical equation is -24.636 kJ/mol

  • <u>For 2:</u>

The expression of K_p for the given chemical equation is:

K_p=\frac{p_{CO}p_{H_2O}}{p_{H_2}p_{CO_2}}

We are given:

p_{CO}=1.18atm\\p_{H_2O}=0.66atm\\p_{CO_2}=0.82atm\\p_{H_2}=0.27atm

Putting values in above equation, we get:

K_p=\frac{1.18\times 0.66}{0.27\times 0.82}\\\\K_p=3.52

Now, calculating the value of \Delta G by using equation 1:

R = Gas constant = 8.314J/K mol

T = temperature = 2000 K

K_p = equilibrium constant in terms of partial pressure = 3.52

Putting values in equation 1, we get:

\Delta G=-(8.314J/Kmol)\times 2000K\times \ln (3.52)\\\\\Delta G=-20925.68J/mol

Converting this into kilo joules, we use the conversion factor:

1 kJ = 1000 J

So, -20925.68 J/mol = -20.925 kJ/mol

Hence, the value of \Delta G for the chemical equation is -20.925 kJ/mol

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What happens after you leave oil and vinegar in a glass for a few minutes?
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Answer:

An emulsion is formed.

Explanation:

An association (emulsion) of two liquids is formed, in this case oil and vinegar, which when stirred, said mixture will separate.

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Answer:

on each side of the salt bridge, which is represented by a double vertical line

Explanation:

While writing a cell notation, the general convention is; anode || cathode. The anode and the cathode are separated by a double line. The anode is written on the lefthand side while the cathode is written on the righthand side.

The cell notation is a shorthand representation of a cell, hence any electrochemical cell can easily be produced based on its cell diagram.

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