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2 years ago

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A 0.580 g sample of a compound containing only carbon and hydrogen contains 0.480 g of carbon and 0.100 g of hydrogen. At STP, 3

3.6 mL of the gas has a mass of 0.087 g. What is the molecular (true) formula for the compound

1 answer:

Sati [7]2 years ago

7 0

Answer:

Molecular formula for the gas is: C₄H₁₀

Explanation:

Let's propose the Ideal Gases Law to determine the moles of gas, that contains 0.087 g

At STP → 1 atm and 273.15K

1 atm . 0.0336 L = n . 0.082 . 273.15 K

n = (1 atm . 0.0336 L) / (0.082 . 273.15 K)

n = 1.500 × 10⁻³ moles

Molar mass of gas = 0.087 g / 1.500 × 10⁻³ moles = 58 g/m

Now we propose rules of three:

If 0.580 g of gas has ____ 0.480 g of C _____ 0.100 g of C

58 g of gas (1mol) would have:

(58 g . 0.480) / 0.580 = 48 g of C

(58 g . 0.100) / 0.580 = 10 g of H

48 g of C / 12 g/mol = 4 mol

10 g of H / 1g/mol = 10 moles

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Suppose, in an experiment to determine the amount of sodium hypochlorite in bleach, you titrated a 26.34 mL sample of 0.0100 M K

Dmitry [639]

Answer:

0.1 M

Explanation:

The overall balanced reaction equation for the process is;

IO3^- (aq)+ 6H^+(aq) + 6S2O3^2-(aq) → I-(aq) + 3S4O6^2-(aq) + 3H2O(l)

Generally, we must note that;

1 mol of IO3^- require 6 moles of S2O3^2-

Thus;

n (iodate) = n(thiosulfate)/6

C(iodate) x V(iodate) = C(thiosulfate) x V(thiosulfate)/6

Concentration of iodate C(iodate)= 0.0100 M

Volume of iodate= V(iodate)= 26.34 ml

Concentration of thiosulphate= C(thiosulfate)= the unknown

Volume of thiosulphate=V(thiosulfate)= 15.51 ml

Hence;

C(iodate) x V(iodate) × 6/V(thiosulfate) = C(thiosulfate)

0.0100 M × 26.34 ml × 6/15.51 ml = 0.1 M

5 0

2 years ago

Albus Dumbledore provides his students with a sample of 19.3 g of sodium sulfate. How many oxygen atoms are in this sample

Dimas [21]

Answer:

<em>3.27·10²³ atoms of O</em>

Explanation:

To figure out the amount of oxygen atoms in this sample, we must first evaluate the sample.

The chemical formula for sodium sulfate is <em>Na₂SO₄, </em>and its molar mass is approximately 142.05 $\frac{g}{mol}$ .

We will use stoichiometry to convert from our mass of <em>Na₂SO₄ </em>to moles of <em>Na₂SO₄</em>, and then from moles of <em>Na₂SO₄ </em>to moles of <em>O </em>using the mole ratio; then finally, we will convert from moles of <em>O </em>to atoms of <em>O </em>using Avogadro's constant.

19.3g <em>Na₂SO₄</em> · $\frac{1 mol Na^2SO^4}{142.05g Na^2SO^4}$ · $\frac{4 mol O}{1 mol Na^2SO^4}$ · $\frac{6.022x10^2^3}{1 mol O}$

After doing the math for this dimensional analysis, you should get a quantity of approximately <em>3.27·10²³ atoms of O</em>.

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2 years ago

Consider the following reaction (X = Cl or Br) which statement s is are correct?

statuscvo [17]

A.S OLOS kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkll

5 0

2 years ago

Olympic cyclist fill their tires with helium to make them lighter. Calculate the mass of air in an air filled tire and the mass

inn [45]

<u>Answer:</u> The mass difference between the two is 7.38 grams.

<u>Explanation:</u>

To calculate the number of moles, we use the equation given by ideal gas follows:

$PV=nRT$

where,

P = pressure = 125 psi = 8.50 atm (Conversion factor: 1 atm = 14.7 psi)

V = Volume = 855 mL = 0.855 L (Conversion factor: 1 L = 1000 mL)

T = Temperature = $25^oC=[25+273]K=298K$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles = ?

Putting values in above equation, we get:

$8.50atm\times 0.855L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 298K\\\\n=\frac{8.50\times 0.855}{0.0821\times 298}=0.297mol$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For air:</u>

Moles of air = 0.297 moles

Average molar mass of air = 28.8 g/mol

Putting values in equation 1, we get:

$0.297mol=\frac{\text{Mass of air}}{28.8g/mol}\\\\\text{Mass of air}=(0.297mol\times 28.8g/mol)=8.56g$

Mass of air, $m_1$ = 8.56 g

<u>For helium gas:</u>

Moles of helium = 0.297 moles

Molar mass of helium = 4 g/mol

Putting values in equation 1, we get:

$0.297mol=\frac{\text{Mass of helium}}{4g/mol}\\\\\text{Mass of helium}=(0.297mol\times 4g/mol)=1.18g$

Mass of helium, $m_2$ = 1.18 g

Calculating the mass difference between the two:

$\Delta m=m_1-m_2$

$\Delta m=(8.56-1.18)g=7.38g$

Hence, the mass difference between the two is 7.38 grams.

5 0

2 years ago

A gas cylinder filled with nitrogen at standard temperature and pressure has a mass of 37.289 g. The same container filled with

andrew-mc [135]

Answer:

Molar mass = 3.9236 g/mol ≅ 4 g/mol

This corresponds to Helium gas.

Explanation:

Let the moles of nitrogen gas = x moles

Moles of carbon dioxide = x moles ( As both are filled at same temperature and pressure conditions )

Given:

$Mass_{Container}+Mass_{Nitrogen\ gas}=37.289\ g$

Molar mass of nitrogen gas, $N_2$ = 28.014 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$x\ moles= \frac{Mass}{28.014\ g/mol}$

Mass of nitrogen gas = 28.014x g

So,

Let, $Mass_{Container}=y$

$y+28.014x=37.289$

Similarly,

$Mass_{Container}+Mass_{Carbon\ dioxide\ gas}=37.440\ g$

Molar mass of nitrogen gas, $CO_2$ = 44.01 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$x\ moles= \frac{Mass{44.01\ g/mol}$

Mass of nitrogen gas = 44.01x g

So,

$y+44.01x=37.440$

Solving the two equations, we get :

$Mass_{Container}=y=37.025\ g$

x = 0.00943 moles

Thus, Given:

$Mass_{Container}+Mass_{Unknown\ gas}=37.062\ g$

$37.025\ g+Mass_{Unknown\ gas}=37.062\ g$

Mass of the gas = 0.037 moles

Moles = 0.00943 moles

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.00943\ moles= \frac{0.037\ g}Molar mass}$

Molar mass = 3.9236 g/mol ≅ 4 g/mol

This corresponds to Helium gas.

7 0

2 years ago