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1 year ago

Determine the empirical formula for a compound that is 64.8% C, 13.6% H, and 21.6% O by mass.

Chemistry

1 answer:

True [87]1 year ago

8 0

Answer:

C4H10O

In case it's not understandable: c four, h ten, no

Explanation:

Molar mass of each component: C = 12.0107, H = 1.00794, O = 15.9994

Convert to moles: C = 5.3951892895502, H = 13.492866638887, O=1.3500506268985

Find the smallest mole value: 1.3500506268985

Divide all components by the smallest value: C = 3.9962866444087, H=9.9943412269543, O = 1

Round to closest whole numbers: C = 4, H = 10, O = 1

Combine to get the empirical formula: C4H10O

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The formula used for determining gas pressure, volume and temperature interaction would be PV=nRT.

• What is the temperature in Kelvins?
You already right at this part. Kelvin temperature formula from celsius should be:
K= C+273.15=
K= 27 +273.15 = 300.15
It is important to remember that the formula in this question is using Kelvin unit at temperature, not Celcius or Fahrenheit.

• Assuming that everything else remains constant, what will happen to the pressure if the temperature decreases to -15 ºC?
In this case, the temperature is decreased from 27C into -15C and you asked the change in the pressure.
Using PV=nRT formula, you can derive that the temperature will be directly related to pressure. If the temperature decreased, the pressure will be decreased too.

 If you increase the number of moles to 6 moles, increase temperature to 400K and reduce the volume to 25 L, what will the new pressure be?
PV=nRT
P= nRT/V
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2 years ago

Which of the following is true regarding atomic radius?

Ymorist [56]

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3 years ago

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2 years ago

A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli

Diano4ka-milaya [45]

Answer:

$\large \boxed{109.17 \, ^{\circ}\text{C}}$

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

ρ(EG) = 1.11 g/mL

ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

$\Delta T_{b} = iK_{b}b$

i is the van’t Hoff factor — the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

$\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}$

2. Kilograms of water

$m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}$

3. Molal concentration of EG

$b = \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}$

4. Increase in boiling point

$\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}$

5. Boiling point

$\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}$

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3 years ago

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2 years ago