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2 years ago

Please help 100 point s

Chemistry

2 answers:

ki77a [65]2 years ago

5 0

Where is the chart?????

Anni [7]2 years ago

5 0

air is NOT a compound - it is a mixture of different gases like O2, N2 n CO2. CO2 is a pure substance but NOT an element. so the CORRECT ans is

C) The chart shows the percentages of pure substances in a mixture.

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How many ATOMS of OXYGEN are there in the following compound

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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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Answer:

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nirvana33 [79]

Answer:

4KNO3 ==> 2K2O + 2N2 + 5O2

Explanation:

It's a decomposition, but not a simple one.

KNO3 ==> K2O + N2 + O2 I don't usually do this, but I think the easiest way to proceed is to balancing the K and N together. That will require a 2 in front of KNO3

4KNO3 ==> 2K2O + 2N2 + 5O2

Now you have (3*4) = 12 oxygens. Two are on the K2O. So the other 10 must be on the O2

That should do it.

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