Balance the following equation:<br> C7H602 + 02 -- CO2 + H20

The nation on average uses around 10 Million tons of salt per year to de-ice roads. How many kilograms of salt is this? (please

Westkost [7]

45359237 kilograms :))))))))))

3 0

2 years ago

Which adaptation is likely to increase the chances of survival of an animal in a rainforest?

LiRa [457]

I know I think it’s a A

6 0

3 years ago

2.00 moles of an ideal gas was found to occupy a volume of 17.4L at a pressure of 3.00 atm and at a temperature of 45 C. Calulat

lidiya [134]

Answer:

$0.082 atm L mol^{-1} K^{-1}$

Explanation:

The pressure, the volume and the temperature of an ideal gas are related to each other by the equation of state:

$pV=nRT$

where

p is the pressure of the gas

V is the volume of the gas

n is the number of moles

R is the gas constant

T is the absolute temperature

For the gas in this problem:

n = 2.00 mol is the number of moles

V = 17.4 L is the gas volume

p = 3.00 atm is the gas pressure

$T=45C+273=318 K$ is the absolute temperature

Solving for R, we find the gas constant:

$R=\frac{pV}{nT}=\frac{(3.00)(17.4)}{(2.00)(318)}=0.082 atm L mol^{-1} K^{-1}$

5 0

3 years ago

Calculate the density of O2(g) at 415 K and 310 bar using the ideal gas and the van der Waals equations of state. Use a numerica

Lera25 [3.4K]

Answer:

Explanation:

From the given information:

The density of O₂ gas = $d_{ideal} = \dfrac{P\times M}{RT}$

here:

P = pressure of the O₂ gas = 310 bar

= $310 \ bar \times \dfrac{0.987 \ atm}{1 \ bar}$

= 305.97 atm

The temperature T = 415 K

The rate R = 0.0821 L.atm/mol.K

molar mass of O₂ gas = 32 g/mol

∴

$d_{ideal} = \dfrac{305.97 \ \times 32}{0.0821 \times 415}$

$d_{ideal}$ = 287.37 g/L

To find the density using the Van der Waal equation

Recall that:

the Van der Waal constant for O₂ is:

a = 1.382 bar. L²/mol² &

b = 0.0319 L/mol

The initial step is to determine the volume = Vm

The Van der Waal equation can be represented as:

$P =\dfrac{RT}{V-b}-\dfrac{a}{V^2}$

where;

R = gas constant (in bar) = 8.314 × 10⁻² L.bar/ K.mol

Replacing our values into the above equation, we have:

$310 =\dfrac{0.08314\times 415}{V-0.0319}-\dfrac{1.382}{V^2}$

$310 =\dfrac{34.5031}{V-0.0319}-\dfrac{1.382}{V^2}$

$310V^3 -44.389V^2+1.382V-0.044=0$

After solving;

V = 0.1152 L

∴

$d_{Van \ der \ Waal} = \dfrac{32}{0.1152}$

$d_{Van \ der \ Waal}$ = 277.77 g/L

We say that the repulsive part of the interaction potential dominates because the results showcase that the density of the Van der Waals is lesser than the density of ideal gas.

5 0

2 years ago

A 1.525g sample of a compound between nitrogen and hydrogen contains 1.333 g of nitrogen. Calculate its empirical formula. The e

vredina [299]

Answer:

NH2.

Explanation:

The mass of hydrogen in the sample = 1.525 - 1.333 = 0.192g.

Dividing the 2 masses by the relative atomic mass of hydrogen and nitrogen:

H: 0.192 / 1.008 = 0.1905

N: 1.333 / 14.007 = 0.09517

The ratio of N to H = 0.09517 : 0.1905

= 1 : 2.

So the empirical formula is NH2.

5 0

2 years ago

Balance the following equation: C7H602 + 02 -- CO2 + H20