All categories

3 years ago

10. What is the atmospheric pressure if the partial pressures of nitrogen, oxygen, and

Chemistry

1 answer:

sashaice [31]3 years ago

7 0

Answer:

1.009 atm is the total pressure for the mixture

Explanation:

To determine the total pressure in atm, of the three gases (N₂, O₂ and Ar) we have to sum all the values.

Sum of partial pressures in a mixture = Total pressure

First of all, we need to convert the values from mmHg to atm

604.5 mmHg . 1atm / 760 mmHg = 0.795 atm

162.8 mmHg . 1atm / 760 mmHg = 0.213 atm

0.500 mmHg . 1atm / 760 mmHg = 6.58×10⁻⁴ atm

Partial pressure N₂ + Partial pressure O₂ + Partial pressure Ar = Total P

0.795 atm + 0.213 atm + 6.58×10⁻⁴ atm = 1.009 atm

You might be interested in

What questions, if we were to answer them, would

Margarita [4]

Answer:

When our bodies are dry and wind blows by, we lose some energy to the air molecules. When are bodies are wet, we have a substance on our skin that likes to absorb heat. So when wind blows by, we lose a LOT of energy to the air molecules. When the body loses heat energy, our body temperature drops.

Explanation:

hope it helps

plzz mark it as brainliest...

7 0

3 years ago

Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit

ser-zykov [4K]

Answer:

For a: The edge length of the unit cell is 314 pm

For b: The radius of the molybdenum atom is 135.9 pm

Explanation:

For a:

To calculate the edge length for given density of metal, we use the equation:

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density = $10.28g/cm^3$

Z = number of atom in unit cell = 2 (BCC)

M = atomic mass of metal (molybdenum) = 95.94 g/mol

$N_{A}$ = Avogadro's number = $6.022\times 10^{23}$

a = edge length of unit cell =?

Putting values in above equation, we get:

$10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm$

Conversion factor used: $1cm=10^{10}pm$

Hence, the edge length of the unit cell is 314 pm

For b:

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

$R=\frac{\sqrt{3}a}{4}$

where,

R = radius of the lattice = ?

a = edge length = 314 pm

Putting values in above equation, we get:

$R=\frac{\sqrt{3}\times 314}{4}=135.9pm$

Hence, the radius of the molybdenum atom is 135.9 pm

4 0

3 years ago

You remove the pot of boiling hot water and set it on a cooler part of the stove. which mechanism is responsible for cooling the

kolbaska11 [484]

All of the mechanisms work together to cool the pot of boiling water. Hope this helps!

3 0

3 years ago

Judging and evaluating advertisements requires critical thinking because

Tom [10]

Judging and evaluating advertisements requires critical thinking because advertisements can contain confusing, misleading information.

3 0

3 years ago

Read 2 more answers

Whats the volume of a 2 year old

musickatia [10]

Answer:

here's a great idea!

take a jug of water by finding the volume of that water and ask the 2 year old to chug it all in... the volume of water drank by the 2 year old is the volume of the two year old

5 0

3 years ago