What is the most important part of solution preparation

Helium is mixed with oxygen gas for deep-sea divers. Calculate the percent by volume of oxygen gas in the mixture if the diver h

kvasek [131]

Answer : The percent by volume of oxygen gas in the mixture is, 5.3 %

Explanation :

According to the Raoult's law,

$p^o_A=X_A\times p_A$

where,

$p_A$ = total partial pressure of solution = 3.8 atm

$p^o_A$ = partial pressure of oxygen = 0.20 atm

$X_A$ = mole fraction of oxygen = ?

Now put all the given values in this formula, we get:

$p^o_A=X_A\times p_A$

$0.20atm=X_A\times 3.8atm$

$X_A=0.0526$

Now we have to calculate the percent by volume of oxygen gas in the mixture.

The mole percent of oxygen gas = $0.0526\times 100=5.3\%$

As we know that, there is a direct relation between the volume of moles.

So, mole percent of oxygen gas = volume percent of oxygen gas

Volume percent of oxygen gas = $0.0526\times 100=5.3\%$

Therefore, the percent by volume of oxygen gas in the mixture is, 5.3 %

5 0

4 years ago

An electron in an atom is known to be in a state with magnetic quantum number . What is the smallest possible value of the princ

nlexa [21]

Answer:

Hi

Each electron in an atom is characterized by four numbers that arise from the resolution of Schrödinger's equations. These numbers are called quantum numbers. Each energy level corresponds to a main known quantum number, which is represented by the letter n. This number gives an idea of the location of an energy level with respect to the nucleus. The higher n, the mayor will be the energy of that level and the farther away from the nucleus is removed.

In each energy level there may be sub-levels. Each of them is specified by another quantum number called secondary, specified with the letter l. The value of this quantum number can vary from zero to n-1. For example, in the first energy level, the quantum number can only take a value that is zero, while in the second level, it can take a value between zero or one. Then, it can be said that the values of the quantum number n indicate the size of the orbital, that is, its proximity to the nucleus; and the values of the quantum number l variables the orbital:

• If l = 0, the orbital is of type s.

• If l = 1, the orbitals are of type p.

• If l = 2, the orbitals are of type d.

• If l = 3, the orbitals are of type f.

Explanation:

8 0

3 years ago

what is the temperature in Kelvin of 5.00 moles of nitrogen gas in a 30 L container with a pressure of 4.00 atm

Marianna [84]

The temperature of the nitrogen gas is 292.5 K.

<u>Explanation:</u>

Given that

Moles of Nitrogen, n = 5 mol

Volume, V = 30 L

Pressure, P = 4 atm

Gas Constant, R = 0.08205 L atm mol⁻¹ K⁻¹

Temperature = ? K

We have to use the ideal gas equation,

PV = nRT

by rearranging the equation, so that the equation becomes,

T = $$\frac{PV}{nR}\\$

Plugin the above values, we will get,

T = $$\frac{4 \times 30}{5 \times 0.08205}$

= 292.5 K

So the temperature of the nitrogen gas is 292.5 K.

7 0

3 years ago

Define valence electrons.

vodka [1.7K]

Answer:

Electrons on the outermost shell of an atom. They are responsible for the chemical properties of an atom.

8 0

3 years ago

How much energy (in JJ) is lost when a sample of iron with a mass of 29.2 gg cools from 79.0 ∘C∘C to 27.0 ∘C∘C?

Anna35 [415]

Answer:

- 0.674 kJ.

Explanation:

The equation used to solve this problem is:

Q = mCΔT

where,

Q = amount of heat

m = mass of the substance

C = specific heat capacity

ΔT = change in temperature

= Temp.f - Temp.i

Given:

m = 29.2 g

Temp.i = 79°C

Temp.f = 27°C

Cp(iron) = 0.444 J/g.K

Q = mCΔT

ΔT = 27.0°C - 79.0°C

= -52°C

= 29.2 * 0.444 * -52

Q = - 674.17 J

= 0.674 kJ.

6 0

3 years ago

What is the most important part of solution preparation​

What is the most important part of solution preparation