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3 years ago

What volume (in liters) does 3.91 moles of nitrogen gas at 5.35 atm and 323 K occupy

Chemistry

1 answer:

ivanzaharov [21]3 years ago

4 0

Data Given:
Moles = n = 3.91 mol

Pressure = P = 5.35 atm

Temperature = T = 323 K

Volume = V = ?

Formula used: Ideal Gas Equation is used,

P V = n R T
Solving for V,
V = n R T / P
Putting Values,
V = (3.91 mol × 0.0825 atm.L.mol⁻¹.K⁻¹ × 323 K) ÷ 5.35 atm

V = 19.36 L

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3.04 Quiz: Chemical Reactions 1 to 10 question

vaieri [72.5K]

Answer:

1. Chemical reactions

2.Substances

3. Properties

4. Precipitate

5. Light

6. Temperature

7. Color

8. Gas

Explanation:

During chemical reactions, new substances that possess new properties are formed. During the process of chemical reactions, evidences are found. Some of the evidences are formation of precipitate or light gel, production of gases and changes in color and temperature.

Chemical reactions start with the substances or compounds which are known as reactants. The reactants react to form new substances known as the products which possess new properties.

I only could help with 1 - 8 sorry.

5 0

3 years ago

Romashka-Z-Leto [24]

Answer:

Avogadro's number or Avogardro’s constant

Explanation:

I’m pretty sure this is correct if it’s not I’m sorry lol.

6 0

2 years ago

How many moles of aluminum are needed to react completely with 1.2 mol of feo

Amanda [17]

Answer:

0.8 mol.

Explanation:

The balanced equation for the reaction between Al and FeO is represented as:

2Al + 3FeO → 3Fe + Al₂O₃,

It is clear that 2 mol of Al react with 3 mol of FeO to produce 3 mol of Fe and 1 mol of Al₂O₃.

Using cross multiplication:

2 mol of Al needs → 3 mol of FeO, from stichiometry.

??? mol of Al needs → 1.2 mol of FeO.

∴ The no. of moles of Al are needed to react completely with 1.2 mol of FeO = (2 mol)(1.2 mol)/(3 mol) = 0.8 mol.

3 0

3 years ago

How many grams of aluminum will react fully with 1.25 moles Cl₂?

Liula [17]

2Al + 3Cl₂ → 2AlCl₃

mol Al = 2/3 x 1.25 = 0.83

mass Al = 0.83 x 27 g/mol = 22.41 g

3 0

2 years ago

The maximum amount of nickel(II) cyanide that will dissolve in a 0.220 M nickel(II) nitrate solution is...?

sweet [91]

Answer : The maximum amount of nickel(II) cyanide is $5.84\times 10^{-12}M$

Explanation :

The solubility equilibrium reaction will be:

$Ni(CN)_2\rightleftharpoons Ni^{2+}+2CN^-$

Initial conc. 0.220 0

At eqm. (0.220+s) 2s

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ni^{2+}][CN^-]^2$

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

$3.0\times 10^{-23}=(0.220+s)\times (2s)^2$

$s=5.84\times 10^{-12}M$

Therefore, the maximum amount of nickel(II) cyanide is $5.84\times 10^{-12}M$

7 0

3 years ago