PLEASE ANSWER

What is the molality of a solution in which 0.42 moles aluminum chloride has been dissolved in 4200 water ?

madreJ [45]

Answer:

0.1 M

<h3>Explanation:</h3>

Molarity refers to the concentration of a solution in moles per liter.
It is calculated by dividing the number of moles of solute by the volume of solvent;

Molarity = Moles of the solute ÷ Volume of the solvent

In this case, we are given;

Number of moles of the solute, NH₄Cl as 0.42 moles
Volume of the solvent, water as 4200 mL or 4.2 L

Therefore;

Molarity = 0.42 moles ÷ 4.2 L

= 0.1 mol/L or 0.1 M

Thus, the molarity of the solution will be 0.1 M

7 0

3 years ago

If 8.50 g of phosphorus reacts with hydrogen gas at 2.00 atm in a 10.0-L container at 298 K, calculate the moles of PH3 produced

ahrayia [7]

Answer:

The moles of PH₃ produced are 0.2742 and the total number of moles of gas present at the end of the reaction is 0.6809.

Explanation:

Phosphorus reacts with H₂ according to the balanced equation:

P₄ (s) + 6 H₂ (g) ⇒ 4 PH₃ (g)

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

P₄: 1 mole
H₂: 6 moles
PH₃:4 moles

Being the molar mass of the compounds:

P₄: 124 g/mole
H₂: 2 g/mole
PH₃: 34 g/mole

The following mass amounts of each compound participate in the reaction:

P₄: 1 mole* 124 g/mole= 124 g
H₂: 6 mole* 2 g/mole= 12 g
PH₃: 4 moles* 34 g/mole= 136 g

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case you know:

P= 2 atm
V= 10 L
n= ?
R= 0.082 $\frac{atm*L}{mol*K}$
T= 298 K

Replacing:

2 atm*10 L= n*0.082 $\frac{atm*L}{mol*K}$ *298 K

and solving you get:

$n=\frac{2 atm*10 L}{0.082\frac{atm*L}{mol*K}*298 K }$

n=0.818 moles

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, you can use a simple rule of three as follows: if 6 moles of H₂ react with 124 g of P₄, 0.818 moles of H₂ with how much mass of P₄ will it react?

$mass of P_{4}=\frac{0.818 moles of H_{2}*124 grams of P_{4}}{6 moles of H_{2}}$

mass of P₄= 16.90 grams

But 16.90 grams of P₄ are not available, 8.50 grams are available. Since you have less mass than you need to react with 0.818 moles of H₂, phosphorus P₄ will be the limiting reagent.

Then you can apply the following rules of three:

If 124 grams of P₄ produce 4 moles of PH₃, 8.50 grams of P₄, how many moles do they produce?

$moles of PH_{3} =\frac{8.5 grams of P_{4}*4 moles of PH_{3} }{124grams of P_{4}}$

moles of PH₃=0.2742

If 124 grams of P₄ react with 6 moles of H₂, 8.50 grams of P₄ with how many moles of H₂ do they react?

$moles of H_{2} =\frac{8.5 grams of P_{4}*6 moles of H_{2} }{124grams of P_{4}}$

moles of H₂= 0.4113

If you have 0.818 moles of H₂, the number of moles of gas H₂ present at the end of the reaction is calculated as:

0.818 - 0.4113= 0.4067

Then the total number of moles of gas present at the end of the reaction will be the sum of the moles of PH₃ gas and H₂ gas that did not react:

0.2742 + 0.4067= 0.6809

Finally, the moles of PH₃ produced are 0.2742 and the total number of moles of gas present at the end of the reaction is 0.6809.

5 0

2 years ago

g If 50.0 mL of a 0.75 M acetic acid solution is titrated with 1.0 M sodium hydroxide, what is the pH after 10.0 mL of NaOH have

V125BC [204]

Answer:

pH = 2.66

Explanation:

Acetic Acid + NaOH → Sodium Acetate + H₂O

First we calculate the number of moles of each reactant, using the given volumes and concentrations:

0.75 M Acetic acid * 50.0 mL = 37.5 mmol acetic acid

1.0 M NaOH * 10.0 mL = 10 mmol NaOH

We calculate how many acetic acid moles remain after the reaction:

37.5 mmol - 10 mmol = 27.5 mmol acetic acid

We now calculate the molar concentration of acetic acid after the reaction:

27.5 mmol / (50.0 mL + 10.0 mL) = 0.458 M

Then we calculate [H⁺], using the following formula for weak acid solutions:

[H⁺] = $\sqrt{C*Ka}=\sqrt{0.458M*1.76x10^{-5}}$

[H⁺] = 0.0028

Finally we calculate the pH:

pH = -log[H⁺]

pH = 2.66

8 0

2 years ago

Look at this graph:

ira [324]

The slope is 3/2
3 over 2

4 0

2 years ago

Which energy source can we can find waste everywhere and shouldn’t run out?

olya-2409 [2.1K]

Answer:

wind

Explanation:

hope this helps

7 0

2 years ago

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