What is the energy of an electron in the first energy level of hydrogen?

s) Suppose we now collect hydrogen gas, H2(g), over water at 21◦C in a vessel with total pressure of 743 Torr. If the hydrogen g

Elenna [48]

This is an incomplete question, here is a complete question.

Suppose we now collect hydrogen gas, H₂(g), over water at 21°C in a vessel with total pressure of 743 Torr. If the hydrogen gas is produced by the reaction of aluminum with hydrochloric acid:

$2Al(s)+6HCl(aq)\rightarrow 2AlCl_3(aq)+3H_2(g)$

what volume of hydrogen gas will be collected if 1.35 g Al(s) reacts with excess HCl(aq)? Express your answer in liters.

Answer : The volume of hydrogen gas that will be collected is 1.85 L

Explanation :

First we have to calculate the number of moles of aluminium.

Given mass of aluminium = 1.35 g

Molar mass of aluminium = 27 g/mol

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

$\text{Moles of aluminium}=\frac{1.35g}{27g/mol}=0.05mol$

The given chemical reaction is:

$2Al(s)+6HCl(aq)\rightarrow 2AlCl_3(aq)+3H_2(g)$

As, hydrochloric acid is present in excess. So, it is considered as an excess reagent.

Thus, aluminium is a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

2 moles of aluminium produces 3 moles of hydrogen gas

So, 0.005 moles of aluminium will produce = $\frac{3}{2}\times 0.05=0.0750mol$ of hydrogen gas

Now we have to calculate the mass of helium gas by using ideal gas equation.

PV = nRT

where,

P = Pressure of hydrogen gas = 743 Torr

V = Volume of the helium gas = ?

n = number of moles of hydrogen gas = 0.075 mol

R = Gas constant = $62.364\text{ L Torr }mol^{-1}K^{-1}$

T = Temperature of hydrogen gas = $21^oC=[21+273]K=294K$

Now put all the given values in above equation, we get:

$743Torr\times V=0.075mol\times 62.364\text{ L Torr }mol^{-1}K^{-1}\times 294K\\\\V=1.85L$

Hence, the volume of hydrogen gas that will be collected is 1.85 L

8 0

4 years ago

Calculate the molarity of a solution of Nach if it contains 7.2.g Nach in 100.0 mL of solution. andver: m Nach . .

Norma-Jean [14]

Answer:

1.23 M

Explanation:

Molarity of a substance , is the number of moles present in a liter of solution .

M = n / V

M = molarity

V = volume of solution in liter ,

n = moles of solute ,

Moles is denoted by given mass divided by the molecular mass ,

Hence ,

n = w / m

n = moles ,

w = given mass ,

m = molecular mass .

From the question ,

w = given mass of NaCl = 7.2 g

As we know , the molecular mass of NaCl = 58.5 g/mol

Moles is calculated as -

n = w / m = 7.2 g / 58.5 g/mol = 0.123 mol

Molarity is calculated as -

V = 100ml = 0.1 L (since , 1 ml = 1/1000L )

M = n / V = 0.123 mol / 0.1 L = 1.23 M

5 0

3 years ago

Calculate the mass, in grams, of Ag2CrO4 that will precipitate when 50.0mL of 0.20M AgNO3 solution is mixed with 40.0mL of 0.10M

Darina [25.2K]

Answer:

1.327 g Ag₂CrO₄

Explanation:

The reaction that takes place is:

2AgNO₃(aq) + K₂CrO₄(aq) → Ag₂CrO₄(s) + 2KNO₃(aq)

First we need to identify the limiting reactant:

We have:

0.20 M * 50.0 mL = 10 mmol of AgNO₃

0.10 M * 40.0 mL = 4 mmol of K₂CrO₄

If 4 mmol of K₂CrO₄ were to react completely, it would require (4*2) 8 mmol of AgNO₃. There's more than 8 mmol of AgNO₃ so AgNO₃ is the excess reactant. That makes K₂CrO₄ the limiting reactant.

Now we calculate the mass of Ag₂CrO₄ formed, using the limiting reactant:

4 mmol K₂CrO₄ * $\frac{1mmolAg_2CrO_4}{1mmolK_2CrO_4} *\frac{331.73mg}{1mmolAg_2CrO_4}$ = 1326.92 mg Ag₂CrO₄

1326.92 mg / 1000 = 1.327 g Ag₂CrO₄

7 0

3 years ago

What fossil helped support Wegener's hypothesis of continental drift?

Dominik [7]

Answer:

Glossopteris

Explanation:

Glossopteris is a fossil fern that helped support Wegener's hypothesis.

Hope this helps! Have a great day!

3 0

3 years ago

of an unknown protein are dissolved in enough solvent to make 5.00mL of solution. The osmotic pressure of this solution is measu

krok68 [10]

The question is incomplete . The complete question is :

100 mg of an unknown protein are dissolved in enough solvent to make 5.00mL of solution. The osmotic pressure of this solution is measured to be 0.107atm at 25.0°C. Calculate the molar mass of the protein. Round your answer to 3 significant digits.

Answer: The molar mass of the protein is $4.57\times 10^3g/mol$

Explanation:

$\pi =CRT$

$\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = 0.107 atm

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (protein) = 100 mg = 0.1 g (Conversion factor: 1 g = 1000 mg)

Volume of solution = 5.00 mL

R = Gas constant = $0.0821\text{ L.atm }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$0.107=1\times \frac{0.1\times 1000}{\text{Molar mass of insulin}\times 5.00}\times 0.0821\text{ Latm }mol^{-1}K^{-1}\times 298K\\\\\text{molar mass of protein}$

$\text{molar mass of protein}=4.57\times 10^3g/mol$

Hence, the molar mass of the protein is $4.57\times 10^3g/mol$

7 0

3 years ago