Carbon is often called "the building block of life". If your body is 18% carbon, what mass

Small quantites of hydrogen gas can be prepared in the laboratory by the addition of aqueous hydrochloric acid to metallic zinc.

ira [324]

<u>Answer:</u> The mass or zinc reacted is 0.624 grams.

<u>Explanation:</u>

We are given:

Total pressure = 1.032 atm

Vapor pressure of water = 32 torr = 0.042 atm (Conversion factor: 1 atm = 760 torr)

To calculate partial pressure of hydrogen gas, we use the equation:

$p_{H_2}=p_T-p_{H_2O}\\\\p_{H_2}=1.032-0.042=0.99atm$

To calculate the number of moles of hydrogen gas, we use the equation given by ideal gas follows:

$PV=nRT$

where,

P = pressure of hydrogen gas = 0.99 atm

V = Volume of hydrogen gas = 240. mL = 0.240 L (Conversion factor: 1 L = 1000 mL)

T = Temperature of hydrogen gas = $30^oC=[30+273]K=303K$

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

n = number of moles of hydrogen gas = ?

Putting values in above equation, we get:

$0.99atm\times 0.240L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 303K\\n=\frac{0.99\times 0.240}{0.0821\times 303}=9.55\times 10^{-3}mol$

The chemical equation for the reaction of zinc and hydrochloric acid follows:

$Zn+2HCl\rightarrow ZnCl_2+H_2$

By Stoichiometry of the reaction:

1 mole of hydrogen gas is produced from 1 mole of zinc metal

So, $9.55\times 10^{-3}mol$ of hydrogen gas is produced from = $\frac{1}{1}\times 9.55\times 10^{-3}=9.55\times 10^{-3}mol$ of zinc metal

To calculate the mass of zinc metal, we use the equation:

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

Molar mass of zinc = 65.38 g/mol

Moles of zinc = $9.55\times 10^{-3}$ moles

Putting values in above equation, we get:

$9.55\times 10^{-3}mol=\frac{\text{Mass of zinc}}{65.38g/mol}\\\\\text{Mass of zinc}=(9.55\times 10^{-3}mol\times 65.38g/mol)=0.624g$

Hence, the mass or zinc reacted is 0.624 grams.

4 0

4 years ago

Which reactant is unlikely to produce the indicated product upon strong heating?

Ilya [14]

2-Methyl-4-oxo-pentanoic acid is unlikely to produce 2-Methyl-3-butanone upon strong heating.

Upon heating, the β ketoacid becomes unstable and decarboxylates, leading to the formation of the methyl ketone.

A carboxylic acid is an organic acid that contains a carboxyl group (C(=O)OH) attached to an R-group. The general formula of a carboxylic acid is R−COOH or R−CO2H, with R referring to the alkyl, alkenyl, aryl, or other group.

Carboxylic acids occur widely. Important examples include the amino acids and fatty acids. Deprotonation of a carboxylic acid gives a carboxylate anion.

Full question :

Q. Which reactant is unlikely to produce the indicated product upon strong heating?

A) 2,2-Dimethylpropanedioic acid 2-methylpropanoic acid
B) 2-Ethylpropanedioic acid Butanoic acid
C) 2-Methyl-3-oxo-pentanoic acid 3-Pentanone
D) 2-Methyl-4-oxo-pentanoic acid 2-Methyl-3-butanone
E) 4-Methyl-3-oxo-heptanoic acid 3-Methyl-2-hexanone

Hence, option (D) is correct.

Learn more about carboxylic acid here : brainly.com/question/26855500

#SPJ4

8 0

2 years ago

What is the balanced chemical equation for the word equation below?

ladessa [460]

Answer:

4Fe + 3O₂ → 2Fe₂O₃

Explanation:

Fe → ²⁺

O → ²⁻

But Iron III is Fe³⁺

So we have Fe³⁺ and O²⁻, the formula for the oxide must be Fe₂O₃ so the equation can be:

4Fe + 3O₂ → 2Fe₂O₃

6 0

3 years ago

If the initial temperature of a movable cylinder was 50 degrees Celsius

slega [8]

Answer:

8.45 L

Explanation:

From the question given above, the following data were obtained:

Initial temperature (T₁) = 50 °C

Initial pressure (P₁) = 2 atm

Initial volume (V₁) = 5 L

Final temperature (T₂) = 0 °C

Final pressure (P₂) = 1 atm

Final volume (V₂) =?

Next, we shall convert celsius temperature to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 50 °C

Initial temperature (T₁) = 50 °C + 273

Initial temperature (T₁) = 323 K

Final temperature (T₂) = 0 °C

Final temperature (T₂) = 0 °C + 273

Final temperature (T₂) = 273 K

Finally, we shall determine the new volume. This can be obtained as follow:

Initial temperature (T₁) = 323 K

Initial pressure (P₁) = 2 atm

Initial volume (V₁) = 5 L

Final temperature (T₂) = 273 k

Final pressure (P₂) = 1 atm

Final volume (V₂) =?

P₁V₁ / T₁ = P₂V₂ / T₂

2 × 5 / 323 = 1 × V₂ / 273

10 / 323 = V₂ / 273

Cross multiply

323 × V₂ = 10 × 273

323 × V₂ = 2730

Divide both side by 323

V₂ = 2730 / 323

V₂ = 8.45 L

Thus, the new volume is 8.45 L

5 0

3 years ago

Instead of 6 M NaOH being added to the solution, 6 M HCl is added. How will this affect the test for the presence of ammonium io

Lera25 [3.4K]

Answer:

Ammonia gas(an alkaline gas with characteristics of choking or irritating smell) is not liberated when 6mole of HCl is added to the solution instead of 6mole of NaOH, to test for the presence of ammonium ion in the solution

Explanation:

As expected, when testing for ammonium ion in a solution (precisely ammonium salt solution), Sodium Hydroxide (NaOH) is required as the test reagent.

When NaOH is added to the solution, A gas with characteristics of choking or irritating smell is liberated.

This gas turn red litmus paper blue.

This liberated gas is an alkaline gas, which is confirmed as an ammonia gas(NH3).

If HCl is added instead of NaOH, the ammonia gas will not be liberated, which indicates that the test reagent used is wrong.

3 0

3 years ago