What the volume of 7 mol of hydrogen at stp

Various members of a class of compounds, alkenes, react with hydrogen to produce a corresponding alkane. Termed hydrogenation, t

Vitek1552 [10]

<u>Answer:</u> The mass of decane produced is $1.743\times 10^2g$

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

Mass of hydrogen gas = 2.45 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1:, we get:

$\text{Moles of }H_2=\frac{2.45g}{2g/mol}=1.225mol$

The chemical equation for the hydrogenation of decene follows:

$C_{10}H_{20}(l)+H_2(g)\rightarrow C_{10}H_{22}(s)$

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

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

By Stoichiometry of the reaction:

1 mole of hydrogen gas produces 1 mole of decane.

So, 1.225 moles of hydrogen gas will produce = $\frac{1}{1}\times 1.225=1.225mol$ of decane

Now, calculating the mass of decane by using equation 1, we get:

Moles of decane = 1.225 mol

Molar mass of decane = 142.30 g/mol

Putting values in equation 1, we get:

$1.225mol=\frac{\text{Mass of decane}}{142.30g/mol}\\\\\text{Mass of carbon dioxide}=(1.225mol\times 142.30g/mol)=174.3g=1.743\times 10^2g$

Hence, the mass of decane produced is $1.743\times 10^2g$

5 0

3 years ago

For each equation , write all possible mole ratios . A. 2HgO(s)>2Hg(l)+O2(g) B. 4NH3(g)+6NO(g)>5N2(g)+6H2O(l)

Dafna11 [192]

A) For balanced chemical equation: 2HgO(s) → 2Hg(l) + O₂(g).

1) Mole ratio 1: n(HgO) : n(Hg) = 2 : 2 (1 : 1).

2) Mole ratio 2: n(HgO) : n(O₂) = 2 : 1.

3) Mole ratio 3: n(Hg) : n(O₂) = 2 : 1.

B) Balanced chemical equation: 4NH₃(g) + 6NO(g) → 5N₂(g) + 6H₂O(l).

1) Mole ratio 1: n(NH₃) : n(NO) = 4 : 6 (2 : 3).

2) Mole ratio 2: n(NH₃) : n(N₂) = 4 : 5.

3) Mole ratio 3: n(NH₃) : n(H₂O) = 4 : 6 (2 : 3).

4) Mole ratio 4: n(NO) : n(N₂) = 6 : 5.

5) Mole ratio 5: n(NO) : n(H₂O) = 6 : 6 (1 :1).

6) Mole ratio 6: n(N₂) : n(H₂O) = 5 : 6.

3 0

3 years ago

Read 2 more answers

If the distance between a point charge and a neutral atom and is multiplied by a factor of 5, by what factor does the force on t

dexar [7]

Given :

The distance between a point charge and a neutral atom and is multiplied by a factor of 5.

To Find :

By what factor does the force on the neutral atom by the point charge change.

Solution :

We know, electrostatic force between two object is directly proportional to product of charge and inversely proportional to distance between them.

Now, charge in neutral atom is 0 C.

So, the electrostatic force between two of them is also 0 N.

Therefore, by changing distance between the charge the forces did no change ( it remains zero).

3 0

3 years ago

Why does the titration of a weak acid with a strong base always have a basic equivalence point? Why does the titration of a weak

CaHeK987 [17]

Answer: Option (c) is the correct answer.

Explanation:

When a weak acid reacts with a strong base then it results into the formation of a basic solution. Hence, the resulting solution will always have a pH greater than 7.

Since, at the equivalence point number of hydrogen ions become equal to the hydroxide ions. Therefore, pH of solution will be about 7.

So at the equivalence point, the weak acid will get neutralized due to the addition of strong base. Therefore, it will lead to the formation of conjugate base.

As a result, the solution will become slightly basic in nature.

Thus, we can conclude that at the equivalence point, the acid has all been converted into its conjugate base, resulting in a weakly acidic solution because at the equivalence point, the acid has all been converted into its conjugate base, resulting in a weakly basic solution.

5 0

3 years ago

Neptunium-237 undergoes a series of α-particle and β-particle productions to end up as thallium-205. How many α particles and β

Fantom [35]

<h3>Answer:</h3>

8 alpha particles

4 beta particles

<h3>Explanation:</h3>

<u>We are given;</u>

Neptunium-237
Thallium-205
Neptunium-237 undergoes beta and alpha decay to form Thallium-205.

We are required to determine the number of beta and alpha particles produced to complete the decay series.

We need to know that when a radioisotope emits an alpha particle the mass number reduces by 4 while the atomic number decreases by 2.
When a beta particle is emitted the mass number of the radioisotope increases by 1 while the atomic number remains the same.

In this case;

Neptunium-237 has an atomic number 93, while,

Thallium-205 has an atomic number 81.

Therefore;

²³⁷₉₃Np → x⁴₂He + y⁰₋₁e + ²⁰⁵₈₁Ti

We can get x and y

237 = 4x + y(0) + 205

237-205 = 4x

4x = 32

x = 8

On the other hand;

93 = 2x + (-y) + 81

but x = 8

93 = 16 -y + 81

y = 4

Therefore, the complete decay equation is;

²³⁷₉₃Np → 8⁴₂He + 4⁰₋₁e + ²⁰⁵₈₁Ti

Thus, Neptunium-237 emits 8 alpha particles and 4 beta particles to become Thallium-205.

5 0

2 years ago