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2 years ago

Element X reacts with element Y to give a product containing X3+ ions and Y2− ions.

Chemistry

1 answer:

Maksim231197 [3]2 years ago

5 0

Answer:

X₂Y₃

Explanation:

X⇒X³⁺ + 3e⁻ /×2

Y + 2e⁻ ⇒ Y²⁻ /×3

2X⇒2X³⁺ 6e⁻

3Y + 6e⁻⇒ 3Y²⁻

2X + 3Y ⇒2X³⁺ + 3Y²⁻ ⇒ X₂Y₃

Electron from one side and from other side can be shortened so we multiply half equations to get equivalent number of electrons on both side. Next step is summing these two half equations. Element Y takes two electrons from element X, so X become positive charged, and Y become negatively charged. It is very likely that element Y have greater electronegativity.

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9 The Haber process is a reversible reaction. N2(g) + 3H2(g) 2NH3(g) The reaction has a 30% yield of ammonia. Which volume of am

Illusion [34]

Answer: 3.36 L of ammonia gas

Explanation:

The balanced chemical reaction is:

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

According to stoichiometry :

3 moles of $H_2$ produce = 2 moles of $NH_3$

Thus 0.75 moles of $H_2$ will producee= $\frac{2}{3}\times 0.75=0.50moles$ of $NH_3$

But as percent yield is 30 %, amount of ammonia produced = $\frac{30}{100}\times 0.50moles=0.15moles$

According to ideal gas equation:

$PV=nRT$

P = pressure = 1 atm

V = Volume = ?

n = number of moles = 0.15

R = gas constant = $0.0821Latm/Kmol$

T =temperature = $273K$

$V=\frac{nRT}{P}$

$V=\frac{0.15\times 0.0820 L atm/K mol\times 273K}{1atm}=3.36L$

Thus 3.36 L of ammonia gas is obtained by reacting 0.75 moles of hydrogen with excess nitrogen.

3 0

1 year ago

For each of the reactions at constant pressure, determine whether the system does work on the surroundings, the surroundings doe

lesantik [10]

Explanation:

When work is done then there will occur change in volume. And, most change in volume occurs when there will be production of gas is taking place. We assume that no work is done when no gas is produced.

(a) For $4Fe(s) + 3O_{2}(g) \rightarrow 2Fe_{2}O_{3}(s)$

Here, 3 moles of gas is producing 0 moles of gas. This means that work is done on the system.

(b) And, more is the production of a gas taking place in a reaction more will be the amount of work done by the system.

For $2H_{2}O_{2}(g) \rightarrow O_{2}(g) + 2H_{2}O(g)$

Here, 2 moles of a gas is producing 3 moles of a gas. Since, gas is increasing so, work will be done by the system.

4 0

2 years ago

A 2.20 g sample of a compound gave 5.63 g CO2 and 2.30 g H2O on combustion in air. The compound is known to contain only C, H, O

ICE Princess25 [194]

The simplest formula is C₅H₁₀O.

We must calculate the masses of C, H, and O from the masses given.

Mass of C =5.63 g CO₂ × (12.01 g C/44.01 g CO₂) = 1.536 g C

Mass of H = 2.30 g H₂O × (2.016 g H/18.02 g H₂O) = 0.2573 g H

Mass of O = Mass of compound - Mass of C - Mass of H

= (2.20 – 1.536 – 0.2573) g = 0.406 g

Now, we must convert these masses to moles and find their ratios.

From here on, I like to summarize the calculations in a table.

Element Mass/g Moles Ratio Integers

C 1.536 0.1279 5.038 5

H 0.2573 0.2553 10.05 10

O 0.406 0.0254 1 1

The empirical formula is C₅H₁₀O.

7 0

2 years ago

When the volume of gas is changed from 3.75 to 6.52 L , the temperature will change from 100.0 K to how many k?

Fed [463]

Hello!

We have an isobaric transformation, that is, when a certain mass under pressure maintains its constant pressure, on the other hand, as we increase the temperature, the volume increases and if we lower the temperature, the volume decreases and vice versa .

We have the following data:

V1 (initial volume) = 3.75 L

V2 (final volume) = 6.52 L

T1 (initial temperature) = 100 K

T2 (final temperature) =? (in Kelvin)

We apply the data to the formula of isobaric transformation (Gay-Lussac), let us see:

$\dfrac{V_1}{T_1} =\dfrac{V_2}{T_2}$