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julia-pushkina [17]
2 years ago
13

Write down the dissolution equation for rubidium chromate dissolving in water. (Chromate is a polyatomic ion with the formula Cr

O42-.) If two moles of the ionic compound are dissolved, then how many moles of the CATION are present in the solution?
Chemistry
1 answer:
ycow [4]2 years ago
5 0

Answer:

Four moles of the cation

Explanation:

2Rb2CrO4(s)<--------> 4Rb^+(aq) + 2CrO4^2-(aq)

Now looking at the reaction equation, it can be seen that one mole of rubidium chromate contains two moles of rubidium ions and one mole of chromate ions.

The dissolution of two moles of rubidium chromate should then yield four moles of rubidium ions and two moles of chromate ions since the ratio of ions present is 2:1.

This explains the reaction equation written above for the dissolution of two moles of rubidium chromate as shown.

You might be interested in
Balance the following redox reaction occurring in an acidic solution. The coefficient of Cr2O72−(aq) is given. Enter the coeffic
ollegr [7]

Answer:

Cr₂O₇²⁻ (aq) + 6Ti³⁺ (aq) + 2 H⁺(aq) → 2 Cr³⁺ (aq) + 6TiO²⁺(aq) + H2O(l)

Explanation:

Cr₂O₇²⁻ (aq) + Ti³⁺ (aq) +  H⁺ (aq)  → Cr³⁺ (aq) + TiO²⁺ (aq) +  H₂O(l)

This is the reaction, without stoichiometry.

We have to notice the oxidation number of each element.

In dicromate, Cr acts with +6 and we have Cr³⁺, so oxidation number has decreased. .- REDUCTION

Ti³⁺ acts with +3, and in TiO²⁺ acts with +4 so oxidation number has increased.  .- OXIDATION

14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ + 7H₂O

To change 6+ to 3+, Cr had to lose 3 e⁻, but as we have two Cr, it has lost 6e⁻. As we have 7 Oxygens in reactant side, we have to add water, as the same amount of oxygens atoms we have, in products side. Finally, we have to add protons in reactant side, to ballance the H.

H₂O  +  Ti³⁺  →  TiO²⁺ + 1e⁻ + 2H⁺

Titanium has to win 1 e⁻ to change 3+ to 4+. We had to add 1 water in reactant, and 2H⁺ in products, to get all the half reaction ballanced.

Now we have to ballance the electrons, so we can cancel them.

(14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ + 7H₂O) .1

(H₂O  +  Ti³⁺  →  TiO²⁺ + 1e⁻ + 2H⁺) .6

Wre multiply, second half reaction .6

14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ + 7H₂O

6H₂O  +  6Ti³⁺  →  6TiO²⁺ + 6e⁻ + 12H⁺

Now we can sum, the half reactions:

14 H⁺ + Cr₂O₇²⁻ + 6e⁻  + 6H₂O  +  6Ti³⁺  → 6TiO²⁺ + 6e⁻ + 12H⁺ + 2Cr³⁺ + 7H₂O

Electrons are cancelled and we can also operate with water and protons

7H₂O - 6H₂O = H₂O

14 H⁺ - 12H⁺ = 2H⁺

The final ballanced equation is:

2H⁺ + Cr₂O₇²⁻ + 6Ti³⁺  → 6TiO²⁺  + 2Cr³⁺ + H₂O

8 0
2 years ago
Salt in crude oil must be removed before the oil undergoes processing in a refinery. The
irina1246 [14]

Answer:

\large \boxed{0.64 \, \%}

Explanation:

Assume you are using 1 L of water.

Then you are washing 4 L of salty oil.

1. Calculate the mass of the salty oil

Assume the oil has a density of 0.86 g/mL.

\text{Mass of oil} = \text{4000 mL} \times \dfrac{\text{0.86 g}}{\text{1 mL}} = \text{3440 g}

2. Calculate the mass of salt in the salty oil

\text{Mass of salt} = \text{3440 g} \times \dfrac{\text{5 g salt}}{\text{100 g oil}} = \text{172 g salt}

3. Calculate the mass of salt in the spent water

\text{Mass of salt} = \text{1000 g water} \times \dfrac{\text{15 g salt}}{\text{100 g water}} = \text{150 g salt}

4. Mass of salt remaining in washed oil

Mass = 172 g - 150 g = 22 g  

5. Concentration of salt in washed oil

\text{Concentration} = \dfrac{\text{22 g}}{\text{3440 g}} \times 100 \, \% = \mathbf{0.64 \, \%}\\\\\text{The concentration of salt in the washed oil is $\large \boxed{\mathbf{0.64 \, \%}}$}

3 0
2 years ago
A human lung at maximum capacity has a volume of 3.0 liters. If the partial pressure of oxygen in the air is 21.1 kilopascals an
MrRa [10]

Answer : 0.026 moles of oxygen are in the lung

Explanation :

We can solve the given question using ideal gas law.

The equation is given below.

PV = nRT

We have been given P = 21.1 kPa

Let us convert pressure from kPa to atm unit.

The conversion factor used here is 1 atm = 101.3 kPa.

21.1 kPa \times \frac{1atm}{101.3kPa}= 0.208 atm

V = 3.0 L

T = 295 K

R = 0.0821 L-atm/mol K

Let us rearrange the equation to solve for n.

n = \frac{PV}{RT}

n = \frac{0.208atm\times 3.0L}{0.0821 L.atm/mol K\times 295 K}

n = 0.026 mol

0.026 moles of oxygen are in the lung

3 0
2 years ago
A student wants to draw a model of an atom. Which statement describes how to find the number of neutrons to include in the model
san4es73 [151]

c is the correct answer

4 0
2 years ago
According to the equation below, how many moles of Ca(OH)2 are required to react with 1.36 mol H3PO4 to produce Ca3(PO4)2? 3Ca(O
allsm [11]

<u>Answer:</u> The amount of calcium hydroxide needed to react is 2.04 moles

<u>Explanation:</u>

We are given:

Moles of phosphoric acid = 1.36 moles

For the given chemical equation:

3Ca(OH)_2+2H_3PO_4\rightarrow Ca_3(PO_4)_2+6H_2O

By Stoichiometry of the reaction:

2 moles of phosphoric acid reacts with 3 moles of calcium hydroxide

So, 1.36 moles of phosphoric acid will react with = \frac{3}{2}\times 1.36=2.04mol of calcium hydroxide

Hence, the amount of calcium hydroxide needed to react is 2.04 moles

3 0
2 years ago
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