All categories

3 years ago

Jimmy uses electrolysis to decompose a hydrochloric acid solution. The hydrochloric acid solution is called the... what?

Chemistry

1 answer:

Anastasy [175]3 years ago

8 0

Answer:

electrolyte

Explanation:

A lliquid which contains ions and can be decomposed by electrolysis

You might be interested in

!!!!HELP PLEASE!!!!!Which of the following molecules has a trigonal pyramidal shape?

aksik [14]

Answer:

No.D is the molecules that has trigonal pyramidal sape

8 0

3 years ago

Read 2 more answers

Calculate the energy required to melt 21 g of ice at 0 oC.

vladimir1956 [14]

We are given:

Mass of ice = 21 grams

The ice is already at 0°c, the temperature at which it melts to form water

Molar heat of fusion of Ice = 6.02 kJ/mol

Finding the energy required:

Number of moles of Ice:

Molar mass of water = 18 g/mol

Number of moles = given mass/ molar mass

Number of moles = 21 / 18 = 7/6 moles

Energy required to melt the given amount of ice:

Energy = number of moles * molar heat of fusion

Energy = (7/6) * (6.02)

Energy = 7.02 kJ OR 7020 joules

7 0

3 years ago

The energy received by Earth from the sun is what type of energy

DedPeter [7]

Answer:

Solar Energy

Explanation:

3 0

3 years ago

Read 2 more answers

1.) The process for converting ammonia to nitric acid involves the conversion of NH3 to

Firdavs [7]

Answer:

a) 1.39 g ; b) O₂ is limiting reactant, NH₃ is excess reactant; c) 0.7 g

Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

MM: 17.03 32.00 30.01

4NH₃ + 5O₂ ⟶ 4NO + 6H₂O

Mass/g: 1.5 1.85

2. Calculate the moles of each reactant

$\text{moles of NH}_{3} = \text{1.5 g NH}_{3} \times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}} = \text{0.0881 mol NH}_{3}\\\\\text{moles of O}_{2} = \text{1.85 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.057 81 mol O}_{2}$

3. Calculate the moles of NO we can obtain from each reactant

From NH₃:

The molar ratio is 4 mol NO:4 mol NH₃

$\text{Moles of NO} = \text{0.0881 mol NH}_{3} \times \dfrac{\text{4 mol NO}}{\text{4 mol NH}_{3}} = \text{0.0881 mol NO}$

From O₂:

The molar ratio is 4 mol NO:5 mol O₂

$\text{Moles of NO} = \text{0.057 81 mol O}_{2}\times \dfrac{\text{4 mol NO}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NO}$

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO.

The excess reactant is NH₃.

5. Calculate the mass of NO formed

$\text{Mass of NO} = \text{0.046 25 mol NO}\times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \textbf{1.39 g NO}$

6. Calculate the moles of NH₃ reacted

The molar ratio is 4 mol NH₃:5 mol O₂

$\text{Moles reacted} = \text{0.057 81 mol O}_{2} \times \dfrac{\text{4 mol NH}_{3}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NH}_{3}$

7. Calculate the mass of NH₃ reacted

$\text{Mass reacted} = \text{0.046 25 mol NH}_{3} \times \dfrac{\text{17.03 g NH}_{3}}{\text{1 mol NH}_{3}} = \text{0.7876 g NH}_{3}$

8. Calculate the mass of NH₃ remaining

Mass remaining = original mass – mass reacted = (1.5 - 0.7876) g = 0.7 g NH₃

8 0

3 years ago

In which of the following cases does the reaction go to farthest to completion ?

Mashcka [7]

Answer:

$k=10^{3}$

Explanation:

k stand for equilibrium constants in terms of reaction

The higher the value of an equilibrium constant the faster the equilibrium reaction comes to completion.

Consider the example below:

$R_{1}\ + R_{2}$ ⇄ $P_{1}\ + P_{2}$

where

$The\ equilibrium\ constant\ K = \frac{P_{1}P_{2}}{R_{1}R_{2}}$

For a faster reaction the numerator i.e. the right hand side of the equation have to be higher than the left hand side (the denominator). therefore the higher the numerator, the higher the value of the equilibrium constant and the faster the reaction get to completion thus option c is correct.

5 0

3 years ago