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2 years ago

Balance this chemical equation. NH4OH AlCl3 → Al(OH)3; NH4Cl

Chemistry

2 answers:

skad [1K]2 years ago

7 0

3NH4OH+AlCl3=Al(OH)3+3NH4Cl

DedPeter [7]2 years ago

5 0

Explanation:

When the number of atoms on both reactant and product side are equal the this chemical equation is known as balanced chemical equation.

For example, $NH_{4}OH + AlCl_{3} \rightarrow Al(OH)_{3} + NH_{4}Cl$

Number of atoms on reactant side are as follows.

N = 1

H = 5

O = 1

Al = 1

Cl = 3

Number of atoms on product side are as follows.

N = 1

H = 7

O = 3

Al = 1

Cl = 1

Therefore, to balance this chemical equation we multiply $NH_{4}OH$ by 3 on reactant side and multiply $NH_{4}Cl$ by 3 on product side.

Hence, the balanced chemical reaction equation is as follows.

$3NH_{4}OH + AlCl_{3} \rightarrow Al(OH)_{3} + 3NH_{4}Cl$

You might be interested in

Convert 2.1 mole of Al2(SO4)3 ionic units to a number of particles.

Semenov [28]

1.26 x 10²⁴particles

Explanation:

Given parameters:

Number of moles of the compound = 2.1 moles

Unknown:

Number of particles contained in the compound = ?

Solution:

A mole can be defined as the amount of a substance that contains Avogadro's number of particles 6.02 x 10²³.

The particles mentioned here can be atoms, molecules, formula units, ions, electrons, protons, neutrons and so forth.

To solve for the number of moles:

Number of particles = number of moles x 6.02 x 10²³

Number of particles = 2.1 x 6.02 x 10²³ = 12.64 x 10²³ = 1.26 x 10²⁴particles

Learn more:

mole calculation brainly.com/question/1841136

#learnwithBrainly

8 0

2 years ago

If you compared 1 m solutions, was a 1 m nacl solution more or less hypertonic than a 1 m sucrose solution? what is your evidenc

igor_vitrenko [27]

When you say the solution is hypertonic, it means that the solution has a higher osmotic pressure. The formula for this is:

P = iMRT,
for strong electrolytes, i = number of ions.
for nonelectrolytes, i = 1

1. The P for sucrose solution which is a nonelectrolyte (assuming room temp):
P = (1)(1m)(8.314 J/mol-K)(298 K)
P = 2477.572 Pa

The P for NaCl solution, which is a strong electrolyte:
P = (2)(1 m)(8.314)(298 K)
P = 4955.144 Pa

So, that means that NaCl is more hypertonic than the sucrose solution.

2. For the second question, the P for the combination of 1 m glucose (nonelectrolyte) and 1 m sucrose is:
P = (1)(1 m)(8.314)(298 K) + (1)(1)(8.314)(298 K) = 4955.144 Pa
In this case, the osmotic pressures are now equal. It is not hypertonic, but isotonic.

7 0

2 years ago

A sample of a solid labeled as NaCl may be impure. a student analyzes the sample and determines that it contains 75 percent chlo

Evgesh-ka [11]

Answer:

Explanation:

The sample must contain impurity that is lower in atomic mass to sodium and since potassium has higher atomic mass to sodium, the answer is the sample contains NaCl and LiCl. We are sure already that the sample is not pure which rules out option a and option b contains sodium iodide which cannot contribute to the increase in chlorine

5 0

2 years ago

What mass of solid NaOH (97.0 % by mass) is required to prepare 1.00 L of a 10.0% solution of NaOH by mass? The density of the 1

Lelechka [254]

Answer: The mass of 97 % of NaOH solution required is 114.33 g

Explanation:

To calculate mass of a substance, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

We are given:

Density of 10 % solution = 1.109 g/mL

Volume of 10% solution = 1 L = 1000 mL (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$1.109g/mL=\frac{\text{Mass of }10\%\text{ solution}}{1000mL}\\\\\text{Mass of }10\%\text{ solution}=1109g$

The mass of 10 % solution is 1109 g.

To calculate the mass of concentrated solution, we use the equation:

$c_1m_1=c_2m_2$

where,

$c_1\text{ and }m_1$ are the concentration and mass of concentrated solution.

$c_2\text{ and }m_2$ are the concentration and mass of diluted solution.

We are given:

$c_1=97\%\\m_1=?g\\c_2=10\%\\m_2=1109g$

Putting values in above equation, we get:

$97\times m_1=10\times 1109\\\\m_1=114.33g$

Hence, the mass of 97 % of NaOH solution required is 114.33 g

3 0

2 years ago

In 200 g of a concentrated solution of 70.4 wt% nitric acid (r = 1.41 g/mL, FW(HNO3) = 63.01 g/mol), how many grams of water are

mars1129 [50]

Answer:

59.2 grams

Explanation:

We are given that 70.4% of the weight of the total 200 g of the concentration is made up of nitric acid, the remaining information is not required to solve the problem. Since water and nitric acid are the only components of the solution, the total weight of water is given by:

$W = 200*(1-0.704)\\W=59.2\ g$

There are 59.2 grams of water in this solution.

5 0

2 years ago