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2 years ago

9

Magnesium hydroxide is used as an antacid in milk of magnesia and reacts with hydrogen chlrode in the stomach to form water and

mageniusm chloride. write out the chemical equation for this reaction with the correct chemical formulas, and balance the equation

1 answer:

8_murik_8 [283]2 years ago

3 0

Mg(OH)2 +2HCl---> MgCl2+2H2O

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The density of an aqueous solution of nitric acid is 1.64 g/mL and the concentration is 1.85 M. What is the concentration of thi

galina1969 [7]

Answer:

Mass % of the solution = 7.1067 %

Explanation:

Given :

Molarity of nitric acid solution = 1.85 M

Density of the solution = 1.64 g/mL

<u>Molarity of a solution is defined as the number of moles of solute present in 1 liter of the solution.</u>

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Lets, consider the volume of the solution = 1 L

Thus,

Moles of nitric acid present in the solution:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Moles of Nitric acid=Molarity \times {Volume\ of\ the\ solution}$

So,

Moles of Nitric acid = 1.85 moles

Molar mass of nitric acid = 63 g/mol

The mass of Nitric acid can be find out by using mole formula as:

$moles=\frac{Mass\ taken}{Molar\ mass}$

Thus,

$Mass\ of\ Nitric\ acid=Moles \times Molar mass}$

$Mass\ of\ Nitric\ acid=1.85 g \times 63 g/mol}$

<u>Mass of Nitric acid = 116.55 g</u>

Also,

$Density=\frac{Mass}{Volume}$

Given : Density = 1.64 g/mL

Also, 1 L = 10³ mL

Volume of the solution is 1000 mL

So, mass of the solution:

$Mass\ of\ the\ solution=Density \times {Volume\ of\ the\ solution}$

$Mass\ of\ the\ solution=1.64 g/mL \times {1000 mL}$

<u>Mass of the solution = 1640 g</u>

Mass % is defined as the mass of solute in 100 g of the solution. The formula for the calculation of mass % is shown below:

$Mass \% =\frac{Mass\ of\ the\ solute}{Mass\ of\ the\ solution} \times {100}$

So,

$Mass \%=\frac{116.55}{1640} \times {100}$

<u>Mass % = 7.1067 %</u>

6 0

2 years ago

31. calculate the amount of energy released as heat by the freezing of 13.3 g of a liquid substance, given that the substance ha

spin [16.1K]

From the calculations, the heat of fusion of the substance is 0.73 kJ

<h3>What is is the heat of freezing?</h3>

The heat of freezing is the energy released when the substance is converted from liquid to solid.

Now we know that the molar mass of the substance is 82.9 g/mol hence the number of moles of the substance is; 13.3 g /82.9 g/mol = 0.16 moles

Now the heat of fusion shall be;

H = 4.60 kj/mol * 0.16 moles

H = 0.73 kJ

Learn more about freezing:brainly.com/question/3121416

#SPJ4

5 0

1 year ago

78.6 grams of O2 and 67.3 grams of F2 are placed in a container with a volume of 40.6 L. Find the total pressure if the gasses a

saul85 [17]

1) List the known and unknown quantities.

<em>Sample: O2.</em>

Mass: 78.6 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K.

<em>Sample: F2.</em>

Mass: 67.3 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K.

2) Find the pressure of O2.

<em>2.1- List the known and unknown quantities.</em>

<em>Sample: O2.</em>

Mass: 78.6 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

<em>2.2- Convert grams of O2 to moles of O2.</em>

The molar mass of O2 is 31.9988 g/mol.

$mol\text{ }O_2=78.6\text{ }g*\frac{1\text{ }mol\text{ }O_2}{31.9988\text{ }g\text{ }O_2}=2.46\text{ }mol\text{ }O_2$

<em>2.3- Set the equation.</em>

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1)

$PV=nRT$

<em>2.4- Plug in the known quantities and solve for P.</em>

$(P)(40.6\text{ }L)=(2.46\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(316.28\text{ }K)$

<em>.</em>

$P_{O_2}=\frac{(2.46\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(316.28\text{ }K)}{40.6\text{ }L}$ $P_{O_2}=1.57\text{ }atm$

<em>The pressure of O2 is 1.57 atm.</em>

3) Find the pressure of F2.

<em>3.1- List the known and unknown quantities.</em>

<em>Sample: F2.</em>

Mass: 67.3 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

3.2- <em>Convert grams of F2 to moles of F2.</em>

The mmolar mass of F2 is 37.9968 g/mol.

$mol\text{ }F_2=67.3\text{ }g\text{ }F_2*\frac{1\text{ }mol\text{ }F_2}{37.9968\text{ }g\text{ }F_2}=1.77\text{ }mol\text{ }F_2$

<em>3.3- Set the equation.</em>

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1)

$PV=nRT$

<em>3.4- Plug in the known quantities and solve for P.</em>

$(P)(40.6\text{ }L)=(1.77\text{ }mol\text{ }F_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(316.28\text{ }K)$

<em>.</em>

$P_{F_2}=\frac{(1.77molF_2)(0.082057L*atm*K^{-1}*mol^{-1})(316.28K)}{40.6\text{ }L}$ $P_{F_2}=1.13\text{ }atm$

<em>The pressure of F2 is 1.13 atm.</em>

4) The total pressure.

Dalton's law - Partial pressure. This law states that the total pressure of a gas is equal to the sum of the individual partial pressures.

<em>4.1- Set the equation.</em>

$P_T=P_A+P_B$

4.2- Plug in the known quantities.

$P_T=1.57\text{ }atm+1.13\text{ }atm$ $P_T=2.7\text{ }atm$

<em>The total pressure in the container is </em>2.7 atm<em>.</em>

5 0

9 months ago

What is the pH of a solution whose H + concentration is 4.0 10 –9?

Dmitrij [34]

Answer:

he pH of a solution is defined as the negative log10 [H+] ... 1 x 10-11. 11. Acidic Solution. 1 x 10-4. 4. 1 x 10-10. 10. 1 x 10-5. 5. 1 x 10-9.

Explanation:

m

5 0

2 years ago

A zinc slug comes from a science supply company with a stated mass of 5.000 g. A student weighs the slugthree times, collecting

soldi70 [24.7K]

Answer: The student’s values are accurate as well as precise.

Explanation:

Precision refers to the closeness of two or more measurements to each other.

For Example: If you weigh a given substance three times and you get same value each time. Then the measurement is very precise.

Accuracy refers to the closeness of a measured value to a standard or known value.

For Example: If the mass of a substance is 50 kg and one person weighed 49 kg and another person weighed 48 kg. Then, the weight measured by first person is more accurate.

Given: Mass = 5.000 g

Mass weighed by A has values 4.891 g , 4.901 g and 4.890. Thus the average value is $\frac{4.891+4.901+4.890}{3}=4.894$

Thus as the measured value is close to the true value, the student’s values are accurate and as the values are close to each other, the measurement is precise.

8 0

3 years ago