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2 years ago

What is the volume, in liters, of a 0.2 M solution containing 8.5 grams of AgNO3?

Chemistry

1 answer:

swat322 years ago

4 0

Answer:

The volume of a 0.2 M solution containing 8.5 grams of AgNO₃ is 0.25 L

Explanation:

Molarity (M) is a common way of expressing the concentration of solutions. It is defined as the number of moles of solute per volume of solution. The molarity of a solution is calculated by dividing the moles of the solute by the volume of the solution.

$Molarity=\frac{number of moles of solute}{volume}$

Molarity is expressed in units $\frac{moles}{liter}$ .

Being the molar mass of AgNO3 169.87 g / mol, then you can apply the following rule of three: if 169.87 grams are present in 1 mole of the compound, 8.5 grams will be present in how many moles?

$amount of moles=\frac{8.5 grams*1 mole}{169.87 grams}$

amount of moles= 0.05

So, being the molarity equal to 0.2 M, replacing in the definition of molarity:

$0.2 M=\frac{0.05 moles}{volume}$

and solving you get:

$volume=\frac{0.05 moles}{0.2 M}$

volume=0.25 L

<u><em>The volume of a 0.2 M solution containing 8.5 grams of AgNO₃ is 0.25 L</em></u>

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What type of bond is between sodium and oxygen

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Answer:

Ionic

Explanation:

Sodium is Metal, Oxygen is Non-metal. Non-metals and metals are automatic Ionic bonds

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3 years ago

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If 4.0 g of helium gas occupies a volume of 22.4 L at 0 o C and a pressure of 1.0 atm, what volume does 3.0 g of He occupy under

WINSTONCH [101]

Answer:

the volume occupied by 3.0 g of the gas is 16.8 L.

Explanation:

Given;

initial reacting mass of the helium gas, m₁ = 4.0 g

volume occupied by the helium gas, V = 22.4 L

pressure of the gas, P = 1 .0 atm

temperature of the gas, T = 0⁰C = 273 K

atomic mass of helium gas, M = 4.0 g/mol

initial number of moles of the gas is calculated as follows;

$n_1 = \frac{m_1}{M} \\\\n_1 = \frac{4}{4} = 1$

The number of moles of the gas when the reacting mass is 3.0 g;

m₂ = 3.0 g

$n_2 = \frac{m_2}{M} \\\\n_2 = \frac{3}{4} \\\\n_2 = 0.75 \ mol$

The volume of the gas at 0.75 mol is determined using ideal gas law;

PV = nRT

$PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\since, \ \frac{RT}{P} \ is \ constant,\ then;\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1n_2}{n_1} \\\\V_2 = \frac{22.4 \times 0.75}{1} \\\\V_2 = 16.8 \ L$

Therefore, the volume occupied by 3.0 g of the gas is 16.8 L.

4 0

3 years ago

How many moles are in 1.05 g of gold (Au)?

Wittaler [7]

Answer:

0.005 mol

Explanation:

Moles is denoted by given mass divided by the molecular mass ,

Hence ,

n = w / m

n = moles ,

w = given mass ,

m = molecular mass .

From the question ,

w = given mass of Gold = 1.05 g ,

m = molecular mass of Gold = 197 g/mol

<u>Hence , moles can be calculated as -</u>

n = w / m = 1.05 g / 197 g/mol = 0.005 mol

7 0

3 years ago

Which of the following shows the conservation of mass during cellular respiration? 3 CO2 3 H2O energy → 3 C6H12O6 3 O2 6 CO2 6 H

Thepotemich [5.8K]

The reaction that has been, following the law of conservation of mass has been $\rm C_6H_1_2O_6\;+\;6\;O_2\;\rightarrow\;6\;CO_2\;+\;6\;H_2O\;+\;Energy$ .

The law of conservation has been given in the chemical reaction that there has been no loss or gain of the mass and energy.

The law of conservation has been evident when there has been an equal number of atoms of each element on the product and the reactant side.

<h3 /><h3>Conservation of mass in Cellular respiration</h3>

The following reactions have been identified as:

$\rm 3\;CO_2\;+\;3\;H_2O\;+\;Energy\;\rightarrow\;3\;C_6H_1_2O_6$

Carbon atoms

Reactant = 3

Product = 18

Oxygen atoms

Reactant = 9

Product = 18

Hydrogen atoms

Reactant = 6

Product = 36

The number of atoms is not equal on the product and reactant side, thus not follows the law of conservation of mass.

$\rm 3\;O_2\;+\;6\;CO_2\;+\;6\;H_2O\;+\;Energy\;\rightarrow\;C_6H_1_2O_6\;+\;6\;O_2$

Carbon atoms

Reactant = 6

Product = 6

Oxygen atoms

Reactant = 24

Product = 18

Hydrogen atoms

Reactant = 12

Product = 12

The number of atoms is not equal on the product and reactant side, thus not follows the law of conservation of mass.

$\rm C_6H_1_2O_6\;+\;6\;O_2\;\rightarrow\;6\;CO_2\;+\;6\;H_2O\;+\;Energy$

Carbon atoms

Reactant = 6

Product = 6

Oxygen atoms

Reactant = 18

Product = 18

Hydrogen atoms

Reactant = 12

Product = 12

The number of atoms is equal on the product and reactant side, thus follows the law of conservation of mass.

$\rm 6\;H_2O\;+\;C_6H_1_2O_6\;\rightarrow\;6\;O_2\;+\;Energy$

Carbon atoms

Reactant = 6

Product = 0

Oxygen atoms

Reactant = 12

Product = 12

Hydrogen atoms

Reactant = 24

Product = 0

The number of atoms is not equal on the product and reactant side, thus not follows the law of conservation of mass.

Thus, the reaction that has been following the law of conservation of mass has been $\rm C_6H_1_2O_6\;+\;6\;O_2\;\rightarrow\;6\;CO_2\;+\;6\;H_2O\;+\;Energy$ .

Learn more about law of conservation, here:

brainly.com/question/2175724

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2 years ago

C10H12O4S(s) + O2(g)  CO2(g) + SO2(g) + H2O(g)

ale4655 [162]

Answer: The coefficient for $O_2(g)$ is 12.

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

$C_{10}H_{12}O_4S(s)+12O_2(g)\rightarrow 10CO_2(g)+SO_2(g)+6H_2O$

Thus in the reactants, there are 12 molecules of oxygen in balanced chemical equation. Thus the coefficient for $O_2(g)$ is 12.

8 0

3 years ago