The mass of an object is 205 gram while the volume was measured to be 82.5 mL. What is this object's density?

Consider the following equilibrium: H2CO3+H2O = H3O+HCO3^-1. What is the correct equilibrium expression?

mylen [45]

Equilibrium expression is $Keq = \frac{[H3O+][HCO3^-]}{[H2CO3]}\\$

<u>Explanation:</u>

Equilibrium expression is denoted by Keq.

Keq is the equilibrium constant that is defined as the ratio of concentration of products to the concentration of reactants each raised to the power its stoichiometric coefficients.

Example -

aA + bB = cC + dD

So, Keq = conc of product/ conc of reactant

$Keq = \frac{[C]^c [D]^d}{[A]^a [B]^b}$

So from the equation, H₂CO₃+H₂O = H₃O+HCO₃⁻¹

$Keq = \frac{[H3O^+]^1 [HCO3^-]^1}{[H2CO3]^1 [H2O]^1}$

The concentration of pure solid and liquid is considered as 1. Therefore, concentration of H2O is 1.

Thus,

$Keq = \frac{[H3O+][HCO3^-]}{[H2CO3]}\\$

Therefore, Equilibrium expression is $Keq = \frac{[H3O+][HCO3^-]}{[H2CO3]}\\$

4 0

3 years ago

I. O2 is chemically bonded together.

oee [108]

THE ANSWER IS B SORRY FOR THE CAPS(:

4 0

3 years ago

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How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos

Alexeev081 [22]

Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of $H_3PO_4$ = 54.219 g

Number of atoms of $Mg$ = $7.179\times 10^{23}$

Molar mass of $H_3PO_4$ = 98 g/mol

First we have to calculate the moles of $H_3PO_4$ and $Mg$ .

$\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}$

$\text{Moles of }H_3PO_4=\frac{54.219g}{98g/mol}=0.553mol$

and,

$\text{Moles of }Mg=\frac{7.179\times 10^{23}}{6.022\times 10^{23}}=1.19mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2$

From the balanced reaction we conclude that

As, 3 mole of $Mg$ react with 2 mole of $H_3PO_4$

So, 0.553 moles of $Mg$ react with $\frac{2}{3}\times 0.553=0.369$ moles of $H_3PO_4$

From this we conclude that, $H_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $Mg$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2$

From the reaction, we conclude that

As, 3 mole of $Mg$ react to give 3 mole of $H_2$

So, 0.553 mole of $Mg$ react to give 0.553 mole of $H_2$

Now we have to calculate the volume of $H_2$ gas at STP.

As we know that, 1 mole of substance occupies 22.4 L volume of gas.

As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas

So, 0.553 mole of hydrogen gas occupies $0.553\times 22.4=12.4L$ volume of hydrogen gas

Therefore, the volume of hydrogen gas produced will be, 12.4 L

4 0

3 years ago

Why atomic mass is an average vàlue

lukranit [14]

Answer:

The atomic mass is the average number of protons and neutrons for all natural isotopes of an element. It is a decimal number.

Explanation:

Atomic Mass and Mass Number Example :

Hydrogen has three natural isotopes: 1H, 2H, and 3H. Each isotope has a different mass number.

1H has 1 proton. Its mass number is 1. 2H has 1 proton and 1 neutron. Its mass number is 2. 3H has 1 proton and 2 neutrons. Its mass number is 3. 99.98% of all hydrogen is 1H 0.018% of all hydrogen is 2H 0.002% of all hydrogen is 3H Together, they give a value of atomic mass of hydrogen equal to 1.0079 g/mol.

4 0

3 years ago

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What happens to the volume of a gas in a closed container if the temperature increases, but the pressure remains the same? Why?

labwork [276]

Answer:

Volume will goes to increase.

Explanation:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

So when the temperature goes to increase the volume of gas also increase. Higher temperature increase the kinetic energy and molecules move randomly every where in given space so volume increase.

Now we will put the suppose values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 4.5 L × 348 K / 298 k

V₂ = 1566 L.K / 298 K

V₂ = 5.3 L

Hence prove that volume increase by increasing the temperature.

3 0

3 years ago