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tigry1 [53]
3 years ago
12

What volume of a 2.5 M stock solution of acetic acid (HC2H3O2) is required to prepare 100.0 milliliters of a 0.50 M acetic acid

solution?. . Use the equation. Mconcentrated*Vconcentrated = Mdilute*Vdilute.
Chemistry
1 answer:
olga55 [171]3 years ago
4 0
We are given with a 2.5 M stock solution of acetic acid and we are required to calculate the volume of the solution needed to prepare 100 milliliters of 0.5 M acetic acid solution. To solve this, we acquire the formula <span>Mconcentrated*Vconcentrated = Mdilute*Vdilute. That is 2.5 M*x=0.5M*100 ml where x is the volume of 2.5 M needed. x is equal to 20 ml. So we need 20 ml of 2.5 M solution and dilute to 100 ml using water as diluent.</span>
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Answer:

Cellular Respiration

Explanation:

Plants convert glucose to energy through a thing called cellular respiration. This process is when glucose + oxygen in the precenses of sunlight = carbon dioxide, water, and ATP(energy for the cells)

3 0
2 years ago
Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M.Express your answer to two significant f
N76 [4]

The question is incomplete, here is the complete question:

Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reaction

A + 2B ⇔ C

whose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:

The table is attached below as an image.

<u>Answer:</u> The initial rate for the formation of C at 25°C is 2.25\times 10^{-2}Ms^{-1}

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

A+2B\rightleftharpoons C

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b

where,

a = order with respect to A

b = order with respect to B

  • Expression for rate law for first trial:

5.4\times 10^{-3}=k(0.30)^a(0.050)^b ....(1)

  • Expression for rate law for second trial:

1.1\times 10^{-2}=k(0.30)^a(0.100)^b ....(2)

  • Expression for rate law for third trial:

2.2\times 10^{-2}=k(0.50)^a(0.050)^b ....(3)

Dividing 2 by 1, we get:

\frac{1.1\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.30)^a(1.00)^b}{(0.30)^a(0.050)^b}\\\\2=2^b\\b=1

Dividing 3 by 1, we get:

\frac{2.2\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.50)^a(0.050)^b}{(0.30)^a(0.050)^b}\\\\4.07=2^a\\a=2

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^1       ......(4)

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

5.4\times 10^{-3}=k[0.30]^2[0.050]^1\\\\k=1.2M^{-2}s^{-1}

Calculating the initial rate of formation of C by using equation 4, we get:

k=1.2M^{-2}s^{-1}

[A] = 0.50 M

[B] = 0.075 M

Putting values in equation 4, we get:

\text{Rate}=1.2\times (0.50)^2\times (0.075)^1\\\\\text{Rate}=2.25\times 10^{-2}Ms^{-1}

Hence, the initial rate for the formation of C at 25°C is 2.25\times 10^{-2}Ms^{-1}

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The balanced equation for combustion in an acetylene torch is shown below:
vichka [17]

Answer:

70mol

Explanation:

The equation of the reaction is given as:

                  2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O

Given parameters:

Number of moles of acetylene = 35.0mol

Number of moles of oxygen in the tank = 84.0mol

Unknown:

Number of moles of CO₂ produced = 35.0mol

Solution:

From the information given about the reaction, we know that the reactant that limits this combustion process is acetylene. Oxygen is given in excess and we don't know the number of moles of this gas that was used up. We know for sure that all the moles of acetylene provided was used to furnish the burning procedure.

To determine the number of moles of CO₂ produced, we use the stoichiometric relationship between the known acetylene and the CO₂ produced from the balanced chemical equation:

From the equation:

         2 moles of acetylene produced 4 moles of CO₂

          ∴ 35.0 mol of acetylene would produced:  

                               \frac{35 x 4}{2} = 70mol

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