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sasho [114]
3 years ago
8

The student titrated 10 ml of standered 0.15 M HCl with his sodium hydroxide solution. When the titration reached the equivalenc

e point, the student found that he had used
10.3 ml of Sodium hydroxide solution. Calculate the molarity of the sodium hydroxide solution.
Chemistry
1 answer:
N76 [4]3 years ago
6 0

Answer:

0.15 M

Explanation:

Step 1: Write the neutralization reaction

NaOH + HCl ⇒ NaCl + H₂O

Step 2: Calculate the moles of HCl that reacted

10 mL of 0.15 M HCl was used. The moles of HCl that reacted are:

0.010L \times \frac{0.15mol}{L} = 1.5 \times 10^{-3} mol

Step 3: Calculate the moles of NaOH that reacted

The molar ratio of NaOH to HCl is 1:1. Then, the moles of NaOH that reacted are 1.5 × 10⁻³ moles.

Step 4: Calculate the concentration of NaOH

1.5 × 10⁻³ moles of NaOH are in 10.3 mL of solution. The molarity of NaOH is:

\frac{1.5 \times 10^{-3} mol}{10.3\times 10^{-3}L} =0.15 M

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All information on a line graph is as precise as the information in the data table. TrueFalse
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Copper has a density of 8.96 g/cm3. If 75.0 g of copper is added to 50.0 mL of water in a graduated cylinder, to what volume rea
Tcecarenko [31]

Answer:

The answer to your question is    Final volume = 58.37 ml

Explanation:

Data

density = 8.96 g/cm³

mass = 75 g

volume of water = 50 ml

Process

1.- Calculate the volume of copper

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Substitution

  Volume = 75/8.96

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  Volume = 8.37cm³    or 8.37 cm³

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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}

8 0
3 years ago
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laila [671]
<h3>Answer:</h3>

C₅H₁₂O(l)+15/2O₂(g)→5CO₂(g)+6H₂O(l)

<h3>Explanation:</h3>

The balanced chemical equation for the combustion of the hydrocarbon in question is;

C₅H₁₂O(l)+15/2O₂(g)→5CO₂(g)+6H₂O(l)

  • A balanced chemical equation is one in which the number of atoms of each element is equal on both sides of the equation.
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5 0
3 years ago
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