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4 years ago

In a titration how much 0.50 M HCl is needed to neutralize 1 liter of a 0.75 M solution of NaOH?

Chemistry

1 answer:

Ymorist [56]4 years ago

6 0

Answer:

1.5 L

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

HCl + NaOH —> NaCl + H2O

From the balanced equation above, The mole ratio of the acid (nA) = 1

The mole ratio of the base (nB) = 1

Step 2:

Data obtained from the question. This includes the following:

Molarity of acid (Ma) = 0.5M

Volume of acid (Va) =..?

Volume of base (Vb) = 1L

Molarity of base (Mb) = 0.75M

Step 3:

Determination of the volume of the acid needed for the reaction.

Using the formula:

MaVa/MbVb = nA/nB

The volume of the acid needed for the reaction can be obtained as follow:

0.5 x Va / 0.75 x 1 = 1

Cross multiply

0.5 x Va = 0.75 x 1

Divide both side by 0.5

Va = 0.75 /0.5

Va = 1.5 L

Therefore, the volume of the acid, HCl needed for the reaction is 1.5L

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Agata [3.3K]

By adding more nutrients to it

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4 years ago

How many molecules are in 3.50 moles of H2O

Debora [2.8K]

Answer:

2.11 x 10²⁴ molecules.

Explanation:

It is known that every 1.0 mole of a molecule contains Avogadro's number of molecules (NA = 6.022 x 10²³).

Using cross multiplication:

1.0 mole of H₂O contains → 6.022 x 10²³ molecules.

3.5 mole of H₂O contains → ??? molecules.

∴ 3.5 mole of H₂O contain = (3.5 mol)(6.022 x 10²³) = 2.11 x 10²⁴ molecules.

5 0

4 years ago

8. Which of the following is TRUE about limiting and excess reactants?

Ivahew [28]

B. The limiting reactant determines the max amount of product that can be formed

5 0

3 years ago

The question is in the picture below

Rus_ich [418]

Answer:

$\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3$

Explanation:

Hess's Law of Constant Heat Summation states that if a chemical equation can be written as the sum of several other chemical equations, the enthalpy change of the first chemical equation is equal to the sum of the enthalpy changes of the other chemical equations. Thus, the reaction that involves the conversion of reactant A to B, for example, has the same enthalpy change even if you convert A to C, before converting it to B. Regardless of how many steps it takes for the reactant to be converted to the product, the enthalpy change of the overall reaction is constant.

With Hess's Law in mind, let's see how A can be converted to 2C +E.

$\bf{\text{A} \rightarrow 2\text{B}}$ (Δ $\text{H}_1$ ) -----(1)

Since we have 2B, multiply the whole of II. by 2:

$\bf{2\text{B} \rightarrow 2\text{C} +2\text{D}}$ (2Δ $\text{H}_2$ ) -----(2)

This step converts all the B intermediates to 2C +2D. This means that the overall reaction at this stage is $\text{A} \rightarrow 2\text{C} +2\text{D}$ .

Reversing III. gives us a negative enthalpy change as such:

$\bf{2\text{D} \rightarrow \text{E}}$ (-Δ $\text{H}_3$ ) -----(3)

This step converts all the D intermediates formed from step (2) to E. This results in the overall equation of $\text{A} \rightarrow 2\text{C} +\text{E}$ , which is also the equation of interest.

Adding all three together:

$\text{A} \rightarrow 2\text{C}+\text{E}$ ( $\bf{\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3 }$ )

Thus, the first option is the correct answer.

Supplementary:

To learn more about Hess's Law, do check out: brainly.com/question/26491956

4 0

2 years ago

A pure copper penny contains approximately 2.9×1022 copper atoms. Use the following definitions to determine how many ______ of

hammer [34]

Complete question is;

A pure copper penny contains approximately 2.9 × 10^(22) copper atoms.

1 doz = 12

1 gross = 144

1 ream = 500

1 mol = 6.022 × 10^(23)

Use these definitions to determine the following:

A) How many dozens of copper atoms are in a penny.

B) How many gross of copper atoms are in a penny

C) How many reams of copper atoms are in a penny.

D) how many moles of copper atoms are in a penny?

All answers can be rounded to two significant figures

Answer:

A) 2.4 × 10^(21) dozens

B) 2.01 × 10^(20) gross

C) 5.8 × 10^(19) reams

D) 0.048 mol

Explanation:

A) A dozen contains 12.

Therefore, 2.9 × 10^(22) copper atoms will contain;

(2.9 × 10^(22))/12 dozens = 2.42 × 10^(21).

In 2 significant figures, we have;

2.4 × 10^(21) dozens

B) 1 gross = 144

Therefore, 2.9 × 10^(22) copper atoms will contain;

(2.9 × 10^(22))/144 gross ≈ 2.01 × 10^(20) gross

C) 1 ream = 500

Therefore, 2.9 × 10^(22) copper atoms will contain;

(2.9 × 10^(22))/500 reams = 5.8 × 10^(19) reams

D) 1 mol = 6.022 × 10^(23)

Therefore, 2.9 × 10^(22) copper atoms will contain;

(2.9 × 10^(22))/(6.022 × 10^(23)) = 0.048 mol

6 0

3 years ago