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3 years ago

How many milliliters of 2.5 M HCl are required to exactly neutralize 1.5 L of 5.0 M NaOH?

Chemistry

1 answer:

Airida [17]3 years ago

7 0

Answer:

3000mL

Explanation:

The following data were obtained from the question:

Volume of acid (Va) =..?

Molarity of acid (Ma) = 2.5M

Volume of base (Vb) = 1.5L

Molarity of base (Mb) = 5M

Next, we shall write the balanced equation for the reaction. This is given below:

HCl + NaOH —> NaCl + H2O

From the balanced equation above,

We obtained the following:

Mole ratio of the acid (nA) = 1

Mole ratio of base (nB) = 1

Next, we shall determine the volume of HCl needed for the reaction. This can be obtained as follow illustrated below:

MaVa / MbVb = nA/nB

2.5 x Va / 5 x 1.5 = 1

Cross multiply

2.5 x Va = 5 x 1.5

Divide both side by 2.5

Va = 5 x 1.5 / 2.5

Va = 3L

The volume of the acid required is 3L.

Finally, we shall convert the volume of the acid from litre (L) to millilitre (mL). This is illustrated below:

1L = 1000mL

Therefore, 3L = 3 x 1000 = 3000mL.

Therefore, the volume of the acid, HCl in mL needed fi4 the reaction is 3000mL

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Convert 7.1x10^25 molecules of water to moles

ruslelena [56]

Answer:

<h2>117.94 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

$n = \frac{7.1 \times {10}^{25} }{6.02 \times {10}^{23} } \\ = 117.940199...$

We have the final answer as

<h3>117.94 moles</h3>

Hope this helps you

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Thepotemich [5.8K]

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