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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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2 years ago

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3 years ago

A 15.0 g sample of nickel metal is heated to 100.0 degrees C and dropped into 55.0 g of water, initially at 23.0 degrees C. Assu

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Answer: The final temperature of nickel and water is $25.2^{o}C$ .

Explanation:

The given data is as follows.

Mass of water, m = 55.0 g,

Initial temp, $(t_{i}) = 23^{o}C$ ,

Final temp, $(t_{f})$ = ?,

Specific heat of water = 4.184 $J/g^{o}C$ ,

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Specific heat of nickel = 0.444 $J/g^{o}C$

Hence, we will calculate the heat energy as follows.

q = $mS \Delta t$

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Therefore, heat energy lost by the alloy is equal to the heat energy gained by the water.

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= $25.2^{o}C$

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