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BaLLatris [955]
3 years ago
14

if it takes 54 mL of 0.1 NaOH to neutralize 125 mL of an HCL solution, what is the concentration of HCL?

Chemistry
1 answer:
OleMash [197]3 years ago
3 0

Answer:

0.0432M

Explanation:

We begin by writing a balanced equation for the reaction. This is illustrated below:

NaOH + HCl —> NaCl + H2O

From the equation above,

The number of mole of the acid (nA) = 1

The number of mole of the base (nB) = 1

Data obtained from the question include:

Vb (volume of the base) = 54mL

Cb (concentration of the base) = 0.1M

Va (volume of the acid) = 125mL

Ca ( concentration of the acid) =?

Using CaVa/CbVb = nA/nB, the concentration of the acid can easily be obtained as shown below:

CaVa/CbVb = nA/nB

Ca x 125 / 0.1 x 54 = 1

Cross multiply to express in linear form:

Ca x 125 = 0.1 x 54

Divide both side by 125

Ca = (0.1 x 54) / 125

Ca = 0.0432M

Therefore, the concentration of the acid is 0.0432M

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Answer:

3.052 × 10^24 particles

Explanation:

To get the number of particles (nA) in a substance, we multiply the number of moles of the substance by Avogadro's number (6.02 × 10^23)

The mass of Li2O given in this question is as follows: 151grams.

To convert this mass value to moles, we use;

moles = mass/molar mass

Molar mass of Li2O = 6.9(2) + 16

= 13.8 + 16

= 29.8g/mol

Mole = 151/29.8g

mole = 5.07moles

number of particles (nA) of Li2O = 5.07 × 6.02 × 10^23

= 30.52 × 10^23

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4 0
3 years ago
what happens when a gas obtained by heating Ammonium Chloride and slaked lime are passed through copper sulphate ?​
gogolik [260]

Answer:

Consequently, what happens when gas obtained by heating slaked lime and ammonium chloride is passed through copper sulphate solution? The HCl in the gas mixture will form hydrochloric and the H+ will react with some of the NH3(aq), forming NH4^+, and with some of the SO4^2-, forming HSO4^-.

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Under certain conditions Argon gas diffuses at a rate of 3.2 cm per second under the same conditions an unknown gas diffuses at
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Answer:

20 g/mol

Explanation:

We can use <em>Graham’s Law of diffusion</em>:

The rate of diffusion (<em>r</em>) of a gas is inversely proportional to the square root of its molar mass (<em>M</em>).

r = \frac{1 }{\sqrt{M}}

If you have two gases, the ratio of their rates of diffusion is

\frac{r_{2}}{r_{1}} = \sqrt{\frac{M_{1}}{M_{2}}}

Squaring both sides, we get

(\frac{r_{2}}{r_{1}})^{2} = \frac{M_{1}}{M_{2}}

Solve for <em>M</em>₂:

M_{2} = M_{1} \times (\frac{r_{1}}{r_{2}})^{2}

M_{2} = \text{39.95 g/mol} \times (\frac{\text{3.2 cm/s}}{\text{4.5 cm/s}})^{2}= \text{39.95 g/mol} \times (0.711 )^{2}

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