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3 years ago

What is the molarity of a hydrochloric acid solution, when 30.0 mL is neutralized by 48.0 mL of 0.100 mol/L

Chemistry

1 answer:

bulgar [2K]3 years ago

3 0

The molarity of a hydrochloric acid solution : 0.32 M

<h3>Further explanation </h3>

Titration is a procedure for determining the concentration of a solution by reacting with another solution which is known to be concentrated (usually a standard solution).

Titrations can be distinguished including acid-base titration, depositional titration, and redox titration. An acid-base titration is the principle of neutralization of acids and bases is used.

Acid-base titration formula

Ma. Va. na = Mb. Vb. nb

Ma, Mb = acid base concentration

Va, Vb = acid base volume

na, nb = acid base valence

1 ⇒HCl (valence=1, HCl ⇒H⁺+Cl⁻, one H⁺)

2⇒Ca(OH)₂(valence=2, Ca(OH)₂⇒Ca²⁺+2OH⁻, two OH⁻)

M₂=0.1 M

V₂=48 ml=0.048 L

V₁=30 ml=0.03 L

$\tt M_1.V_1.n_1=M_2.V_2.n_2\\\\M_1\times 0.03\times 1=0.1\times 0.048\times 2\\\\M_1=0.32$

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3 years ago

Predict the aldol product when the following ketone undergoes self-condensation in the presence of NaOH. Do not show the dehydra

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Explanation:

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3 years ago

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Nitella [24]

Answer:

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Explanation:

First find the substance formula.

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Magnesium is a +2 charge.

Nitrate is a -1 charge.

So to balance the chemical formula,

We need 1 magnesium atom for every nitrate atom.

2(1) + 1(-2) = 0

So the substance formula is Mg(NO4)2.

Now find the molar mass of Mg(NO4)2.

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N = 14.0 amu

O = 16.0 amu

They are three nitrogen and twelve oxygen atoms.

So you do this: 24.3 + 14.0(2) + 16.0(8) = 180.3 g/mol

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3 years ago

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2 years ago

5. A beam of photons with a minimum energy of 222 kJ/mol can eject electrons from a potassium surface. Estimate the range of wav

torisob [31]

Answer: The range of wavelengths of light that can be used to cause given phenomenon is $8.953 \times 10^{21} m$ .

Explanation:

Given: 222 kJ/mol (1 kJ = 1000 J) = 222000 J

Formula used is as follows.

$E = \frac{hc}{\lambda}$

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Substitute the values into above formula as follows.

$E = \frac{hc}{\lambda}\\222000 J = \frac{6.625 \times 10^{-34}Js \times 3 \times 10^{8} m/s}{\lambda}\\\lambda = 8.953 \times 10^{21} m$

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3 years ago