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Step2247 [10]
2 years ago
13

. Citric acid, which can be obtained from lemon juice, has the molecular formula C6H8O7. A 0.250-g sample of citric acid dissolv

ed in 25.0 mL of water requires 37.2 mL of 0.105 M NaOH for complete neutralization. What number of acidic hydrogens per molecule does citric acid have?
Chemistry
1 answer:
marysya [2.9K]2 years ago
3 0

Answer:

3 acidic hydrogens per molecule of citric acid

Explanation:

In a sample of 37.2 mL(0.0372 L) of 0.105 mol/L of NaOH, will have:

n = 0.105x0.0372 = 0.0039 mol of NaOH

The dissociation of NaOH will give the same number of moles of Na⁺ and OH⁻.

The molar mass of citric acid is:

C: 12g/mol x 6 = 72 g/mol

H: 1g/mol x 8 = 8g/mol

O: 16 g/mol x 7 = 112 g/mol

192 g/mol

So, 0.250g of the acid has

n = mass/molar mass

n = 0.250/192

n = 0.0013 mol.

To be neutralized, it will be necessary 0.0039 mol of acidic hydrogens to react with the 0.0039 mol of OH⁻.

The dissociation reaction of one molecule of  the acid will give the stoichiometry:

1 mol of acid ----------------------- x mol of acidic hydrogens

0.0013 mol --------------------------- 0.0039

For a simple direct three rule:

0.0013x = 0.0039

x = 3 acidic hydrogens per molecule of citric acid.

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Answer:

Thus, when the volume of the gas is exposed to a temperature above -273.15 K, the volume increases linearly with the temperature.

Explanation:

The expression for Charles's Law is shown below:

\frac {V_1}{T_1}=\frac {V_2}{T_2}

This states that the volume of the gas is directly proportional to the absolute temperature keeping the pressure conditions and the moles of the gas constant.

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1 year ago
1.50 × 104 J of energy is transferred thermally into a huge tank filled with liquid water. The water temperature remains constan
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Answer:

53j/k

Explanation:

ΔH = TΔS => ΔS = ΔH/T = 1.5 x 10⁴ joules/283 Kelvin = 53 joules/K

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1 year ago
Calculate the mass in grams of 0.800 mole of H2CO3. g
PilotLPTM [1.2K]
MH₂CO₃: (1g×2) + 12g + (16g×3) = 62 g/mol

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2 years ago
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Methane, CH4, reacts with I2 according to the reaction CH4(g)+I2(g)⇌CH3I(g)+HI(g)
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Answer:

pCH₄ = 105.1 - 0.42 = 104.68 torr

pI₂ = 7.96 -0.42 = 7.54 torr

pCH₃I = 0.42 torr

pHI = 0.42 torr

Explanation:

Kp is the equilibrium constant for the partial pressure of the gases in the reaction, and it is calculated for a general equation:

aA(g) + bB(g) ⇄ cC(g) + dD(g)

Kp = \frac{(pC)^cx(pD)^d}{(pA)^ax(pB)^b}, where p is the partial pressure in the equilibrium. By the reaction given:

CH₄(g) + I₂(g) ⇄ CH₃I(g) + HI(g)

105.1 torr   7.96 torr  0       0            <em> initial partial pressure</em>

-x                  -x            +x     +x          <em> react</em>

105.1-x       7.96-x      x        x            <em>equilibrium</em>

Then:

Kp = \frac{pCH3IxpHI}{pCH4xpI2} = \frac{x^2}{(105.1-x)(7.96-x)}

2.26x10^{-4} = \frac{x^2}{836.596 - 113.06x -x^2}

x² = 0.1891 - 0.0255x -2.26x10⁻⁴x²

0.9997x² + 0.0255x - 0.1891 = 0

Using Bhaskara's rule:

Δ = (0.0255)² - 4x(0.9997)x(-0.1891)

Δ = 0.7568

x = \frac{-b+/-\sqrt{0.7568} }{2a} = \frac{-0.0255 +/-0.8699}{1.9994}

Using only the positive term, x = 0.42 torr.

So,

pCH₄ = 105.1 - 0.42 = 104.68 torr

pI₂ = 7.96 -0.42 = 7.54 torr

pCH₃I = 0.42 torr

pHI = 0.42 torr

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