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2 years ago

What is the citric acid concentration in a soda if it requires 32. 27 ml of 0. 0148 m naoh to titrate 25. 00 ml of soda?.

Chemistry

2 answers:

sveticcg [70]2 years ago

8 0

Answer:

soln

32.27ml-25.00

7.27 ÷ 0.0148

7248 answer

Rudiy272 years ago

7 0

Titration

citric acid: CH₃COOH, c = 1

caustic soda: NaOH c = 1

V₁M₁c₁=V₂M₂c₂ (c = valence)

$\tt 32.27\times 0.0148\times 1=25\times M_2\times 1\\\\M_2=0.019\:M$

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3 years ago

Nitrogen dioxide and water react to form nitric acid and nitrogen monoxide, like this:

posledela

Answer: The value of the equilibrium constant Kc for this reaction is 3.72

Explanation:

Equilibrium concentration of $HNO_3$ = $\frac{15.5g}{63g/mol\times 9.5L}=0.026M$

Equilibrium concentration of $NO$ = $\frac{16.6g}{30g/mol\times 9.5L}=0.058M$

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Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_c$

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The expression for $K_c$ is written as:

$K_c=\frac{[NO_2]^3\times [H_2O]^1}{[HNO_3]^2\times [NO]^1}$

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2 years ago

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3 years ago

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3 years ago

If 71.5 moles of an ideal gas is at 5.03 atm at 6.80 °C, what is the volume of the gas?

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Use the clapeyron equation:

T in kelvin : 6.80 + 273 => 279.8 K

R = 0.082

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Therefore:

P x V = n x R x T

5.03 x V = 71.5 x 0.082 x 279.8

5.03 x V = 1640.4674

V = 1640.4674 / 5.03

V = 326.13 L

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3 years ago