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3 years ago

Write the acid-base equilibrium reaction between clo– and h2o. ignore phases.

2 answers:

Crank3 years ago

7 0

<span>CLO- + H2O </span>→<span>HCLO + OH-.

The equilibrium is being attained even without going through complete conversion.</span>

m_a_m_a [10]3 years ago

4 0

Answer : The acid-base equilibrium reaction between $ClO^-$ and $H_2O$ will be,

$ClO^-+H_2O\rightleftharpoons HClO+OH^-$

Explanation :

Acid-base equilibrium reaction : It is defined as a state of chemical equilibrium that exists when the both species of conjugate acid-base pair are present in the solution.

In the reaction, an acid donates a proton to a base and to produce a conjugate acid and a conjugate base in the solution.

The acid-base equilibrium reaction between $ClO^-$ and $H_2O$ will be,

$ClO^-+H_2O\rightleftharpoons HClO+OH^-$

In this reaction, $ClO^-$ is a base and $H_2O$ is an acid that donates a proton to a base to produce $HClO$ as a conjugate acid and $OH^-$ as a conjugate base.

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In the citric acid cycle, malate is dehydrogenated to oxaloacetate in a highly endergonic reaction with a ΔG’o of +30 kJ mol-1:

Doss [256]

Answer : The value of $K_{eq}$ of this reaction is, $5.51\times 10^{-6}$

At equilibrium, [L-malate] > [oxaloacetate]

Explanation :

The relation between the equilibrium constant and standard Gibbs free energy is:

$\Delta G^o=-RT\times \ln K_{eq}$

where,

$\Delta G^o$ = standard Gibbs free energy = +30 kJ/mol = +30000 J/mol

R = gas constant = 8.314 J/K.mol

T = temperature = $25^oC=273+25=298K$

$K_{eq}$ = equilibrium constant = ?

The given reaction is:

$\text{L-malate}+NAD^+\rightleftharpoons \text{oxaloacetate}+NADH+H^+$

$\Delta G^o=-RT\times \ln K_{eq}$

$+30000J/mol=-(8.314J/K.mol)\times (298K)\times \ln K_{eq}$

$K_{eq}=5.51\times 10^{-6}$

Therefore, the value of $K_{eq}$ of this reaction is, $5.51\times 10^{-6}$

As, the value of $K_{eq}$ < 1 that means the reaction mixture contains reactants.

At equilibrium, [L-malate] > [oxaloacetate]

7 0

4 years ago

How many protons and neutrons does nitrogen-14 have?

zhenek [66]

Answer:

7 protons

Isotopes are atoms that have the same number of protons but different numbers of neutrons in the nucleus. You know that nitrogen-14 has 7 protons in the nucleus because it is an isotope of nitrogen, which has an atomic number equal to 7 .

4 0

4 years ago

ASAAAP HURRRY PLS!!!

Setler79 [48]

C or D! When reduced coenzymes produce in the process of the cycle, are oxidized in the process of oxidative phosphorylation in electron transport chain! I hope this helps :(

3 0

3 years ago

Read 2 more answers

Determine the number of moles of each type of monatomic or polyatomic ion in one mole of the following compounds. For each polya

Andrei [34K]

Answer:

KNO₃ → 1 mol of K⁺ ; 1 mol of nitrate (1 mol of oxygen + 3 mol of nitrogen)

Na₂SO₄ → 2 mol of Na⁺ ; 1 mol of sulfate ( 1 mol of sulfur + 4 mol of oxygen)

Ca(OH)₂ → 1 mol of calcium; 2 moles of hydroxide

(NH₄)₂SO₃ → 2 mol of ammonium; 2 mol of sulfite (1 mol of sulfur + 3 mol of oxygen)

Ca₃(PO₄)₂ → 3 mol of calcium: 2 mol of phosphate (1 mol of phosphorus + 4 mol of oxygen)

Al₂(CrO₄)₃ → 2 mol of aluminum ; 3 mol of cromate (1 mol of Cr + 4 mol of oxygen)

Explanation:

Let's verify all the dissociations:

KNO₃ → K⁺ + NO₃⁻

Na₂SO₄ → 2Na⁺ + SO₄⁻²

Ca(OH)₂ → Ca²⁺ + 2OH⁻

(NH₄)₂SO₃ → 2NH₄⁺ + 2SO₃⁻²

Ca₃(PO₄)₂ → 3Ca²⁺ + 2PO₄⁻³

Al₂(CrO₄)₃ → 2Al³⁺ + 3CrO₄⁻²

6 0

4 years ago

Please help me solve this!

yulyashka [42]

Answer : The image is attached below.

Explanation :

For $O_3$ :

Molar mass, M = 48 g/mol

Mass, m = 24 g

Moles, n = $\frac{m}{M}=\frac{24g}{48g/mol}=0.5mol$

Number of particles, N = $n\times 6.022\times 10^{23}=0.5\times 6.022\times 10^{23}=3.0\times 10^{23}$

For $NH_3$ :

Molar mass, M = 17 g/mol

Mass, m = 170 g

Moles, n = $\frac{m}{M}=\frac{170g}{17g/mol}=10mol$

Number of particles, N = $n\times 6.022\times 10^{23}=10\times 6.022\times 10^{23}=6.0\times 10^{24}$

For $F_2$ :

Molar mass, M = 38 g/mol

Mass, m = 38 g

Moles, n = $\frac{m}{M}=\frac{38g}{38g/mol}=1mol$

Number of particles, N = $n\times 6.022\times 10^{23}=1\times 6.022\times 10^{23}=6.0\times 10^{23}$

For $CO_2$ :

Molar mass, M = 44 g/mol

Moles, n = 0.10 mol

Mass, m = $n\times M=0.10mol\times 44g/mol=4.4g$

Number of particles, N = $n\times 6.022\times 10^{23}=0.10\times 6.022\times 10^{23}=6.0\times 10^{22}$

For $NO_2$ :

Molar mass, M = 46 g/mol

Moles, n = 0.20 mol

Mass, m = $n\times M=0.20mol\times 46g/mol=9.2g$

Number of particles, N = $n\times 6.022\times 10^{23}=0.20\times 6.022\times 10^{23}=1.2\times 10^{23}$

For $Ne$ :

Molar mass, M = 20 g/mol

Number of particles = $1.5\times 10^{23}$

Moles, n = $\frac{N}{6.022\times 10^{23}}=\frac{1.5\times 10^{23}}{6.022\times 10^{23}}=0.25mol$

Mass, m = $n\times M=0.25mol\times 20g/mol=5g$

For $N_2O$ :

Molar mass, M = 44 g/mol

Number of particles = $1.2\times 10^{24}$

Moles, n = $\frac{N}{6.022\times 10^{23}}=\frac{1.2\times 10^{24}}{6.022\times 10^{23}}=1.9mol$

Mass, m = $n\times M=1.9mol\times 44g/mol=83.6g$

For unknown substance:

Number of particles = $3.0\times 10^{23}$

Mass, m = 8.5 g

Moles, n = $\frac{N}{6.022\times 10^{23}}=\frac{3.0\times 10^{23}}{6.022\times 10^{23}}=0.50mol$

Molar mass, M = $\frac{m}{n}=\frac{8.5g}{0.50mol}=17g/mol$

The substance is $NH_3$ .

3 0

3 years ago