Nh4i(aq)+koh(aq)→ express your answer as a chemical equation. identify all of the phases in your answer.

2 answers:

5 0

NH4I (aq) + KOH (aq) in chemical equation gives

NH4I (aq) + KOH (aq) = KI (aq) + H2O(l) + NH3 (l)

Ki is in aqueous state H2o is in liquid state while NH3 is in liquid state

from the equation above 1 mole of NH4I (aq) react with 1 mole of KOH(aq) to form 1mole of KI(aq) , 1mole of H2O(l) and 1 Mole of NH3(l)

Norma-Jean [14]2 years ago

4 0

Answer:

NH4I(aq) + KOH(aq) -------->KI(aq) + NH3(g) + H2O(l)

Explanation:

NH4I(aq) + KOH(aq) -------->KI(aq) + NH3(g) + H2O(l)

This is a reaction of aqueous ammonium iodide and aqueous potassium hydroxide to yield aqueous potassium iodide, gaseous ammonia and liquid water as shown above.

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For a school event 1/6 of the athletic field is reversed for the fifth -grade classes the reserved part of the field is divided

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Answer:

$\frac{1}{24}$

Explanation:

Given:

For a school event, 1/6 of the athletic field is reserved for the fifth -grade classes and the reserved part of the field is divided equally among the 4 fifth grade classes in the school.

To find: fraction of the whole athletic field reserved for each fifth class

Solution:

Fraction of the whole athletic field reserved for four fifth classes = $\frac{1}{6}$

So, fraction of the whole athletic field reserved for each fifth class = $\frac{1}{4}(\frac{1}{6})=\frac{1}{24}$

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1 year ago

A 3.96x10^-24 M solution of compound A exhibited an absorbance of 0.624 at 238 nm in a 1.000-cm cuvet; a blank solution containi

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Actual question from source:-

A 3.96x10-4 M solution of compound A exhibited an absorbance of 0.624 at 238 nm in a 1.000 cm cuvette. A blank had an absorbance of 0.029. The absorbance of an unknown solution of compound A was 0.375. Find the concentration of A in the unknown.

Answer:

Molar absorptivity of compound A = $1502.53\ {Ms}^{-1}$