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2 years ago

58.5g of sodium iodide are dissolved in 0.5dm3. what concentration would this be?

1 answer:

7 0

percent by mass=(mass of solute/ mass of solution)*100 %

mass of solute=58.5 g

density (H₂O)=1 g/cm³*(1000 cm³/1dm³)=1000 g/dm³

mass of solvent (H₂O)=0.5 dm³ * (1000 g/dm³)=500 g
mass of solution=mass of solvent + mass of solut
mass of solution=58.5 g+500 g=558.5 g

% mass=(58.5 g/558.5 g) * 100%=10.47% of Na.

solution: 10.47% of Na.

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2 years ago

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Sodium carbonate (Na2CO3) reacts with acetic acid (CH3COOH) to form sodium acetate (NaCH3COO), carbon dioxide (CO2), and water (

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Use the formula q=m×Cp×delta T

m=1.500 kg=1,500 g
Co=2.52 J/g·k
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q=(1,500g)(2.52 J/g·k)(0.985k)

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2 years ago

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At the equivalence point of a KHP/NaOH titration, you have added enough OH- to react with all of the HP- such that the only spec

Nataly_w [17]

Explanation:

The given data is as follows.

$[P^{2-}]$ = 0.042 M, $K_{a}$ for $HP^{-} = 3.9 \times 10^{-6}$

According to the given situation $P^{2-}$ acts as a base.The reaction equation will be as follows.

$P^{2-} + H_{2}O \rightleftharpoons HP^{-} + OH^{-}$

Relation between $K_{b}$ and $K_{a}$ are as follows.

$K_{a} \times K_{b} = K_{w}$

$K_{b} = \frac{1 \times 10^{-14}}{K_{a}}$

= $\frac{1 \times 10^{-14}}{3.9 \times 10^{-6}}}$

= $2.6 \times 10^{-9}$

Also, $K_{b} = \frac{[HP^{-}][OH^{-}]}{P^{2-}}$

Let us take $[OH^{-}] = [HP^{-}]$ = x

So, $K_{b} = \frac{[HP^{-}][OH^{-}]}{P^{2-}}$

$2.6 \times 10^{-9} = \frac{x \times x}{0.042}$

x = $1.04 \times 10^{-5}$

$[OH^{-}] = [HP^{-}]$ = $1.04 \times 10^{-5}$

pOH = - log $[OH^{-}]$

= - log ( $1.04 \times 10^{-5}$ )

= 4.99

As it is known that pH + pOH = 14

so, pH + 4.99 = 14

pH = 9.01

Thus, we can conclude that pH of the solution is 9.01.

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2 years ago

Be sure to answer all parts. ΔH o f of hydrogen chloride [HCl(g)] is −92.3 kJ/mol. Given the following data, determine the ident

Akimi4 [234]

Answer:

NH₄Cl(s) → NH₃(g) + HCl(g)

ΔH°rxn 74,89 kJ/mol

Explanation:

The change in enthalpy of formation (ΔHf) is defined as the change in enthalpy in the formation of a substance from its constituent elements. For HCl(g):

(1) ¹/₂H₂(g) + ¹/₂ Cl₂ → HCl(g) ΔH = -92,3 kJ/mol

It is possible to sum ΔH of different reactions to obtain ΔH of a global reaction (Hess's law).

For the reactions:

(2) N₂(g) + 4H₂(g) + Cl₂(g) → 2NH₄Cl(s) ΔH°rxn = −630.78 kJ/mol

(3) N₂(g) + 3H₂(g) → 2NH₃(g) ΔH°rxn = −296.4 kJ/mol

The sum of -(2) + (3) gives:

-(2) 2NH₄Cl(s) → N₂(g) + 4H₂(g) + Cl₂(g) ΔH°rxn = +630.78 kJ/mol

(3) N₂(g) + 3H₂(g) → 2NH₃(g) ΔH°rxn = −296.4 kJ/mol

-(2) + (3) 2NH₄Cl(s) → 2NH₃(g) + H₂(g) + Cl₂(g)

ΔH°rxn = +630.78 kJ/mol −296.4 kJ/mol = +334,38 kJ/mol

Now, the sum of -(2) + (3) + 2×(1)

-(2) + (3) 2NH₄Cl(s) → 2NH₃(g) + H₂(g) + Cl₂(g) ΔH°rxn = +334,38 kJ/mol

2×(1) H₂(g) + Cl₂(g)→ 2HCl(g) ΔH = 2×-92,3 kJ/mol

-(2) + (3) + 2×(1) 2NH₄Cl(s) → 2NH₃(g) + 2HCl(g)

ΔH°rxn = +334,38 kJ/mol + 2×-92,3 kJ/mol = 149,78 kJ/mol

The reaction of:

NH₄Cl(s) → NH₃(g) + HCl(g)

Has ΔH°rxn = 149,78kJ/mol / 2 = 74,89 kJ/mol

I hope it helps!

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2 years ago

A researcher suspects that the pressure gauge on a 54.3-L gas cylinder containing nitric oxide is broken. An empty gas cylinder

mixer [17]

Answer:

Number of moles nitric acid in the cylinder is 400.539g/mol.

Explanation:

From the given,

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Number of moles of nitric acid =

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Therefore, number of moles nitric acid in the cylinder is 400.539g/mol.

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2 years ago