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Slav-nsk [51]
3 years ago
11

When HCl(g) reacts with NH3(g) to form NH4Cl(s) , 176 kJ of energy are evolved for each mole of HCl(g) that reacts. Write a bala

nced thermochemical equation for the reaction with an energy term in kJ as part of the equation.
Chemistry
1 answer:
Orlov [11]3 years ago
8 0

Answer:

q = -176kJ

Explanation:

HCl and NH3 reacts as following to NH4Cl

HCl(g) + NH3(g)=========>NH4Cl(s)   : ΔH = -176 KJ

Clearly,

ENERGY IS EVOLVED MEANING IT IS A EXOTHERMIC REACTION .

therefore, the value of heat evolved as q = -176kJ

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25 g of a compound is added to 500 mL of water if the freezing point of the resulting solution is
vlabodo [156]

Answer:

a)   119 g/mol

Explanation:

-We apply the formula for freezing point depression to obtain the molality of the solution:

\bigtriangleup T_f=K_fm, \  \ K_f=1.36\textdegree C/m\\\\\therefore m=\frac{\bigtriangleup T_f}{K_f}\\\\=\frac{0.57\textdegree C}{1.36\textdegree C}\\\\=0.4191\ mol/Kg\\\\

#We use the molality above to calculate the molar mass:

m=\frac{0.4191\ mol}{1\ Kg}=\frac{25\ g}{0.5\ Kg}\\\\\therefore 1 \ mol=\frac{25\ g}{0.5\ Kg}\times\frac{1\ Kg}{ 0.4191}\\\\=119.3033\approx 119\ g/mol

Hence, the molar mass of the compound is 119 g/mol

5 0
3 years ago
Why is secondary growth important to plants
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3 years ago
What is the volume of 3.89 grams of helium gas?
lukranit [14]

Answer: There are 971.77 millimoles in 3.89 grams of Helium

Explanation:

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7 0
3 years ago
In which case would recalibrating a thermometer be an important next step in an experiment dealing with boiling points?
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7 0
3 years ago
Will give lots of points if answered correctly. Determine the kb for chloroform when 0.793 moles of solute in 0.758 kg changes t
Liono4ka [1.6K]

Answer: The value of K_{b} for chloroform is 3.62^{o}C/m when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

Explanation:

Given: Moles of solute = 0.793 mol

Mass of solvent = 0.758

\Delta T_{b} = 3.80^{o}C

As molality is the number of moles of solute present in kg of solvent. Hence, molality of given solution is calculated as follows.

Molality = \frac{no. of moles}{mass of solvent (in kg)}\\= \frac{0.793 mol}{0.758 kg}\\= 1.05 m

Now, the values of K_b is calculated as follows.

\Delta T_{b} = i\times K_{b} \times m

where,

i = Van't Hoff factor = 1 (for chloroform)

m = molality

K_{b} = molal boiling point elevation constant

Substitute the values into above formula as follows.

\Delta T_{b} = i\times K_{b} \times m\\3.80^{o}C = 1 \times K_{b} \times 1.05 m\\K_{b} = 3.62^{o}C/m

Thus, we can conclude that the value of K_{b} for chloroform is 3.62^{o}C/m when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

7 0
3 years ago
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