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saveliy_v [14]
3 years ago
8

When 14.8 g KOH is dissolved in 85.6 g of water in a coffee-cup calorimeter, the temperature rises from 19.3 °C to 32.76 °C. Wha

t is the enthalpy change per gram of potassium hydroxide dissolved in the water?
Chemistry
1 answer:
KonstantinChe [14]3 years ago
6 0

Answer:

This question is incomplete, the complete question is; assuming that the solution has a specific heat of 4.18 J/g°C

The answer is 381.67 J/g

Explanation:

Enthalpy change denoted by ΔH can be calculated using the formula;

ΔH = m × c × ΔT

Where; m= mass of reactants

c= specific heat

ΔT = change in temperature

ΔT = T2 - T1

ΔT = 32.76 - 19.3

ΔT = 13.46 °C

mass of reactants= 85.6 + 14.8 = 100.4g, c = 4.18J/g°C

Hence; ΔH = m × c × ΔT

ΔH = 100.4 × 4.18 × 13.46

ΔH = 5648.78J

Enthalpy change per gram of potassium hydroxide dissolved in the water is;

ΔH = 5648.78/14.8

ΔH = 381.67 J/g

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Answer:

\begin{array}{cccll}\textbf{Element 1} & \textbf{ Element 2} &\textbf{Compound?} &\textbf{Formula} &\textbf{Type}\\\text{Ar}&\text{Xe} &\text{No} &\text{None}&\text{Neither}\\\text{F}& \text{Cs} &\text{Yes} &\text{CsF} &\text{Ionic}\\\text{N} &\text{Br} &\text{Yes} & \text{NBr}_{3}&\text{molecular} \\\end{array}

Explanation:

You look at the type of atom and their electronegativity difference.

If ΔEN <1.6, covalent; if ΔEN >1.6, ionic

Ar/Xe: Noble gases; no reaction

F/Cs: Non-metal + metal; ΔEN = |3.98 – 0.79| = 3.19; Ionic

N/Br: Two nonmetals; ΔEN = |3.04 - 2.98| = 0.

4 0
3 years ago
In the laboratory you are asked to make a 0.175 m barium iodide solution using 13.9 grams of barium iodide. How much water shoul
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<u>Answer:</u> The mass of water that should be added in 203.07 grams

<u>Explanation:</u>

To calculate the molality of solution, we use the equation:

\text{Molality}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

Where,

m = molality of barium iodide solution = 0.175 m

m_{solute} = Given mass of solute (barium iodide) = 13.9 g

M_{solute} = Molar mass of solute (barium iodide) = 391.14 g/mol

W_{solvent} = Mass of solvent (water) = ? g

Putting values in above equation, we get:

0.175=\frac{13.9\times 1000}{391.14\times W_{solvent}}\\\\W_{solvent}=\frac{13.9\times 1000}{391.14\times 0.175}=203.07g

Hence, the mass of water that should be added in 203.07 grams

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How many moles of N are in 0.221 g of N2O?
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4 0
3 years ago
Read 2 more answers
4. Identify the different parts to this word problem. A bowling ball of 35.2kg, generates 218 kg.m/s units of momentum. What is
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<u>Given:</u>

Mass of the ball = 35.2 kg

Momentum =  218 kg m/s

<u>To determine:</u>

The velocity of the ball

<u>Explanation:</u>

Momentum (p) of an object is the product of its mass  (m) and velocity (v)

p = m*v

v = p/m = 218 kg.ms-1/35.2 kg = 6.19 m/s

Ans: The velocity of the ball is 6.19 m/s

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