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3 years ago

A certain mass of water was heated with 41,840 Joules, raising its temperature from 22.0°C to 28.5 °C. Find the

Chemistry

1 answer:

Westkost [7]3 years ago

4 0

Answer:

1.5 × 10³ g

Explanation:

Step 1: Given and required data

Transferred heat (Q): 41,840 J

Initial temperature: 22.0 °C

Final temperature: 28.5 °C

Specific heat capacity of water (c): 4.184 J/g.°C

Step 2: Calculate the temperature change

ΔT = 28.5°C - 22.0 °C = 6.5 °C

Step 3: Calculate the mass (m) of water

We will use the following expression.

Q = c × m × ΔT

m = Q / c × ΔT

m = 41,840 J / (4.184 J/g.°C) × 6.5 °C = 1.5 × 10³ g

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3 years ago

Read 2 more answers

A chemist titrates 160.0mL of a 0.3403M aniline C6H5NH2 solution with 0.0501M HNO3 solution at 25°C . Calculate the pH at equiva

maxonik [38]

Answer : The pH of the solution is, 5.24

Explanation :

First we have to calculate the volume of $HNO_3$

Formula used :

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the initial molarity and volume of $C_6H_5NH_2$ .

$M_2\text{ and }V_2$ are the final molarity and volume of $HNO_3$ .

We are given:

$M_1=0.3403M\\V_1=160.0mL\\M_2=0.0501M\\V_2=?$

Putting values in above equation, we get:

$0.3403M\times 160.0mL=0.0501M\times V_2\\\\V_2=1086.79mL$

Now we have to calculate the total volume of solution.

Total volume of solution = Volume of $C_6H_5NH_2$ + Volume of $HNO_3$

Total volume of solution = 160.0 mL + 1086.79 mL

Total volume of solution = 1246.79 mL

Now we have to calculate the Concentration of salt.

$\text{Concentration of salt}=\frac{0.3403M}{1246.79mL}\times 160.0mL=0.0437M$

Now we have to calculate the pH of the solution.

At equivalence point,

$pOH=\frac{1}{2}[pK_w+pK_b+\log C]$

$pOH=\frac{1}{2}[14+4.87+\log (0.0437)]$

$pOH=8.76$

$pH+pOH=14\\\\pH=14-pOH\\\\pH=14-8.76\\\\pH=5.24$

Thus, the pH of the solution is, 5.24

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jolli1 [7]

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