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Wittaler [7]
3 years ago
15

Calculate the amount of energy , in Joules, required to raise the temperature of 15.5 g of liquid water from 0.00o C to 75.0 oC.

Chemistry
1 answer:
deff fn [24]3 years ago
8 0

Answer:

10043.225 J

Explanation:

We'll begin by calculating the amount of heat needed to change ice to water since water at 0°C is ice. This is illustrated below:

Mass (m) = 15.5g

Latent heat of fussion of water (L) = 334J/g

Heat (Q1) =..?

Q1 = mL

Q1 = 15.5 x 334

Q1 = 5177 J

Next, we shall calculate the amount of heat needed to raise the temperature of water from 0°C to 75°C.

This is illustrated below:

Mass = 15.5g

Initial temperature (T1) = 0°C

Final temperature (T2) = 75°C

Change in temperature (ΔT) = T2 – T1 = 75 – 0 = 75°C

Specific heat capacity (C) of water = 4.186J/g°C

Heat (Q2) =?

Q2 = MCΔT

Q2 = 15.5 x 4.186 x 75

Q2 = 4866.225 J

The overall heat energy needed is given by:

QT = Q1 + Q2

QT = 5177 + 4866.225

QT = 10043.225 J

Therefore, the amount of energy required is 10043.225 J

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Given chemical equation in balanced form ;

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5 0
2 years ago
2Pb(s) + O2(aq) + 4H+(aq) → 2H2O(l) + 2Pb2+(aq)
defon

Answer:

The answer to your question is 0.269 g of Pb

Explanation:

Data

Lead solution = 0.000013 M

Volume = 100 L

mass = 0.269 g

atomic mass Pb = 207.2 g

Chemical reaction

                        2Pb(s) + O₂(aq) + 4H⁺(aq) → 2H₂O(l) + 2Pb₂⁺(aq)

Process

1.- Calculate the mass of Pb in solution

Formula

Molarity = \frac{number of moles}{volume}

Solve for number of moles

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