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3 years ago

Fe2o3(s)+3h2(g)→2fe(s)+3h2o(l) Calculate the mass of H2O

Chemistry

1 answer:

777dan777 [17]3 years ago

7 0

Answer:

mass of H₂O = 54.06 g

Explanation:

balanced chemical equation

Fe₂O₃(s) + 3H₂(g) → 2Fe(s) + 3H₂O(l)

given data

moles of H₂O = 3 mol

mas of H₂O = ?

Molar mass of H₂O = 18.02 g/mol

we find the mass of H₂O by:

mass of H₂O = moles × molar mass

mass of H₂O = 3 mol × 18.02 g/mol

mass of H₂O = 54.06 g

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4 years ago

How much energy is released by the decay of 3 grams of 230Th in the following reaction 230 Th - 226Ra + "He (230 Th = 229.9837 g

Serhud [2]

<u>Answer:</u> The energy released for the decay of 3 grams of 230-Thorium is $2.728\times 10^{-15}J$

<u>Explanation:</u>

First we have to calculate the mass defect $(\Delta m)$ .

The equation for the alpha decay of thorium nucleus follows:

$_{90}^{230}\textrm{Th}\rightarrow _{88}^{226}\textrm{Ra}+_2^{4}\textrm{He}$

To calculate the mass defect, we use the equation:

Mass defect = Sum of mass of product - Sum of mass of reactant

$\Delta m=(m_{Ra}+m_{He})-(m_{Th})$

$\Delta m=(225.9771+4.008)-(229.9837)=1.4\times 10^{-3}amu=2.324\times 10^{-30}kg$

(Conversion factor: $1amu=1.66\times 10^{-27}kg$ )

To calculate the energy released, we use Einstein equation, which is:

$E=\Delta mc^2$

$E=(2.324\times 10^{-30}kg)\times (3\times 10^8m/s)^2$

$E=2.0916\times 10^{-13}J$

The energy released for 230 grams of decay of thorium is $2.0916\times 10^{-13}J$

We need to calculate the energy released for the decay of 3 grams of thorium. By applying unitary method, we get:

As, 230 grams of Th release energy of = $2.0916\times 10^{-13}J$

Then, 3 grams of Th will release energy of = $\frac{2.0916\times 10^{-13}}{230}\times 3=2.728\times 10^{-15}J$

Hence, the energy released for the decay of 3 grams of 230-Thorium is $2.728\times 10^{-15}J$

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3 years ago

Which of the following is an accurate statement

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What are the statements?

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3 years ago

Calculate the amount of time required to raise temperature of 0.50 liter of water from

Burka [1]

<h3>Answer:</h3>

19 seconds

<h3>Explanation:</h3>

Heat energy was supplied and used to heat 0.50 liters of water from 0°C to 10°C.

This means water gained heat energy that was supplied.

Therefore;

Heat supplied = Heat gained by water

We are required to calculate the time taken to raise the temperature of water from 0°C to 10°C.

Step 1: Calculate the heat gained by water

Quantity of heat = Mass × specific heat capacity × change in temperature

Density of water 1 g/ml

Volume of water is 500 mL

Therefore, since Mass = density × volume

The mass of water = 500 g

Change in temperature = 10° C

Specific heat capacity of water = 4.18 J/g/°C

Thus;

Quantity of heat = 500 g × 4.18 J/g/°C × 10

= 20,900 Joules

Step 2: Heat supplied by a heater

Heat supplied = Power × time

Power = 1100 J/s

Assuming the time required is x

Heat supplied = 1100x Joules

Step 3: Time required

Remember; heat supplied = heat gained by water

Therefore;

1100x joules = 20,900 Joules

x = 20,900/1100

= 19 seconds

Therefore, the time required to raise the temperature of water from 0°C to 10°C is 19 seconds.

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3 years ago