The total quantity of heat evolved in converting the steam to ice is determined as -12,928.68 J.
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Heat evolved in converting the steam to ice</h3>
The total heat evolved is calculated as follows;
Q(tot) = Q1(steam to boiling point) + Q2(boiling point to ice) +Q3(freezing to -42 ⁰C)
where;
Q = = mcΔθ
where;
- m is mass, (mass of water = 18 g/mol)
- c is specific heat capacity,
- Δθ is change in temperature
Q(tot) = 2(18)(2.01)(100 - 135) + 2(18)(2.01)(0 - 100) + 2(18)(2.09)(-42 - 0)
Q(tot) = -12,928.68 J
Thus, the total quantity of heat evolved in converting the steam to ice is determined as -12,928.68 J.
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Mass plus atomic and number
Answer:
3,855.532 grams
Explanation:
1 pound = 453.592 grams
8.50 = ? grams
--> 8.50 * 453.592 = 3,855.532 grams.
Answer:
52.8 g of O2.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
4Al + 3O2 —> 2Al2O3
From the balanced equation above,
4 moles of Al reacted with 3 moles of O2 to produce 2 moles of Al2O3
Next, we shall determine the number of mole of O2 needed to react with 2.2 moles of Al. This can be obtained as follow:
From the balanced equation above,
4 moles of Al reacted with 3 moles of O2.
Therefore, 2.2 moles of Al will react with = (2.2 × 3)/4 = 1.65 moles of O2.
Thus, 1.65 moles of O2 is needed for the reaction.
Finally, we shall determine the mass of O2 needed as shown below:
Mole of O2 = 1.65 moles
Molar mass of O2 = 2 × 16= 32 g/mol
Mass of O2 =?
Mole = mass/Molar mass
1.65 = mass of O2 /32
Cross multiply
Mass of O2 = 1.65 × 32
Mass of O2 = 52.8 g
Therefore, 52.8 g of O2 is needed for the reaction.
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