The moles of gas in the bottle has been 0.021 mol.
The ideal gas has been given as the gas where there has been negligible amount of interatomic collisions. The ideal gas equation has been given as:
<h3>Computation for the moles of gas</h3>
The gi<em>ve</em>n gas has standard pressure, 
The volume of the gas has been, 
The temperature of the gas has been, 
Substituting the values for the moles of gas, <em>n:</em>
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The moles of gas in the bottle has been 0.021 mol.
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To find the amount of energy need to raise the temperature of 125 grams of water from 25.0° C to 35.0° C, we will need to use the formula: q = mcΔt.
In this formula, q is the heat absorbed, m is the mass, c is the specific heat, and Δt is the change in temperature, which is found by final temperature minus the initial temperature.
Firstly, we can find the change in temperature. We are given the initial temperature, which is 25.0° C and the final temperature, which is 35.0° C. It is found by subtract the final temperature from the initial temperature.
35.0° C - 25.0° C = 10.0° C
We are also given the specific heat and the grams of water. With that, we can substitute the given values into the equation and multiply.
q = 125 g × 4.184 J/g °C × 10.0° C
q = 523 J/°C × 10.0° C
q = 5230 J
Therefore, it will take 5230 joules (J) to raise the temperature of the water.
Explanation:
The structure of Ferrarrisite Ca5(HAs O4)2(AsO4)2
Answer:
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Answer:
The percent yield of the reaction is 82%
Explanation:
First step: make the chemist equation.
2 Al (s) + Fe2O3 (s) → 2 Fe (s) + Al2O3 (s)
As the statement says that aluminun is in excess, the limiting reactant is the Fe2O3
Second step: Find out the moles in the reactant.
Molar weight Fe2O3: 159.7 g/m
Mass / Molar weight = moles
50 g /159.7 g/m = 0.313 moles
Third step: Analyse the reaction. 1 mol of Fe2O3 makes 2 moles of Fe.
1 mol Fe2O3 ____ 2Fe
0.313 mol Fe2O3 ____ 0.626 moles
Molar weight Fe = 55.85 g/m
Moles . molar weight = mass
55.85g/m . 0.626m = 34.9 grams
This will be the 100% yield of the reaction but we only made 28.65 g
34.9 g ____ 100%
28.65 g ____ 82.09 %
