Answer: 2
Explanation:
To calculate the moles :
According to stoichiometry :
1 mole of 
are produced by = 4 moles of 
Thus 1.18 moles of will be produced by=
of
Mass of 
Thus 85.0 g of will be required and 2 steps are required to get the answer.
The volume of Helium gas needed for storage is 2.00 L (answer C)
<u><em> calculation</em></u>
The volume of Helium is calculated using ideal gas equation
That is Pv =nRT
where;
P( pressure) = 203 KPa
V(volume)=?
n(number of moles) = 0.122 moles
R(gas constant) = 8.314 L.Kpa/mol.K
T(temperature)= 401 K
make V the subject of the formula by diving both side by P
V=nRT/p
V={[0.122 moles x 8.314 L. KPa/mol.K x 401 K] / 203 KPa} = 2.00 L
1 2 3 4 5 and how the question ask
Problem One
You will use both m * c * deltaT and H = m * heat of fusion.
Givens
m = 12.4 grams
c = 0.1291
t1 = 26oC
t2 = 1204
heat of fusion (H_f) = 63.5 J/grams.
Equation
H = m * c * deltaT + m * H_f
Solution
H = 12.4 * 0.1291 * (1063 - 26) + 12.4 * 63.5
H = 1660.1 + 787.4
H = 2447.5 or 2447.47 is the exact answer. I have to leave the rounding to you. I have no idea where to round it although I suspect 2450 would be right for 3 sig digs.
Problem Two
Formula and Givens
t1 = 14.5
t2 = 50.0
E = 5680
c = 4.186
m = ??
E = m c * deltaT
Solution
5680 = m * 4.186 * (50 - 14.5)
5680 = m * 4.186 * (35.5)
5680 = m * 148.603 * m
m = 5680 / 148.603
m = 38.22 grams That isn't very much. Be very sure you are working in joules. You'd leave that many grams in the kettle after drying it thoroughly.
m = 38.2 to 3 sig digs.
The answer is AIM NOT POSTIVE
