Answer:
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Explanation:
You need the specif heat capacities of both cobalt and lead.
- Specific heat of cobalt: 0.42 J/g.ºC
- Specific heat of lead: 0.13 J/g.ºC
When the two sheets reach the thermal equilibrium their temperatures are equal.
You can use the equations for the thermal heat to find the equilibrium temperature:
Thermal heat released by the hot sheet, lead:
- Q = 16.6 kg × 0.13J/g.ºC × (63ºC - T)
Thermal heat absorbed by the cold sheet, cobalt:
- Q = 5.78 kg × 0.42J/g.ºC × (T - 11ºC)
Equal the two equations to solve for T:
- 16.6 kg × 0.13J/g.ºC × (63ºC - T) = 5.78kg × 0.42J/g.ºC × (T - 11ºC)
I remove the units for easier handling:
- 135.954 - 2.158T = 2.4276T - 26.7036
Round to 2 significant figures: 35ºC ← answer
I think it is around 95.211
Answer: a. 3.0 moles of water
Explanation:
According to stoichiometry,
2 moles of hydrogen combine with 1 mole of oxygen.
Thus 3 moles of hydrogen combine = moles of oxygen.
Thus hydrogen is the limiting reagent as it limits the formation of product and oxygen is the excess reagent as it is present in excess. (1.8-1.5)= 0.3 moles of oxygen will be left as such.
If 2 moles of hydrogen produce = 2 moles of water
3 moles of hydrogen will produce =moles of water.
1) Chemical equation
2Al + 6 HCl ---> 2Al Cl3 + 3 H2
2) molar ratios
2 mol Al : 3 moles H2
3) Proportion
2 mol Al / 3mol H2 = x / 9 mol H2
4) Solve for x
x = 9 mol H2 * 2 mol Al / 3 mol H2 = 6 mol Ag
Answer: 6 moles