Given:
128g sample of titanium
2808J of heat energy
specific heat of titanium is 0.523 J/ g °C.
Required:
Change in temperature
Solution:
This can be solved
through the equation H = mCpT
where H is the heat, m is the mass, Cp is the specific heat and T is the change in temperature.
Plugging in the
values into the equation
H = mCpT
2808J = (128g) (0.523
J /g °C) T
T
= 41.9 °C
Moles Li = 3.50 g / 6.941 g/mol= 0.504
the ratio between Li and N2 is 6 : 1
moles N2 required = 0.504 /6=0.0840
we have 3.50 g / 28.0134 g/mol=0.125 moles of N2 so N2 is in excess
the ratio between Li and Li3N is 6 : 2
moles Li3N = 0.504 x 2 /6=0.168
mass Li3N = 0.168 mol x 34.8297 g/mol=5.85 g
Answer:
392 g
Explanation:
The given concentration tells us that<em> in 100 g of solution, there would be 15.3 g of 2-ethyltoluene</em>.
With that in mind we can<u> calculate how many grams of solution would contain 60.0 g of 2-ethyltoluene</u>:
- Mass of solution * 15.3 / 100 = 60.0 g 2-ethyltoluene
Answer: I think it’s atomic radius
Explanation:
