Answer:
Concentration of ethanol required = 48.476 M
Explanation:
Given that:
the absorption intensity = 1.00
Molarity of ethanol = 1M
NMR instrument used = 160 MHz
Temperature used = 300 K
The required concentration of ethanol can be determined as follows:


= 48.476 M
When the reaction equation is:
HgBr2(s) ↔ Hg2+(aq) + 2Br-(aq)
So Ksp expression = [Hg2+] [Br-]^2
assume the solubility S = X = 2.66 x 10^-7 M
and from the reaction equation :
we can see that [Hg2+] = X
and the [Br-] = 2 X
so by substitution in Ksp formula will can get the Ksp value:
∴ Ksp = X * (2X)^2
= 2.66 x 10^-7 * (2*2.66 x 10^-7)^2
= 7.53 x 10^-20
70.0 g. The decomposition of 125 g CaCO3 produces 700 g CaO.
MM = 100.09 56.08
CaCO3 → CaO + CO2
Mass 125 g
a) Moles of CaCO3 = 125 g CaCO3 x (1 mol CaCO3/100.09 g CaCO3)
= 1.249 mol CaCO3
b) Moles of CaO = 1.249 mol CaCO3 x (1 mol CaO/1 mol CaCO3)
= 1.249 mol CaO
c) Mass of CaO = 1.249 mol CaO x (56.08 g CaO/1 mol CaO) = 70.0 g
Answer:
Final temperature of water is 
Explanation:
1 mol of LiBr releases 48.83 kJ of heat upon dissolution in water.
So, 2 moles of LiBr release
or 97.66 kJ of heat upon dissolution in water.
This amount of heat is consumed by 1000.0 g of water. Hence temperature of water will increase.
Let's say final temperature of water is
.
So, change in temperature (
) of water is
or (t-25) K
Heat capacity (C) of water is 
Hence, 
where m is mass
So, 
or, 
Hence final temperature of water is 