Answer:
well it might be 1.)Cadmium Fluoride CdF2 150.4078
2.)Cadmium(II) Perfluorate Cd(FO4) 2278.403
3.)Cadmium Ferrocyanide Cd2Fe(CN) 6436.7714
Explanation:
I think the question is incomplete as we need initial temperature to get the new temperature. but you can use equation Q=mc0 for guidelines
Q= energy
m= mass of solution
c= specific heat capacity of. solution
0= final temperature- initial temperature
Answer:
[HI] = 0.097 M
Explanation:
Let's consider the following reaction.
2 HI(g) ⇄ H₂(g) + I₂(g)
The order of reaction for HI is 1. Thus, we can calculate the concentration of HI ([HI]) at certain time using the following expression:
ln [HI] = ln [HI]₀ - k. t
where,
[HI]₀ is the initial concentration of HI
k is the rate constant
t is the time elapsed
When [HI]₀ = 0.440 M and t = 0.210 s, the concentration of HI is
ln [HI] = ln (0.440) - 7.21 s⁻¹ × 0.210 s
ln [HI] = -2.33
[HI] = 0.097 M
Answer:
53.4 % is the percent yield
Explanation:
This is the reaction:
C₆H₆ + Cl₂ → C₆H₅Cl + HCl
First of all we need to know the moles of benzene we used
39 g . 1 mol / 78 g = 0.5 moles
Ratio is 1:1 so 1 mol of benzene produces 1 mol of chloride
0.5 moles of chloride were produced by 0.5 moles of benzene
We must calculate the mass of chloride we produced
0.5 mol . 112.45 g / 1 mol = 56.2g
Let's calculate the percent yield
(Yield produced / Theoretical yield ) . 100
(30 g / 56.2 g) . 100 = 53.4 %
MHCl: 1+35,5 = 36,5 g/mol
36,5g ---- 1 mol
21g ------- X
X = 0,575 mol
C = n/V
V = n/C
V = 0,575/4,3
V = 0,1337 L = 133,7 mL
:•)