<u>Answer:</u> The initial concentration of 
are 0.0192 M and 0.0192 M respectively.
<u>Explanation:</u>
We are given:
Equilibrium concentration of HI = 0.030 M
Moles of hydrogen gas = Moles of iodine gas (concentration will also be the same)
For the given chemical equation:
<u>Initial:</u> x x -
<u>At eqllm:</u> x-c x-c 2c
Calculating the value of 'c'
The expression of 
for above reaction follows:
![K_{eq}=\frac{[HI]^2}{[H_2]\times [I_2]}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BHI%5D%5E2%7D%7B%5BH_2%5D%5Ctimes%20%5BI_2%5D%7D)
We are given:
![[H_2]=(x-c)=(x-0.015)](https://tex.z-dn.net/?f=%5BH_2%5D%3D%28x-c%29%3D%28x-0.015%29)
![[I_2]=(x-c)=(x-0.015)](https://tex.z-dn.net/?f=%5BI_2%5D%3D%28x-c%29%3D%28x-0.015%29)
Putting values in above equation, we get:
Neglecting the value of x = 0.0108 M, because the initial concentration cannot be less than the equilibrium concentration.
x = 0.0192 M
Hence, the initial concentration of are 0.0192 M and 0.0192 M respectively.
Answer:
mass CaI2 = 23.424 Kg
Explanation:
From the periodic table we obtain for CaI2:
⇒ molecular mass CaI2: 40.078 + ((2)(126.90)) = 293.878 g/mol
∴ mol CaI2 = (4.80 E25 units )×(mol/6.022 E23 units) = 79.708 mol CaI2
⇒ mass CaI2 = (79.708 mol CaI2)×(293.878 g/mol) = 23424.43 g
⇒ mass CaI2 = 23.424 Kg
MAY BE THE ANSWER IS (2) BECAUSE GLYCOSIS IS THE PROCESS IN WHICH THE CONVERSION OF 6 C GLUCOSE INTO PYRUVATE INSIDE THE CYTOPLASM
Answer:
0.022 mol O
Explanation:
Mg3(Si2O5)2(OH)2
We can see that 1 mol of this substance has 3 mol of Mg.
Oxygen altogether is 5*2 (from (Si2O5)2) + 2(from(OH)2) = 10 +2 = 12
So, 1 mol of this substance has 12 mol oxygen.
So, 1 mol of this substance contains 3 mol Mg and 12 mol O, or
ratio Mg : O = 3 : 12 = 1 : 4
1 mol Mg ----- 4 mol O
0.055 mol Mg ---x mol O
x = 0.055*4/1 = 0.220 mol O
