Water and hydrogen peroxide are different compounds although they have the same kind of atoms. The molecular formula of water is H2O (two atoms of oxygen chemically bonded to one atom of oxygen). The molecular formula of hydrogen peroxide is H2O2, (each atom of oxygen is chemical bonded to one atom of H and other atom of O). So,<span> the presence of different chemical bonds leads to different products with different chemical properties.</span><span />
Answer:
24x10³
Explanation:
2CO₂(g) + 4H₂O(g) → 2CH₃OH(l) + 3O₂ (g)
The equilibrium constant for this reaction is:
Kc =
The expression of [CH₃OH] is left out as it is a pure liquid.
Now we <u>convert the given masses of the relevant species into moles</u>, using their <em>respective molar masses</em>:
- CO₂ ⇒ 3.28 g ÷ 44 g/mol = 0.0745 mol CO₂
- H₂O ⇒ 3.86 g ÷ 18 g/mol = 0.214 mol H₂O
- O₂ ⇒ 2.80 g ÷ 32 g/mol = 0.0875 mol O₂
Then we calculate the concentrations:
- [CO₂] = 0.0745 mol / 7.5 L = 0.0099 M
- [H₂O] = 0.214 mol / 7.5 L = 0.0285 M
- [O₂] = 0.0875 mol / 7.5 L = 0.0117 M
Finally we <u>calculate Kc</u>:
- Kc = = 24x10³
3A=6 mol
A = 6/3
A = 2 mol
E= 2 mol
thats what I learned from school, but its been 3 months since I graduated, I hope I'm not making any mistake
the final concentration of NaI solution in 60 grams/litre.
Explanation:
Given that:
Initial concentration of NaI solution M1 = 0.2 M
initial volume of NaI V1 = 2 L
Final volume V2 = 1 Litre
Final molarity=?
concentration in grams/litre = ?
molar mass of NaI = 150 gram/mole
For dilution following formula is used:
M1 V1 = M2V2
putting the values in the equation
0.2 X 2 = 1 X M2
M2 = 0.4
For concentration in grams/litre, formula used
molarity =
mass = 0.4 x 150
= 60 grams
So, 60 grams of NaI will be present in final solution of NaI after evaporation.
The concentration is 60 grams/ L (as volume got reduced to 1 litre from 2 litres)
Answer:
The concentration of the H2SO4 solution is 0.0858 M
Explanation:
Step 1: Data given
Volume of H2SO4 = 30.00 mL = 0.030 L
Volume of NaOH= 37.85 mL = 0.03785 L
Concentration of NaOH= 0.1361 M
Step 2: The balanced equation
H2SO4 + 2NaOH → Na2SO4 + 2H2O
Step 3: Calculate the concentration of the H2SO4 solution
b*Ca*Va = a*Cb*Vb
⇒with b = the coefficient of NaOH = 2
⇒with Ca = the concentration of H2SO4 = TO BE DETERMINED
⇒with Va = the volume of H2SO4= 30.0 mL = 0.030 L
⇒with a = the coefficient of H2SO4 = 1
⇒with Cb = the concentration of NaOH = 0.1361 M
⇒with Vb = the volume of NaOH = 37.85 mL = 0.03785 L
2*Ca*0.030 = 1* 0.1361 * 0.03785
0.060 * Ca = 0.00515
Ca = 0.0858 M
The concentration of the H2SO4 solution is 0.0858 M