Answer:
The formula for volume of a rectangle is length multiply by width multiply thus, 0.25 m multiply 6.1 m multiply by 4.9 m = 7.5m^3.
Explanation:
the least number of significant figures is 2 thus the final answer will have the same number of significant figures. 7.5m^3
Answer : 0.0392 grams of Zn metal would be required to completely reduced the vanadium.
Explanation :
Let us rewrite the given equations again.
On adding above equations, we get the following combined equation.
We have 12.1 mL of 0.033 M solution of VO₂⁺.
Let us find the moles of VO₂⁺ from this information.
From the combined equation, we can see that the mole ratio of VO₂⁺ to Zn is 2:3.
Let us use this as a conversion factor to find the moles of Zn.
Let us convert the moles of Zn to grams of Zn using molar mass of Zn.
Molar mass of Zn is 65.38 g/mol.
We need 0.0392 grams of Zn metal to completely reduce vanadium.
Whewww!! What class is this for
Answer:
The answer to your question is 8.21 g of H₂O
Explanation:
Data
mas of water = ?
mass of hydrogen = 4.6 g
mass of oxygen = 7.3 g
Balanced chemical reaction
2H₂ + O₂ ⇒ 2H₂O
Process
1.- Calculate the atomic mass of the reactants
Hydrogen = 4 x 1 = 4 g
Oxygen = 16 x 2 = 32 g
2.- Calculate the limiting reactant
Theoretical yield = H₂/O₂ = 4 / 32 = 0.125
Experimental yield = H₂/ O₂ = 4.6/7.3 = 0.630
From the results, we conclude that the limiting reactant is Oxygen because the experimental yield was higher than the theoretical yield.
3.- Calculate the mass of water
32 g of O₂ ---------------- 36 g of water
7.3 g of O₂ --------------- x
x = (7.3 x 36) / 32
x = 262.8 / 32
x = 8.21 g of H₂O
When it comes to equilibrium reactions in chemistry, there are a lot of equilibrium constants that can be used. In the case of solubility, the appropriate one to use is the equilibrium constant of solubility product denotes as Ksp. This is the concentration of products raised to their coefficients. For example,
cC ⇔ aA + bB
Ksp = {[A^a][B^b]}
Now, for the this problem, the reaction is
BaSO₄ ⇔ Ba²⁺ + SO₄²⁻
The reaction is already balanced. Since we don't know the value of Ba²⁺ and SO₄²⁻, let's denote this at x.
1.1 × 10⁻¹⁰ = [x][x] =[x²]
[x] = [Ba²⁺] = [SO₄²⁻] = [BaSO₄] = 1.049 × 10⁻⁵ M
