Answer:
Baking Soda is NaHCO3 ---Sodium hydrogen Carbonate
Excessive use of Fertilizer can cause plants to develop shallow root systems,Decrease soil fertility and reduce Quality of Crops.
Answer:
%age Yield = 34.21 %
Explanation:
The balance chemical equation for the decomposition of KClO₃ is as follow;
3 KOH + H₃PO₄ → K₃PO₄ + 3 H₂O
Step 1: Calculate moles of H₃PO₄ as;
Moles = Mass / M/Mass
Moles = 334.6 g / 97.99 g/mol
Moles = 3.414 moles
Step 2: Find moles of K₃PO₄ as;
According to equation,
1 moles of H₃PO₄ produces = 1 moles of K₃PO₄
So,
3.414 moles of H₃PO₄ will produce = X moles of K₃PO₄
Solving for X,
X = 1 mol × 3.414 mol / 1 mol
X = 3.414 mol of K₃PO₄
Step 3: Calculate Theoretical yield of K₃PO₄ as,
Mass = Moles × M.Mass
Mass = 3.414 mol × 212.26 g/mol
Mass = 724.79 g of K₃PO₄
Also,
%age Yield = Actual Yield / Theoretical Yield × 100
%age Yield = 248 g / 724.79 × 100
%age Yield = 34.21 %
Answer:
Kc for this reaction is 0.06825
Explanation:
Step 1: Data given
Number of moles formaldehyde CH2O = 0.055 moles
Volume = 500 mL = 0.500 L
At equilibrium, the CH2O(g) concentration = 0.051 mol
Step 2: The balanced equation
CH2O <=> H2 + CO
Step 3: Calculate the initial concentrations
Concentration = moles / volume
[CH2O] = 0.055 moles . 0.500 L
[CH2O] = 0.11 M
[H2] = 0M
[CO] = 0M
Step 4: The concentration at the equilibrium
[CH2O] = 0.11 - X M = 0.051 M
[H2] = XM
[CO] = XM
[CH2O] = 0.11 - X M = 0.051 M
X = 0.11 - 0.051 = 0.059
[H2] = XM = 0.059 M
[CO] = XM = 0.059 M
Step 5: Calculate Kc
Kc = [H2][CO]/[CHO]
Kc = (0.059 * 0.059) / 0.051
Kc = 0.06825
Kc for this reaction is 0.06825
Answer:
I think the answer is 368.544 centigrams
Explanation:
I hope I helped
Answer:
6.4 × 10^-10 M
Explanation:
The molar solubility of the ions in a compound can be calculated from the Ksp (solubility constant).
CaF2 will dissociate as follows:
CaF2 ⇌Ca2+ + 2F-
1 mole of Calcium ion (x)
2 moles of fluorine ion (2x)
NaF will also dissociate as follows:
NaF ⇌ Na+ + F-
Where Na+ = 0.25M
F- = 0.25M
The total concentration of fluoride ion in the solution is (2x + 0.25M), however, due to common ion effect i.e. 2x<0.25, 2x can be neglected. This means that concentration of fluoride ion will be 0.25M
Ksp = {Ca2+}{F-}^2
Ksp = {x}{0.25}^2
4.0 × 10^-11 = 0.25^2 × x
4.0 × 10^-11 = 0.0625x
x = 4.0 × 10^-11 ÷ 6.25 × 10^-2
x = 4/6.25 × 10^ (-11+2)
x = 0.64 × 10^-9
x = 6.4 × 10^-10
Therefore, the molar solubility of CaF2 in NaF solution is 6.4 × 10^-10M
