Answer:
Reaction is already balanced
Explanation:
The equation in reaction 1 is given as;
CH3COOH + NaHCO3 --> CO2 + H2O + Na+ + CH3COO-
The reaction is already balanced. This is because the umber of atoms of elements in the reactant is equal to that of the products
Carbon
Reactant = 2 + 1 = 3
Product = 2 + 2 = 3
Hydrogen
Reactant = 3 + 1 + 1 = 5
Product = 2 + 3 = 5
Oxygen
Reactant = 2 + 3 = 5
Product = 2 +1 + 2 = 5
Sodium
Reactant = 1
Product = 1
The answer is Liquid iron.
NaH(s)+ H2O (l)=>NaOH(aq)+H2(g)
You want to calculate the mass of NaH, I assume. Otherwise, the question isn't clear. It simply says calculate the mass(??)
So, calculate the moles of H2 gas that satisfy the conditions of 982 ml at 28ºC and 765 torr. But you must subtract the vapor pressure of water at 28º to get the actual pressure of the H2 gas. So, the actual conditions are 982 ml (0.982 L) and 301 K and 765-28 = 737 torr.
PV = nRT
n = PV/RT = (737 torr)(0.982 L)/(62.4 L-torr/Kmol)(301 K)
n = 0.0385 moles H2
moles NaH needed = 0.0385 moles H2 x 1 mole NaH/mole H2 = 0.0385 moles NaH required
mass of NaH needed = 0.0385 moles x 24 g/mole = 0.925 g NaH
Brainliest Please :)
Answer: I think it's C
Explanation: I hope this helps (Sorry if it doesn't)
Answer:
59.8%
Explanation:
First find the Mr of manganese (III) nitrate.
Mr of Mn(NO₃)₃ = 54.9 + (14 × 3) + (16 × 3 × 3) = <u>240.9</u>
Since we have to find the percentage composition of oxygen, we need to find the Mr of oxygen in the compound, which is:
Mr of (O₃)₃ = (16 × 3) × 3 = <u>144</u>
Now we can find percentage composition / percentage by mass of oxygen.
% composition = 
× 100
% composition = 
× 100 = <u>59.776%</u>
∴ % compostion of oxygen in maganese(III)nitrate is 59.8% (to 3 significant figures).
