Answer:
False
Explanation:
2CH₄ + 2O₂ ⟶ 2CO₂ + 2H₂O
Let’s count the atoms on each side of the reaction arrow.
<u>Atom</u> <u>On the left</u> <u>On the right</u>
C 2 2
H 8 4
O 4 6
C is balanced, but there are different numbers of H and O on each side.
The equation is not balanced
.
The number of solid precipitate that will be formed is 37.08 g
calculation
write the equation for reaction
=Hg(NO3)2 +Na2SO4 = HgSO4(s) +2NaNO3(aq)
find the moles of each reactant
moles ofHg(NO3)2=126.27/324.6= 0.389 moles
moles of Na2SO4=17.796/142=0.125 moles
NaSO4 is the limiting reagent and by use of mole ratio of NaSO4:HgSO4 which is 1:1 therefore the moles of H2SO4 is also= 0.125 moles
mass HgSO4=moles x molar mass
=0.125 x296.65= 37.08g
C. The thermometer reads 100
Answer:
a. 300 Kg of fertilizer
b. 225 Kg of fertilizer
c. 400 Kg of fertilizer
d. 600 Kg of fertilizer
Explanation:
The percentage composition ratio of Nitrogen, Phosphorus and Potassium in a 1 Kg bag of the given fertilizer is 40:15:10.
Therefore a 1 Kg bag contains;
40/100 * 1 Kg = 0.4 Kg of Nitrogen;
15/100 * 1 Kg = 0.15 Kg of phosphorus;
10/100 * 1 Kg = 0.1 Kg of potassium
Quantity of fertilizer required to add to a hectare to supply;
a. Nitrogen at 120 kg/ha = 120/0.4 = 300 Kg of fertilizer
b.. Nitrogen at 90 Kg/ha = 90/0.4 = 225 Kg of fertilizer
c. Phosphorus at a rate of 60 kg/ha = 60/0.15 = 400 Kg of fertilizer
d. Potassium at a rate of 60 kg/ha = 60/0.1 = 600 Kg of fertilizer
