Answer : The number of moles present in ammonia is, 70.459 moles.
Solution : Given,
Mass of ammonia = 
Molar mass of ammonia = 17.031 g/mole
Formula used :
Therefore, the number of moles present in ammonia is, 70.459 moles.
Answer:
1461.7 g of AgI
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
CaI₂ + 2AgNO₃ —> 2AgI + Ca(NO₃)₂
From the balanced equation above,
1 mole of CaI₂ reacted to produce 2 moles of AgI.
Next, we shall determine the number of mole AgI produced by the reaction of 3.11 moles of CaI₂. This can be obtained as follow:
From the balanced equation above,
1 mole of CaI₂ reacted to produce 2 moles of AgI.
Therefore, 3.11 moles of CaI₂ will react to produce = 3.11 × 2 = 6.22 moles of AgI
Finally, we shall determine the mass of 6.22 moles of AgI. This can be obtained as follow:
Mole of AgI = 6.22 moles
Molar mass of AgI = 108 + 127
= 235 g/mol
Mass of AgI =?
Mass = mole × molar mass
Mass of AgI = 6.22 × 235
Mass of AgI = 1461.7 g
Therefore, 1461.7 g of AgI were obtained from the reaction.
Answer:
if my calculation are correct, it's 295 grams
Explanation:
because liters converted to grams is .1=100 so if you take 2.95 times 100, it equals 295
Missing question: Express the salt concentration in kg/m³.
Answer is: the salt concentration is 9.8 kg/m³.
m(NaCl) = 9.8 g ÷ 1000 g/kg.
m(NaCl) = 0.0098 kg.
V(solution) = 1 L = 1 dm³.
V(solution) = 1 dm³ ÷ 1000 dm³/m³.
V(solution) = 0.001 m³.
d(solution) = m(NaCl) ÷ V(solution).
d(solution) = 0.0098 kg ÷ 0.001 m³.
d(solution) = 9.8 kg/m³.
