Answer:
16.02 g
Explanation:
the balanced equation for the decomposition of CuCO₃ is as follows
CuCO₃ --> CuO + CO₂
molar ratio of CuCO₃ to CO₂ is 1:1
number of CuCO₃ moles decomposed - 45 g / 123.5 g/mol = 0.364 mol
according to the molar ratio
1 mol of CuCO₃ decomposes to form 1 mol of CO₂
therefore 0.364 mol of CuCO₃ decomposes to form 0.364 mol of CO₂
number of CO₂ moles produced - 0.364 mol
therefore mass of CO₂ produced - 0.364 mol x 44 g/mol = 16.02 g
16.02 g of CO₂ produced
Answer:
Explanation:
It often helps to write the heat as if it were a reactant or a product in the thermochemical equation.
Then you can consider it to be 11018 "moles" of "kJ"
We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.
M_r: 32.00
2C₈H₁₈ + 25O₂ ⟶ 16CO₂ + 8H₂O + 11 018 kJ
n/mol: 7280
1. Moles of O₂
The molar ratio is 25 mol O₂:11 018 kJ
2. Mass of O₂
The volume of 0.160 m Li2S solution required to completely react with 130 ml of 0.160 CO(NO3)2 is calculated as below
write the reacting equation
Co(NO3)2 + Li2S = 2LiNO3 + COS
find the moles of CO(NO3)2 = molarity x volume
= 130 ml x 0.160=20.8 moles
since the reacting moles between CO(NO3)2 to LiS is 1:1 the moles of LiS is also 20.8 moles
volume of Lis is therefore = moles of Lis/ molarity of LiS
= 20.8/0.160 = 130 Ml
Answer:
How the incident happened
Any chemicals involved in an incident
Any other hazards present in the lab
Explanation:
Above are the types of information that are necessary to communicate with emergency responders. The emergency responders ask the first question that how the incident happened. After that they ask that is there any harmful chemicals are present in the laboratory or what types of chemicals present in the laboratory. These questions were asked by the emergency responders in order to give the patient a suitable treatment.
