According to the law of conservation of mass, the amount of BARIUM present of the reactants is the same as the amount present in the products (the precipitate).
(11.21 g BaSO4) / (233.4 g/mol BaSO4) = 0.0480 mol BaSO4 and original barium salt
(10.0 g) / (0.0480 mol) = 208.3 g/mol
So it must have been BaCl2, because the molar mass of Barium is 137 which leave 71 grams left. Since Barium is a +2 charge, it means the atom next to it must be twice. Chlorine mass is 35, which twice is 71
Answer:
4.92 L
Explanation:
Rearrange ideal gas law and solve.
Change C to K.
- Hope that helps! Please let me know if you need further explanation.
Answer:
= 9.28 g CO₂
Explanation:
First write a balanced equation:
CH₄ + 2O₂ -> 2H₂O + CO₂
Convert the information to moles
7.50g CH₄ = 0.46875 mol CH₄
13.5g O₂ = 0.421875 mol O₂
Theoretical molar ratio CH₄:O₂ -> 1:2
Actual ratio is 0.46875 : 0.421875 ≈ 1:1
If all CH₄ is used up, there would need to be more O₂
So O₂ is the limiting reactant and we use this in our equation
Use molar ratio to find moles of CO₂
0.421875 mol O₂ * 1 mol CO₂/2 mol O₂=0.2109375 mol CO₂
Then convert to grams
0.2109375 mol CO₂ = 9.28114 g CO₂
round to 3 sig figs
= 9.28 g CO₂
Answer:
4200ml
Explanation:
Converting 3.1kg to g
3.1*1000= 3100g
Since density = mass/volume, then
volume = mass/ density
Therefore volume = 3100/0.74
= 4189.2ml
converting it to two significant figures
= 4200ml
Answer:
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