I would say A and D. A is right for sure, as for D, I'm not so sure. Hope this helps!
Answer: 30. 7 moles SO3
31. 3 moles SO2 and 3 moles SO3
Explanation: To solve for this problem use the mole ratio of the substances involved in the reaction.
Solution for number 30:
3.5 moles O2 x 2 moles SO3 / 1 mole O2
= 7 moles SO3
31. 192 g SO2 x 1 mole SO2 / 64 g/ mol SO2
= 3 moles SO2
3 moles SO2 x 2 moles SO3 / 2 moles SO2
= 3 moles SO3
Answer:
1.263 moles of HF
Explanation:
The balance chemical equation for given single replacement reaction is;
Sn + 2 HF → SnF₂ + H₂
Step 1: <u>Calculate Moles of Tin as;</u>
As we know,
Moles = Mass / A.Mass ----- (1)
Where;
Mass of Tin = 75.0 g
A.Mass of Tin = 118.71 g/mol
Putting values in eq. 1;
Moles = 75.0 g / 118.71 g/mol
Moles = 0.6318 moles of Sn
Step 2: <u>Find out moles of Hydrogen Fluoride as;</u>
According to balance chemical equation,
1 mole of Sn reacted with = 2 moles of HF
So,
0.6318 moles of Sn will react with = X moles of HF
Solving for X,
X = 0.6318 mol × 2 mol / 1 mol
X = 1.263 moles of HF
Answer:
% (COOK)2H2O = 37.826 %
Explanation:
mix: (COOK)2H2O + Ca(OH)2 → CaC2O4 + H2O
∴ mass mix = 4.00 g
∴ mass (CaC2O4)H2O = 1.20 g
∴ Mw (COOK)2H2O = 184.24 g/mol
∴ Mw (CaC2O4)H2O = 146.12 g/mol
∴ r = mol (COOK)2H2O / mol (CaC2O4)H2O = 1
- % (COOK)2H2O = (mass (COOK)2H2O / mass Mix) × 100
⇒ mass (COOK)2H2O = (1.20 g (CaC2O4)H2O)×(mol (CaC2O4)H2O / 146.12 g (CaC2O4)H2O)×(mol (COOK)2H2O/mol (CaC2O4)H2O)×(184.24 g (COOK)2H2O/mol (COOK)2H2O)
⇒ mass (COOK)2H2O = 1.513 g
⇒ % (COOK)2H2O = ( 1.513 g / 4 g )×100
⇒ % (COOK)2H2O = 37.826 %
The only liquid elements at standard temperature and pressure are bromine (Br) and mercury (Hg). Although, elements caesium (Cs), rubidium (Rb), Francium (Fr) and Gallium (Ga) become liquid at or just above room temperature.
