The answer is 2.37 x10^22 atoms
Answer: an iceberg
Explanation: the reason for this is because it has much more mass then a hot cup of coffee even the the temputure of coffee is much warmer
Lets take 100 g of this compound,
so it is going to be 2.00 g H, 32.7 g S and 65.3 g O.
2.00 g H *1 mol H/1.01 g H ≈ 1.98 mol H
32.7 g S *1 mol S/ 32.1 g S ≈ 1.02 mol S
65.3 g O * 1 mol O/16.0 g O ≈ 4.08 mol O
1.98 mol H : 1.02 mol S : 4.08 mol O = 2 mol H : 1 mol S : 4 mol O
Empirical formula
H2SO4
Answer:
a) ΔHvap=35.3395 kJ/mol
b) Tb=98.62 °C
Explanation:
Given the reaction:
C₇H₁₆ (l) ⇔ C₇H₁₆ (g)
Kp=P(C₇H₁₆) since the concentration ratio for a pure liquid is equal to 1.
When
T₁=50°C=323.15K ⇒P₁=0.179
T₂=86°C=359.15K ⇒P₂=0.669
The Clasius-Clapeyron equation is:
ΔHvap=35339.5 J/mol=35.3395 KJ/mol
Normal boiling point ⇒ P=1 atm
Hence, we find the normal boiling point where:
T₁=323.15K
P₁=0.179 atm
P₂=1 atm
T₂=371.77 K= 98.62 °C
Answer:
7.32g of HNO3 are required.
Explanation:
1st) From the balanced reaction we know that 2 moles of HNO3 react with 1 mole of Ca(OH)2 to produce 2 moles of H2O and 1 mole of Ca(NO3)2.
From this, we find that the relation between HNO3 and Ca(OH)2 is that 2 moles of HNO3 react with 1 mole of Ca(OH)2.
2nd) This is the order of the relations that we have to use in the equation to calculate the grams of nitric acid:
• starting with the 4.30 grams of Ca(OH)2.
,
• using the molar mass of Ca(OH)2 (74g/mol).
,
• relation of the 2 moles of HNO3 that react with 1 mole of Ca(OH)2 .
,
• using the molar mass of HNO3 (63.02g/mol).
So, 7.32g of HNO3 are required.