Answer:
1.2 atm
Explanation:
Given data
- Volume of the gas in the tank (V₁): 200.0 L
- Pressure of ethylene gas in the tank (P₁): ?
- Volume of the gas in the torch (V₂): 300 L
- Pressure of the gas in the torch (P₂): 0.8 atm
If we consider ethylene gas to be an ideal gas, we can find the pressure of ethylene gas in the tank using Boyle's law.
Answer:
Explanation: The strengths of the inter molecular forces varies as follows -
The normal boiling point of CSe2 is 125°C and that of CS2 is 116°C, which explains the trend that as we move down the group, the boiling point of e compound increases as the size increases.
This usually happens because larger and heavier atoms have a tendency to exhibit greater inter molecular strengths due to the increase in size . As the size increases, the valence shell electrons move far away from the nucleus, thus has a greater tendency to attract the temporary dipoles.
And larger the inter molecular forces, more tightly the electrons will be held to each other and thus more thermal energy would be required to break the bonds between them.
First, we need to get the molar mass of:
KClO3 = 39.1 + 35.5 + 3*16 = 122.6 g/mol
KCl =39.1 + 35.5 = 74.6 g/mol
O2 = 16*2 = 32 g/mol
From the given equation we can see that:
every 2 moles of KClO3 gives 3 moles of O2
when mass = moles * molar mass
∴ the mass of KClO3 = (2mol of KClO3*122.6g/mol) = 245.2 g
and the mass of O2 then = 3 mol * 32g/mol = 96 g
so, 245.2 g of KClO3 gives 96 g of O2
A) 2.72 g of KClO3:
when 245.2 KClO3 gives → 96 g O2
2.72 g KClO3 gives → X
X = 2.72 g KClO3 * 96 g O2/245.2 KClO3
= 1.06 g of O2
B) 0.361 g KClO3:
when 245.2 g KClO3 gives → 96 g O2
0.361 g KClO3 gives → X
∴ X = 0.361g KClO3 * 96 g / 245.2 g
= 0.141 g of O2
C) 83.6 Kg KClO3:
when 245.2 g KClO3 gives → 96 g O2
83.6 Kg KClO3 gives → X
∴X = 83.6 Kg* 96 g O2 /245.2 g KClO3
= 32.7 Kg of O2
D) 22.4 mg of KClO3:
when 245.2 g KClO3 gives → 96 g O2
22.4 mg KClO3 gives → X
∴X = 22.4 mg * 96 g O2 / 245.2 g KClO3
= 8.8 mg of O2
Answer:
The answer to your question is Q = 355.64 J
Explanation:
Data
Heat = Q = ?
Temperature 1 = T1 = 20°C
Temperature 2 = T2 = 37°C
mass = m = 5 g
Specific heat = Cp = 4.184 J/g°C
Formula
Q = mCp(T2 - T1)
-Substitution
Q = (5)(4.184)(37 - 20)
-Simplification
Q = (5)(4.184)(17)
-Result
Q = 355.64 J
