Answer:
The volume of the gas will be 573.52 mL.
Explanation:
Boyle’s law states that <u>the pressure of a fixed amount of gas at a constant temperature is inversely proportional to the volume of the gas</u>.
The equation for Boyle's law is:
P₁V₁ = P₂V₂
We want to know V₂, that is, the volume of the container after the compression. We rearrange the equation and calculate:
V₂ = P₁V₁ ÷ P₂
V₂ = 1.14 atm × 815 mL ÷ 1.62 atm = 573.52 mL
Answer:

Explanation:
Hello there!
In this case, according to the given information, it is possible to recall the equation to calculate the heat, Q, in these calorimetry problems as shown below:

Thus, given the absorbed heat, mass and temperatures, we can easily calculate the specific heat of the metal as shown below:

Then, by plugging in we obtain:

Best regards!
unit coversation
1.429 atm - 1086mmhg
9361 pa-9.36 KPa - 70.21 mmhg
725 mmhg -0.95 atm- 96.26 kpa
calculation
(a) 1 atm = 760 mmhg
1.429 atm = ?
1.429 x760/1 = 1086.34 mm hg
(B) 1 mmhg = 101.325 kpa
? =9361 KPa
9361 x1 /101.25 =70.21 mmhg
760 mm hg= 101.325 KPa
70.21 mm hg=?
70.21 x101.325/760 = 9.36 Kpa
(C ) 1 atm = 760 mmhg
? = 725
= 725 x1/ 760=0.95 atm
1 atm = 101.325 kpa
0.95 =?
0.95 x101.325/1 = 96.26 KPa
Complete question:
A spirit burner used 1.00 g methanol to raise the temperature of 100.0 g water in a metal can from 28.00C to 58.0C. Calculate the heat of combustion of methanol in kJ/mol.
Answer:
the heat of combustion of the methanol is 402.31 kJ/mol
Explanation:
Given;
mass of water,
= 100 g
initial temperature of water, t₁ = 28 ⁰C
final temperature of water, t₂ = 58 ⁰C
specific heat capacity of water = 4.184 J/g⁰C
reacting mass of the methanol, m = 1.00 g
molecular mass of methanol = 32.04 g/mol
number of moles = 1 / 32.04
= 0.0312 mol
Apply the principle of conservation of energy;

Therefore, the heat of combustion of the methanol is 402.31 kJ/mol