Answer:
Q = 7898.38 J = 7.898 KJ
Explanation:
∴ ΔHv H2O = 47 KJ/mol = 47000 J/mol
∴ molar mass H2O = 18.015 g/mol
⇒ ΔHv H2O = (47000 J/mol)*(mol/18.015 g) = 2608.94 J/g
∴ m H2O = 2.73 g
⇒ ΔHv H2O = (2.73 g)(2608.94 J/g) = 7122.406 J
∴ C H2O = 4.18 J/g°C
∴ ΔT = 100 - 32 = 68°C
⇒ Q = (2.73 g)(4.18 J/g°C)(68°C) + 7122.406 J
⇒ Q = 775.975 J + 7122.406 J
⇒ Q = 7898.38 J
1.08 atm is the pressure for a certain tire in atmosphere.
<u>Explanation:</u>
One kilo pascal (1 kPa) corresponds to 1000 pascal. Another common unit used for pressure is atmosphere (symbolised as ‘atm’). 1 atm refers the standard atmospheric pressures and corresponds to 760 mm Hg and 101.3 kPa. Atmospheric pressures are commonly referred as square inches (psi)/ pounds.
Given:
The air pressure for a certain tire = 109 kPa
We need to find pressure in atmospheres
So, we know,
1 atm = 101.3 kPa
Hence,
1.08 atm is the pressure for a certain tire in atmosphere.
0.0062985632
6.30 x 10^ -3
( Its 6.29 but if you are rounding to the nearest thousandths then its 6.30 x 10 to the power of -3 since jumped 3 spaces back. Remember that when it starts with a 0 , you are moving to the first digit which in this case is 6)
- you're welcome
mass = volume times density
2.00 x 0.160 = 0.32 grams