Answer:
The initial temperature of the metal is 84.149 °C.
Explanation:
The heat lost by the metal will be equivalent to the heat gain by the water.
- (msΔT)metal = (msΔT)water
-32.5 grams × 0.365 J/g°C × ΔT = 105.3 grams × 4.18 J/g °C × (17.3 -15.4)°C
-ΔT = 836.29/12.51 °C
-ΔT = 66.89 °C
-(T final - T initial) = 66.89 °C
T initial = 66.89 °C + T final
T initial = 66.89 °C + 17.3 °C
T initial = 84.149 °C.
It allows electrons to flow from the anode to the cathode.
Answer:
The volume of the gas sample at standard pressure is <u>819.5ml</u>
Explanation:
Solution Given:
let volume be V and temperature be T and pressure be P.



1 torr= 1 mmhg
42.2 torr=42.2 mmhg
so,


Now
firstly we need to find the pressure due to gas along by subtracting the vapor pressure of water.

=735-42.2=692.8 mmhg
Now
By using combined gas law equation:



Here
are standard pressure and temperature respectively.
we have

Substituting value, we get


Compared to the charge and mass of a proton an electron has......
A proton has approximately the same mass as..........
Answer:
9.29 mol
Explanation:
Given data:
Number of moles = ?
Mass = 148.6 g
Solution:
Number of moles = mass/ molar mass
Molar mass of CH₄ = 16 g/mol
Now we will put the values in formula.
Number of moles = 148.6 g/ 16 g/mol
Number of moles = 9.29 mol
Thus 148.6 g have 9.29 moles.