Answer:
2.676e22 atoms of Hydrogen
Explanation:
Knowing the density of water is 1g/ml, 1(g/ml)*0.4ml=0.4 grams of H2O. Knowing that there are 16 grams of H2O in 1 mole of H2O, we can set up a unit conversion of 0.4 Grams H2O * 
=
2.676e22 atoms of Hydrogen.
Answer:
14175 j heat released.
Explanation:
Given data:
Mass of aluminium = 350.0 g
Initial temperature = 70.0°C
Final temperature = 25.0°C
Specific heat capacity of Aluminium = 0.9 j/g.°C
Heat changed = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Heat change:
ΔT = Final temperature - initial temperature
ΔT = 25.0°C - 70°C
ΔT = -45°C
Q = m.c. ΔT
Q = 350 g × 0.9 j/g.°C × -45°C
Q = -14175 j
With the information given most likely in order to find the partial pressure of the gas produced you have to subtract the total air pressure in the collection flask by the atmospheric pressure since you assume that the flask started with atmospheric pressure when it was sealed and then the gas was added as the reaction took place increasing the pressure.
1.44atm-0.95 atm=0.49atm
The second volume : V₂= 0.922 L
<h3>
Further explanation
</h3><h3>Given
</h3>
7.03 Liters at 31 C and 111 Torr
Required
The second volume
Solution
T₁ = 31 + 273 = 304 K
P₁ = 111 torr = 0,146 atm
V₁ = 7.03 L
At STP :
P₂ = 1 atm
T₂ = 273 K
Use combine gas law :
P₁V₁/T₁ = P₂V₂/T₂
Input the value :
0.146 x 7.03 / 304 = 1 x V₂/273
V₂= 0.922 L
