Molar mass NH3 = 17.031 g/mol
1 mole NH3 -------------- 17.031 g
?? moles NH3 ---------- 346 g
346 x 1 / 17.031
=> 20.31 moles of NH3
Heat required to vaporize 1 mol of water from water at 100C to steam at 100C = 40.7 kJ
<span>1 mol of water weighs = 18.015g
</span>1.55 kg = <span>1550/18.015 mol = 86.03 mol
</span><span>Heat required to vaporize :
</span>= 86.03 mol x <span>40.7 kJ
</span>
= 3501.421 kJ
Answer:
312 g of O₂
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
2KClO₃ —> 2KCl + 3O₂
From the balanced equation above,
2 mole of KClO₃ decomposed to 3 moles of O₂.
Next, we shall determine the number of mole of O₂ produced by the reaction of 6.5 moles of KClO₃. This can be obtained as follow:
From the balanced equation above,
2 mole of KClO₃ decomposed to 3 moles of O₂.
Therefore, 6.5 moles of KClO₃ will decompose to produce = (6.5 × 3)/2 = 9.75 moles of O₂.
Finally, we shall determine the mass of 9.75 moles of O₂. This can be obtained as follow:
Mole of O₂ = 9.75 moles
Molar mass of O₂ = 2 × 16 = 32 g/mol
Mass of O₂ =?
Mole = mass / Molar mass
9.75 = Mass of O₂ / 32
Cross multiply
Mass of O₂ = 9.75 × 32
Mass of O₂ = 312 g
Thus, 312 g of O₂ were obtained from the reaction.
Answer:
9.91 mL
Explanation:
Using the combined gas law equation as follows;
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (torr)
P2 = final pressure (torr)
V1 = initial volume (mL)
V2 = final volume (mL)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the information provided in this question;
V1 = 15.0mL
V2 = ?
P1 = 760 torr
P2 = 1252 torr
T1 = 10°C = 10 + 273 = 283K
T2 = 35°C = 35 + 273 = 308K
Using P1V1/T1 = P2V2/T2
760 × 15/283 = 1252 × V2/308
11400/283 = 1252V2/308
Cross multiply
11400 × 308 = 283 × 1252V2
3511200 = 354316V2
V2 = 3511200 ÷ 354316
V2 = 9.91 mL
Answer:
S(metal) = 0.66J/g°C
Explanation:
We can find specific heat of a material, S, using the equation:
q = m*S*ΔT
<em>Where q is change in heat, m is the mass of the substance, S specific heat and ΔT change in temperature.</em>
The heat given by the metal is equal to the heat that water absorbs, that is:
m(Metal)*S(metal)*ΔT(Metal) = m(Water)*S(water)*ΔT(water)
<em>Where:</em>
m(Metal) = 76.0g
S(metal) = ?
ΔT(Metal) = 96.0°C-31.0°C = 65.0°C
m(Water) = 120.0g
S(water) = 4.184J/g°C
ΔT(water) = 31.0°C-24.5°C = 6.5°C
Replacing:
76.0g*S(metal)*65.0°C = 120.0g*4.184J/g°C*6.5°C
S(metal) = 0.66J/g°C
<em />
The law of conservation applies because the energy is not been created or destroyed. The energy that the metal gives is absorbed by the water.
