C) change to water at the same temperature
Explanation:
Adding 334Joules of heat to one gram of ice at STP will cause ice to change to water at the same temperature.
- The heat of fusion is the amount of energy needed to melt a given mass of a solid
- It is also conversely the amount of energy removed from a substance to freeze it.
- The addition of this energy does not cause a decrease or increase in temperature.
- Only a phase change occurs.
Learn more:
Heat of fusion brainly.com/question/4050938
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Answer:
<h2>
The equilibrium constant Kc for this reaction is 19.4760</h2>
Explanation:
The volume of vessel used= 
ml
Initial moles of NO= 
moles
Initial moles of H2= 
moles
Concentration of NO at equilibrium= 
M
Moles of NO at equilibrium= 
=
moles
2H2 (g) + 2NO(g) <—> 2H2O (g) + N2 (g)
<u>Initial</u> :1.3*10^-2 2.6*10^-2 0 0 moles
<u>Equilibrium</u>:1.3*10^-2 - x 2.6*10^-2-x x x/2 moles
∴
⇒
![Kc=\frac{[H2O]^2[N2]}{[H2]^2[NO]^2} (volume of vesselin litre)](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BH2O%5D%5E2%5BN2%5D%7D%7B%5BH2%5D%5E2%5BNO%5D%5E2%7D%20%28volume%20of%20vesselin%20litre%29)
<u>Equilibrium</u>:0.31*10^-2 1.61*10^-2 0.99*10^-2 0.495*10^-2 moles
⇒
⇒
<span>decomposition of SrCO3 to SrO and CO2 =change in mass
moles of CO2 =(1.850 g - 1.445 g).
</span>Mass of <span>C<span>O2</span></span><span> in mixture: 1.850-1.445 = 0.405g
</span>0.405g/44.01 g/mol <span>C<span>O2</span></span><span> = 0.0092 moles </span><span>C<span>O2</span></span><span>.
</span>ratio of <span>C<span>O2</span></span><span> to SrO in Sr</span><span>C<span>O3</span></span><span> is 1:1
</span><span> mass ratio = 1.358/1.850 = 0.7341, </span>
or 73.41% Sr<span>C<span>O3</span></span><span>.
</span>hope this helps
First calculate the moles of N2 and H2 reacted.
moles N2 = 27.7 g / (28 g/mol) = 0.9893 mol
moles H2 = 4.45 g / (2 g/mol) = 2.225 mol
We can see that N2 is the limiting reactant, therefore we
base our calculation from that.
Calculating for mass of N2H4 formed:
mass N2H4 = 0.9893 mol N2 * (1 mole N2H4 / 1 mole N2) * 32
g / mol * 0.775
<span>mass N2H4 = 24.53 grams</span>
Answer:
The standard heat of formation of Compound X at 25°C is -3095.75 kJ/mol.
Explanation:
Mass of compound X = 7.00 g
Moles of compound X = 
Mass of water in calorimeter ,m= 35.00 kg = 35000 g
Change in temperature of the water in calorimeter = ΔT
ΔT = 2.113°C
Specific heat capacity of water ,c= 4.186 J/g °C
Q = m × c × ΔT
Heat gained by 35 kg of water is equal to the heat released on burning of 0.100 moles of compound X.
Heat of formation of Compound X at 25°C:
= -3095.75 kJ/mol
