Answer:
(i) specific heat
(ii) latent heat of vaporization
(iii) latent heat of fusion
Explanation:
i. Q = mcΔT; identify c.
Here, Q is heat, m is the mass, c is the specific heat and ΔT is the change in temperature.
The amount of heat required to raise the temperature of substance of mass 1 kg by 1 degree C is known as the specific heat.
ii. Q = mLvapor; identify Lvapor
Here, Q is the heat, m is the mass and L is the latent heat of vaporization.
The amount of heat required to convert the 1 kg liquid into 1 kg vapor at constant temperature.
iii. Q = mLfusion; identify Lfusion
Here, Q is the heat, m is the mass and L is the latent heat of fusion.
Here, Q is the heat, m is the mass and L is the latent heat of vaporization.
The amount of heat required to convert the 1 kg solid into 1 kg liquid at constant temperature.
Answer:
6626 g
Explanation:
Given that:
Density of water = 1.00 g/ml, volume of water = 42800 ml.
Since density = mass/ volume
mass of water = volume of water * density of water = 42800 ml * 1 g/ml = 42800 g
Initial temperature of water = 22°C and final temperature of water = 45°C.
specific heat capacity for water = 4.184 J/g°C
ΔT water = 45 - 22 = 23°C
For iron:
mass = m,
specific heat capacity for iron = 0.444 J/g°C
Initial temperature of iron = 1445°C and final temperature of water = 45°C.
ΔT iron = 45 - 1445 = -1400°C
Quantity of heat (Q) to raised the temperature of a body is given as:
Q = mCΔT
The quantity of heat required to raise the temperature of water is equal to the temperature loss by the iron.
Q water (gain) + Q iron (loss) = 0
Q water = - Q iron
42800 g × 4.184 J/g°C × 23°C = -m × 0.444 J/g°C × -1400°C
m = 4118729.6/621.6
m = 6626 g
32.8 g of Butane is required and 99.3 g of CO₂ is produced
<u>Explanation:</u>
The above mentioned reaction can be written as,
C₄H₁₀(g) + 13 O₂(g) → 4CO₂(g) + 5 H₂O(g) where ΔH (rxn)= -2658 kJ
It is given that 1.5 × 10³ kJ of energy is produced, the original reaction says that 2658 kJ of heat is produced, which means that less than one mole of butane is used in the reaction.
That is
of butane reacted
Now this moles is converted into mass by multiplying it with its molar mass = 0.564 mol × 58.122 g / mol
= 32.8 g of butane.
Mass of CO₂ produced = 0.564 ×44.01 g /mol × 4 mol
= 99.3 g of CO₂
Thus 32.8 g of Butane is required and 99.3 g of CO₂ is produced
Answer:
the reaction will come to a halt and the other reactant will still be present.
Explanation:
For the given reaction:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
![Rate=k[CO]^x[H_2]^y](https://tex.z-dn.net/?f=Rate%3Dk%5BCO%5D%5Ex%5BH_2%5D%5Ey)
where x and y are order wrt to 
and 
According to collision theory , the molecules must collide for a reaction to take place. According to collision theory , the rate of a reaction is proportional to rate of collision of reactants.
Thus with an increase in concentration of reactants , the rate of reaction also increases. This is because if the concentration of reactants increases , the chances of collision between molecules also increases and thus more products wil be formed which in turn increases the rate of reaction.
