Answer:
The temperature would be reduced by half
Explanation:
Charles' Law => V ∝ T => V = kT => k = V/T
For two sets of T vs V conditions, the system constant (k) remains unchanged and k₁ = k₂ => V₁/T₁ = V₂/T₂.
Therefore, if V₁ is reduced to 1/2V₁ = V₂ => V₁/T₁ = 1/2V₁/T₂ => V₁/T₁ = V₁/2T₂
and solving for T₂ => 1/T₁ = 1/2T₂ => 2T₂ = T₁ => T₂ = 1/2T₁
∴ The initial temperature (T₁) would be reduced by half or, T₂ = 1/2T₁
You multiply 6.50 by the molar mass of H2SO4.
Δmc
2
For one reaction:
Mass Defect =Δm
=2(m
H
)−m
He
−m
n
=2(2.015)−3.017−1.009
=0.004 amu
1 amu=931.5 MeV/c
2
Hence,
E=0.004×931.5 MeV=3.724 MeV
E=3.726×1.6×10
−13
J=5.96×10
−13
J
For 1 kg of Deuterium available,
moles=
2g
1000g
=500
N=500N
A
=3.01×10
26
Energy released =
2
N
×5.95×10
−13
J
=8.95×10
13
Answer:
the answer is 6
Explanation:
there is 3 hydrogen molecules in NH3 and there's 2 molecules of NH3 so in total, there are 6 hydrogen molecules on the products side.
Since the density of water is 1 g /mL, hence there is 100
g of H2O. So total mass is:
m = 100 g + 5 g = 105 g
=> The heat of reaction can be calculated using the
formula:
δhrxn = m C ΔT
where m is mass, C is heap capacity and ΔT is change in
temperature = negative since there is a decrease
δhrxn = 105 g * 4.18 J/g°C * (-2.30°C)
δhrxn = -1,009.47 J
=> However this is still in units of J, so calculate
the number of moles of NaCl.
moles NaCl = 5 g / (58.44 g / mol)
moles NaCl = 0.0856 mol
=> So the heat of reaction per mole is:
δhrxn = -1,009.47 J / 0.0856 mol
δhrxn = -11,798.69 J/mol = -11.8 kJ/mol
