Answer:
40% of the energy release by the peanut is 3500 calories
Explanation:
One calorie is defined as the amount of energy required to increase the temperature of one gram of water for one degree Celsius (or one Kelvin)
Equation for energy gain by water is
Q = mcΔT
where, m is the mass of the object
c is the specific heat capacity
ΔT is the change in temperature
c = 1.0 cal/g?°C.
m = 50 g
ΔT = 50°C - 22°C
= 28°C
Q = (50)× (1)× (28)
= 1400calories
The peanut contain 1400calories of energy .
amount that 40% of energy is released to water ,
so,
Q = 1400 calories / 0.4
= 3500 calories
Therefore, 40% of the energy release by the peanut is 3500 calories
Answer:
B. double replacement
Explanation:
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Answer:
1) 2Al + 6HCl ⟶ 2AlCl₃ + 3H₂
Fe + 2HCl ⟶ FeCl₂ + H₂
2) Cu = 2.5 g; Al = 3.5 g; Fe = 4.0 g
Explanation:
1) Possible reactions
2Al + 6HCl ⟶ 2AlCl₃ + 3H₂
Fe + 2HCl ⟶ FeCl₂ + H₂
2) Mass of each metal
a) Mass of Cu
The waste was the unreacted copper.
Mass of Cu = 2.5 g
b) Masses of Al and Fe
We have two relations
:
Mass of Al + mass of Fe = 10 g - 2.5 g = 7.5 g
H₂ from Al + H₂ from Fe = 6.38 L at NTP
i) Calculate the moles of H₂
NTP is 20 °C and 1 atm.

(ii) Solve the relationship
Let x = mass of Al. Then
7.5 - x = mass of Fe
Moles of Al = x/27
Moles of Fe = (7.5 - x)/56
Moles of H₂ from Al = (3/2) × Moles of Al = (3/2) × (x/27) = x
/18
Moles of H₂ from Fe = (1/1) × Moles of Fe = (7.5 - x)/56
∴ x/18 + (7.5 - x)/56 = 0.2652
56x + 18(7.5 - x) = 267.3
56x + 135 - 18x = 267.3
38x = 132.3
x = 3.5 g
Mass of Al = 3.5 g
Mass of Fe = 7.5 g - 3.5 g = 4.0 g
The masses of the metals are Cu = 2.5 g; Al = 3.5 g; Fe = 4.0 g
Answer:
(a) 7.11x10⁻⁴ M/s
(b) 2.56 mol.L⁻¹.h⁻¹
Explanation:
(a) The reaction is:
O₃(g) + NO(g) → O₂(g) + NO₂(g) (1)
The reaction rate of equation (1) is given by:
(2)
<u>We have:</u>
k: is the rate constant of reaction = 3.91x10⁶ M⁻¹.s⁻¹
[O₃]₀ = 2.35x10⁻⁶ M
[NO]₀ = 7.74x10⁻⁵ M
Hence, to find the inital reacion rate we will use equation (2):
Therefore, the inital reaction rate is 7.11x10⁻⁴ M/s
(b) The number of moles of NO₂(g) produced per hour per liter of air is:
t = 1 h
V = 1 L
![\frac{\Delta[NO_{2}]}{\Delta t} = rate](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%5BNO_%7B2%7D%5D%7D%7B%5CDelta%20t%7D%20%3D%20rate)
![\frac{\Delta[NO_{2}]}{\Delta t} = 7.11 \cdot 10^{-4} M/s*\frac{3600 s}{1 h} = 2.56 mol.L^{-1}.h{-1}](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%5BNO_%7B2%7D%5D%7D%7B%5CDelta%20t%7D%20%3D%207.11%20%5Ccdot%2010%5E%7B-4%7D%20M%2Fs%2A%5Cfrac%7B3600%20s%7D%7B1%20h%7D%20%3D%202.56%20mol.L%5E%7B-1%7D.h%7B-1%7D)
Hence, the number of moles of NO₂(g) produced per hour per liter of air is 2.56 mol.L⁻¹.h⁻¹
I hope it helps you!