Answer:
Reversible reactions exhibit the same reaction rate for forward and reverse reactions at equilibrium.
Reversible reactions exhibit constant concentrations of reactants and products at equilibrium
Explanation:
A reversible reaction is a reaction that can proceed in both forward and backward direction.
Equilibrium is attained in a chemical system when there is no observable change in the properties of the system.
At equilibrium, a reversible reaction is occurring in at same rate. That is, the forward and backward reaction is occurring at the same rate. As the rate of the forward and backward reaction remains the same, the concentrations of the reactants and products will also be the same in order for the equilibrium to be maintained.
The total mass of the products is 10.76 g + 204.44 g = 215.20 g.
The masses of all the reactants but one are known so,
215.20 g - 120.00 g - 8.15 g - 75.00 g = 12.05 g
12.05 g is the mass of the unweighed barium nitrate.
A student measures the volume of a solution to be 0.01370 have 5 significant digits in this measurement.
<h3>What are significant digits?</h3>
The significant digits are the minimum number from zero to nine for reporting any measurement where the digits are uncertain.
The significant digits starting from zero are not significant digits, decimal is not a significant digit, and ending zero after the decimal are significant digits.
Therefore, the student measures the volume of a solution to be 0.01370 5 significant digits are in this measurement.
Learn more about significant digits, here:
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Answer:
Part A is just T2 = 58.3 K
Part B ∆U = 10967.6 x C
You can work out C
Part C
Part D
Part E
Part F
Explanation:
P = n (RT/V)
V = (nR/P) T
P1V1 = P2V2
P1/T1 = P2/T2
V1/T1 = V2/T2
P = Pressure(atm)
n = Moles
T = Temperature(K)
V = Volume(L)
R = 8.314 Joule or 0.08206 L·atm·mol−1·K−1.
bar = 0.986923 atm
N = 14g/mol
N2 Molar Mass 28g
n = 3.5 mol N2
T1 = 350K
P1 = 1.5 bar = 1.4803845 atm
P2 = 0.25 bar = 0.24673075 atm
Heat Capacity at Constant Volume
Q = nCVΔT
Polyatomic gas: CV = 3R
P = n (RT/V)
0.986923 atm x 1.5 = 3.5 mol x ((0.08206 L atm mol -1 K-1 x 350 K) / V))
V = (nR/P) T
V = ((3.5 mol x 0.08206 L atm mol -1 K-1)/(1.5 x 0.986923 atm) )x 350K
V = (0.28721/1.4803845) x 350
V = 0.194 x 350
V = 67.9036 L
So V1 = 67.9036 L
P1V1 = P2V2
1.4803845 atm x 67.9036 L = 0.24673075 x V2
100.52343693 = 0.24673075 x V2
V2 = P1V1/P2
V2 = 100.52343693/0.24673075
V2 = 407.4216 L
P1/T1 = P2/T2
1.4803845 atm / 350 K = 0.24673075 atm / T2
0.00422967 = 0.24673075 /T2
T2 = 0.24673075/0.00422967
T2 = 58.3 K
∆U= nC ∆T
Polyatomic gas: C = 3R
∆U= nC ∆T
∆U= 28g x C x (350K - 58.3K)
∆U = 28C x 291.7
∆U = 10967.6 x C
The answer is D. Heterogeneous mixture
