Answer:
3.47 ×10^-10
Explanation:
The equation of the reaction is 2Cr3+(aq) + Pb(s)------->2Cr2+(aq) + Pb2+(aq)
A total of two moles of electrons were transferred in the process. The chromium was reduced while the lead was oxidized. Hence the lead species will constitute the oxidation half equation and the chromium will constitute the reduction half equation.
E°cell = E°cathode - E°anode
E°cathode = -0.41 V
E°anode = -0.13 V
E°cell = -0.41 -(-0.13) = -0.28 V
From
E°cell = 0.0592/n log K
n= 2, K= the unknown
-0.28 = 0.0592/2 log K
log K = -0.28/0.0296
log K = -9.4595
K = Antilog ( -9.4595)
K= 3.47 ×10^-10
Answer:
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Theory
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Explanation:
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Answer : Option A) HF.
Explanation : Assuming the complete question as per the attachment.
H-F bonds are formed because of hydrogen bonding between H atom and F atom and is considered to be strongest amongst the rest other options given.
As given in the table it is clear that the molar mass of HF molecule is the lowest. It shows that, when dispersion increases in the molecule molecular weight also increases.
8 moles of H 2O are produced.
First, we need to figure out the chemical equation for producing water with oxygen which is H 2 + O2 = H 2O. Then, we need to balance the equation, resulting in 2H 2 + O2 = 2H 2O.
<h3>How many moles of H2 are required to make one mole of NH3?</h3>
Calculate 0.88074 mol H2's mass. If N2 is too much, 1.776 g H2 is needed to create 10.00 g of NH3. To create 8.2 moles of ammonia, 2 moles of NH3 are created when 1 mole of N2 and 3 moles of H2 mix. 4.1 moles of N2 Fast are consequently needed to make 8.2 moles of NH3.
<h3>
How many moles of h2 are needed to produce a solution?</h3>
An O-H bond has a bond energy of 1 09 Kcal. 3.6. A 38.0mL 0.026M HCl solution and a 0.032M NaOH solution react. Thus, 10 moles of NH 3 are obtained by dividing 15 moles of H2 by the 1.5 moles of H2 required for the product. and 9.3 x 10-3 moles of bromobutane (1.27/137 =.00927moles).
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