Solution :
From the balanced chemical equation, we can say that 1 moles of KBr will produce 1 moles of KCl .
Moles of KBr in 102 g of potassium bromide.
n = 102/119.002
n = 0.86 mole.
So, number of miles of KCl produced are also 0.86 mole.
Mass of KCl produced :
Hence, this is the required solution.
Answer:
0.0957 g
Explanation:
Half life is the time required by the substance to decay to half of its initial mass. Most often exponential decays of radioactive substances are explained by half life. Half life remains constant throughout the life of decay.
It is given as;
<em> N</em>(t) = N₀ (1/2)^t/t1/2
Where;
<em>N</em>(t) = Amount remaining after certain time
N₀ = The starting initial amount
t1/2 = Half life of given substance
Putting values,
<em> N</em>(t) = 98 × (1/2)^215/21.5
<em> N</em>(t) = 0.0957 g
I think the correct answer is B I am so sorry if it’s incorrect
Answer:- 
for the reaction is -58 kJ/mol.
Solution:- Oxidation half reaction taking place at the anode is--
= 0.14 V
Reduction half equation taking place at the cathode is---
= -0.04V
for the cell = 
for the cell = -0.04V + 0.14V = 0.10V
Now we could easily calculate by using the below formula--
= -nF
where n is the number of electrons transferred in the over all reaction. Looking at two half equations the value of n is 6.
F is Farady constant and it's value is 96500 C/mol.
Plugging in the values in the formula...
= -(6)(96500)(0.10)
= -57900 J
Since, 1000 J = 1 kJ
So, = -58 kJ/mol
1:3 hope DAT helps #ZedTheZom
