Answer:
a. Minimum 1.70 V
b. There is no maximum.
Explanation:
We can solve this question by remembering that the cell potential is given by the formula
ε⁰ cell = ε⁰ reduction - ε⁰ oxidation
Now the problem states the cell must provide at least 0.9 V and that the reduction potential of the oxidized species 0.80 V, thus
ε⁰ reduction - ε⁰ oxidation ≥ ε⁰ cell
Since ε⁰ oxidation is by definition the negative of ε⁰ reduction , we have
ε⁰ reduction - ( 0.80 V ) ≥ 0.90 V
⇒ ε⁰ reduction ≥ 1.70 V
Therefore,
(a) The minimum standard reduction potential is 1.70 V
(b) There is no maximum standard reduction potential since it is stated in the question that we want to have a cell that provides at leat 0.9 V
Answer:
We assume you are converting between moles CaCl2 and gram. You can view more details on each measurement unit: molecular weight of CaCl2 or grams This compound is also known as Calcium Chloride. The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles CaCl2, or 110.984 grams.
Molar mass :
NaBr = 103 g/mol
Pb(NO3)2 = 331.20 g/mol
<span><span /><span>Balanced chemical equation :
</span></span>2 NaBr + 1 Pb(NO3)2 = 2 NaNO3 + 1 PbBr<span>2
</span><span>
2*103 g NaBr ------------> 1 * 331.20 g Pb(NO3)2
g NaBr -------------------> 311 g Pb(NO3)2
331.20 g = 2*103*311
331.20 g = 64066
mass ( NaBr ) = 64066 / 331.20
mass ( naBr) = 193,43 g of NaBr
hope this helps!.
</span>
Answer: The empirical formula for the given compound is 
Explanation : Given,
Percentage of C = 38.8 %
Percentage of H = 16.2 %
Percentage of N = 45.1 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of C = 38.8 g
Mass of H = 16.2 g
Mass of N = 45.4 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.
Moles of Carbon =
Moles of Hydrogen = 
Moles of Nitrogen = 
Step 2: Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3.23 moles.
For Carbon = 
For Hydrogen = 
For Oxygen = 
Step 3: Taking the mole ratio as their subscripts.
The ratio of C : H : N = 1 : 5 : 1
Hence, the empirical formula for the given compound is 
