Answer: 16.32 g of 
as excess reagent are left.
Explanation:
To calculate the moles :
According to stoichiometry :
2 moles of 
require = 1 mole of
Thus 0.34 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
Moles of left = (0.68-0.17) mol = 0.51 mol
Mass of 
Thus 16.32 g of as excess reagent are left.
Answer:
see explanation below
Explanation:
Question is incomplete, so in picture 1, you have a sample of this question with the missing data.
Now, in general terms, the absorbance of a substance can be calculated using the beer's law which is the following:
A = εlc
Where:
ε: molar absortivity
l: distance of the light in solution
c: concentration of solution
However, in this case, we have a plot line and a equation for this plot, so all we have to do is replace the given data into the equation and solve for x, which is the concentration.
the equation according to the plot is:
A = 15200c - 0.018
So solving for C for an absorbance of 0.25 is:
0.25 = 15200c - 0.018
0.25 + 0.018 = 15200c
0.268 = 15200c
c = 0.268/15200
c = 1.76x10⁻⁵ M
Answer:
What is the question? It looks like a fact.
Explanation:
The reaction between K₂SO₄(aq) and SrI₂(aq) produces KI(aq) and SrSO₄(s) as products.
The reaction is
K₂SO₄(aq) + SrI₂(aq) → KI(aq)+ SrSO₄(s)
To balance the equation both side of the reaction should have same number of atoms in each element.
Right hand side of the reaction has 1 K, 1 I, 1 Sr, 1 S and 4 O atoms while 2 K, 2 I, 1 Sr,1 S and 4 O present in left hand side of the reaction.
Hence, number of I atoms and number of K atoms are not balanced.
To balance the K atoms we should add 2 before KI. Then I atoms will be 2 at the right hand side.
Hence, the balanced reaction equation is
K₂SO₄(aq) + SrI₂(aq) → 2KI(aq)+ SrSO₄(s)
The answer to this problem is quite simple, it’s 9
