You're off to a good start, now find the mass of H2O and put it under I mol,
then multiply 1 mol over the mass of H2O by 215 grams
Given:3.40g sample of the steel used to produce 250.0 mLSolution containing Cr2O72−
Assuming all the Cr is contained in the BaCrO4 at the end.
(0.145 g BaCrO4) / (253.3216 g BaCrO4/mol) x (250.0 mL / 10.0 mL) x (1 mol Cr / 1 mol BaCrO4) x (51.99616 g Cr/mol / (3.40 g) = 0.219 = 21.9% Cr
The equilibrium constant of the reaction is 282. Option D
<h3>What is equilibrium constant?</h3>
The term equilibrium constant refers to the number that often depict how much the process is able to turn the reactants in to products. In other words, if the reactants are readily turned into products, then it follows that the equilibrium constant will be large and positive.
Concentration of bromine = 0.600 mol /1.000-L = 0.600 M
Concentration of iodine = 1.600 mol/1.000-L = 1.600M
In this case, we must set up the ICE table as shown;
Br2(g) + I2(g) ↔ 2IBr(g)
I 0.6 1.6 0
C -x -x +2x
E 0.6 - x 1.6 - x 1.190
If 2x = 1.190
x = 1.190/2
x = 0.595
The concentrations at equilibrium are;
[Br2] = 0.6 - 0.595 = 0.005
[I2] = 1.6 - 0.595 = 1.005
Hence;
Kc = [IBr]^2/[Br2] [I2]
Kc = ( 1.190)^2/(0.005) (1.005)
Kc = 282
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Answer: 0.172 M
Explanation:
a) To calculate theconcentration of base, we use the equation given by neutralization reaction:

where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base which is NaOH.
We are given:

Putting values in above equation, we get:

b) To calculate the concentration of acid, we use the equation given by neutralization reaction:

where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base which is NaOH.
We are given:

Putting values in above equation, we get:

The concentration of the phosphoric acid solution is 1.172 M
Answer:
the error could have been the fact that the unit for volume wasn't changed from cm³ to dm³
hence the calculation error
the solution to this would be first dividing the volume by 1000 to get that same amount in dm³ which is the standard unit to be used for volume-density calculations