Answer:
11.0 L
Explanation:
The equation for this reaction is given as;
2H2 + O2 --> 2H2O
2 mol of H2 reacts with 1 mol of O2 to form 2 mol of H2O
At STP;
1 mol = 22.4 L
This means;
44.8 L of H2 reacts with 22.4 L of O2 to form 44.8 L of H2O
In this reaction, the limiting reactant is H2 as O2 is in excess.
The relationship between H2 and H2O;
44.8 L = 44.8 L
11.0 L would produce x
Solving for x;
x = 11 * 44.8 / 44.8
x = 11.0 L
(05.01)
Which statement best explains if the graph correctly represents the proportional relationship y = 0.5x? (4 points)
<span>Yes, the points shown on the line would be part of y = 0.5xYes, all proportions can be shown on a graph of this lineNo, the points shown would not be part of y = 0.5xNo, proportions cannot be represented on a graph</span>
pH value of a 0. 011 m naf solution is 8.57.
Solution
This problem uses the relationship between Kb and the dissociation constants which is expressed as Kw = KaKb. Calculations are as follows:
Kb = KaKb
1.00 x 10^-14 = 7.2 x 10^-4(x)
x = 1.39 x 10^-11
We now need to calculate the [OH¯] using the Kb expression:
1.39 x 10^-11 = x^2 / (0.30 - x)
The denominator can be neglected. Thus, x is 3.73 x 10^-6.
pOH = -log 3.73 x 10^-6 = 5.43
pH = 14-5.43 = 8.57
To learn more ph of naf solution refer here:
brainly.com/question/1411794
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To solve this problem. use the formula
m1 v1 = m2 v2
where m1 is the molarity of the stock soln
v1 is the volume of the stock soln
m2 is the molarity of the new soln
v2 is the volume of the new soln
m1 v1 = m2 v2
13.5(v2) = (10.5)(1.8)
v2 = (10.5)(1.8) / 13.5
v2 =1.4 L is needed from the 13.5 M HCl stock soln