A catalyst
A catalyst can be in many forms
Answer:
Molarity =5.32 M
Explanation:
Given data:
Mass of glucose = 239 g
Volume = 250 mL (250 /1000 = 0.25 L)
Molarity = ?
Solution;
Formula:
Molarity = number of moles / volume in litter
Number of moles:
Number of moles = mass/ molar mass
Number of moles = 239 g / 180.2 g/mol
Number of moles = 1.33 mol
Molarity:
Molarity = number of moles / volume in litter
Molarity = 1.33 mol / 0.25 L
Molarity =5.32 M
Answer:
0.302L
Explanation:
<em>...97.1mL of 1.21m M aqueous magnesium fluoride solution</em>
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In this problem the chemist is disolving a solution from 1.21mM = 1.21x10⁻³M, to 389μM = 389x10⁻⁶M. That means the solution must be diluted:
1.21x10⁻³M / 389x10⁻⁶M = 3.11 times
As the initial volume of the original concentration is 97.1mL, the final volume must be:
97.1mL * 3.11 = 302.0mL =
0.302L
Answer:
The answer is (H30+) =3,55e-8M and (OH-)=2,82e-7M
Explanation:
We use the formulas:
pH= - log(H30+) and Kwater=(H30+)x(OH-)
pH= - log(H30+) ----< (H30+)= antilog- pH=antilog- 7,45=3,55E-8M
Kwater=(H30+)x(OH-)
(OH-)=Kwater/(H30+)= 1,00e-14/3,55e-8 = 2,82e-7
Answer:
10 L of CO₂.
Explanation:
The balanced equation for the reaction is given below:
2CO + O₂ —> 2CO₂
From the balanced equation above,
2 L of CO reacted to produce 2 L of CO₂.
Finally, we shall determine the volume of CO₂ produced by the reaction of 10 L CO. This can be obtained as follow:
From the balanced equation above,
2 L of CO reacted to produce 2 L of CO₂.
Therefore, 10 L of CO will also react to produce 10 L of CO₂.
Thus, 10 L of CO₂ were obtained from the reaction.
