Answer:
0.0400M of KI
Explanation:
Molarity is an unit of concentration defined as the ratio between moles of solute and liters of solution.
When you add 10.0 mL of 0.10M KI and 15.0mL, total volume is:
25.0mL = <em>0.025L of solution</em>
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And moles of KI are:
0.0100L × 0.10M = <em>0.00100 moles of KI</em>
<em />
Thus, molarity is:
0.00100 moles / 0.025L = <em>0.0400M of KI</em>
Stoichiomety:
1 moles of C + 1 mol of O2 = 1 mol of CO2
multiply each # of moles times the atomic molar mass of the compund to find the relation is weights
Atomic or molar weights:
C: 12 g/mol
O2: 2 * 16 g/mol = 32 g/mol
CO2 = 12 g/mol + 2* 16 g/mol = 44 g/mol
Stoichiometry:
12 g of C react with 32 g of O2 to produce 44 g of CO2
Then 18 g of C will react with: 18 * 32/ 12 g of Oxygen = 48 g of Oxygen
And the result will be 12 g of C + 48 g of O2 = 60 g of CO2.
You cannot obtain 72 g of CO2 from 18 g of C.
May be they just pretended that you use the law of consrvation of mass and say that you need 72 g - 18g = 54 g. But it violates the proportion of C and O2 in the CO2 and is not possible.
Answer:
B
Explanation:
Because when you go to the doctors they say, "You need to get your x-rays done" not microwave
Ummmmmmmmmmmmmmmmmmmmmmmmm, 21
Question is incomplete, complete question is;
A 34.8 mL solution of
(aq) of an unknown concentration was titrated with 0.15 M of NaOH(aq).

If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of
? For your answer, only type in the numerical value with two significant figures. Do NOT include the unit.
Answer:
0.044 M is the molarity of
(aq).
Explanation:
The reaction taking place here is in between acid and base which means that it is a neutralization reaction .
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:

0.044 M is the molarity of
(aq).