What do you call a large body of air? wind thunderstorm cold front air mass
2 answers:
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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).
Answer:
D) 1.61 times faster
Explanation:
= √(3)RTM
R constant= 0.08206
T=constant, so in this problem we dont need a value for it
M=17.031 g/mol
√(3)(0.08206)(17.031)= 2.047
= √(3)RTM
R constant= 0.08206
T=constant, so in this problem we dont need a value for it
M= 44.01 g/mol
√(3)(0.08206)(44.01)= 3.29
Since we are trying to measure how much faster NH3 will be, we have to realize that mass and speed have an inverse relationship.
So instead of doing (2.047)/(3.29) = 6.22
we have to flip the values to get (3.29)/(2.047)= 1.61