Answer:
.081 g of O2
Explanation:
4Cr + 3O2 -----> 2Cr2O3
.175 g Cr x [1 mole / 52.0 g] x [2 moles Cr2O3 / 4 moles Cr] x [152 g / 1 mole] = .256 g of Cr2O3
.175 g Cr x [1 mole / 52.0 g] x [3 moles O2 / 4 moles Cr] x [32 g / 1 mole] = .081 g of O2
Non metal atoms got this from google btw
Given that
1 skvarnick = 45 quibs
3 quibs = 7 sleps
Now if we have 45 quibs it means we have you have one skvarnick
three quibs = 7 sleps
so one quib = 7/3 sleps
so 45 quibs = 7 X 45 / 3 = 105 sleps
The question is incomplete, here is the complete question:
What volume (mL) of the partially neutralized stomach acid having concentration 2 M was neutralized by 0.1 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)
<u>Answer:</u> The volume of HCl neutralized is 1.25 mL
<u>Explanation:</u>
To calculate the volume of acid, we use the equation given by neutralization reaction:

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

Putting values in above equation, we get:

Hence, the volume of HCl neutralized is 1.25 mL
Answer:
the last period. i think has the largest energy level
Explanation: