The volume of oxygen at STP required would be 252.0 mL.
<h3>Stoichiometic problem</h3>
The equation for the complete combustion of C2H2 is as below:

The mole ratio of C2H2 to O2 is 2:5.
1 mole of a gas at STP is 22.4 L.
At STP, 100.50 mL of C2H2 will be:
100.50 x 1/22400 = 0.0045 mole
Equivalent mole of O2 according to the balanced equation = 5/2 x 0.0045 = 0.01125 moles
0.01125 moles of O2 at STP = 0.01125 x 22400 = 252.0 mL
Thus, 252.0 mL of O2 gas will be required at STP.
More on stoichiometric problems can be found here: brainly.com/question/14465605
#SPJ1
Do not trust people that share links these people hack and put viruses. to your phone/computer’s
DO NOT TRUST THESE LINKS
Answer:
94,200 milligrams in 94.2 grams
To get answer multiply 94.2 by 1000.
688x
Explanation- Your welcome
Answer:
2.11 ×10²⁴ atoms
Explanation:
Given data:
Number of moles of NaCl = 3.5 mol
Number of atoms = ?
Solution:
It is known that 1 mole contain 6.022×10²³ atoms. The number 6.022×10²³ is called Avogadro number.
3.5 mol ×6.022×10²³ atoms / 1 mol
21.1 ×10²³ atoms
2.11 ×10²⁴ atoms