The number of moles for co2=mass(g)/molar mass
n=.22/44=.005 mole of CO2
from the equation we see the relationship between the moles of co2 and O2 and we find that they have the same number of moles
So we need .005mole of O2
Multiple the number of moles with avogadro’s number to know the number of O2molecules
.005x6.022 x10^23
Question 9. The first one is the smallest. Anything with a negative exponent is going to be less than 1, the .00000241. The exponent tells you the number of zeroes to the right of the decimal point. Farther to right gets smaller and smaller.
Question 10. The last one is true. If the last digit is smaller than 5, drop the digit, and do not change. (If it is a 5 or larger, the digit before it would round up)
Answer:
The mass of NaCl is 0.029 grams
Explanation:
Step 1: Data given
Molecular weight of NaCl = 58.44 g/mol
Volume of solution = 100 mL = 0.100 L
Molarity = 0.0050 M
Step 2: Calculate moles NaCl
Moles NaCl = molarity * volume
Moles NaCl = 0.0050 M * 0.100 L
Moles NaCl = 0.00050 moles
Step 3: Calculate mass NaCl
Mass NaCl = moles NaCl * molar mass NaCl
Mass NaCl = 0.00050 moles * 58.44 g/mol
Mass NaCl = 0.029 grams
The mass of NaCl is 0.029 grams
Answer:
a) After the balloon inflated after 440 uL of dropwise due to the reaction of 1-Decene and the solution in the conical vial. b) 
⇒ 16
c) No 
was not the limiting reactant.
Explanation:
Generally, hydrogenation is the chemical reaction between a compound or element and molecular hydrogen in the presence of catalysts such as platinum.
a) After the balloon inflated after 440 uL of dropwise 1-Decene solution was added due to the reaction between 1-Decene and the solution in the conical vial.
b) ⇒ 16
c) was not the limiting reactant based on the mol to mol ratio of and decane which is 1:1. Therefore, if 0.8 mol of decane was produced then 0.8 mol of would also be produced.
Answer
2.1 x 10 ^23
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