The correct option is (C) 6.02 X 10²³
A sample of CH₄O with a mass of 32.0 g contains <u>6.02 X 10²³</u> molecules of CH₄O.
To calculate the number of moles;
Molar mass of CH₄O = C + 4(H) + O
= 12.01 + 4(1.008) + 16
= 32.04 g/mol
So, 1 mol of CH₄O = 32.04 g of CH₄O
Given, 32.0 g of CH₄O
According to Avagadro's constant 1 mole of a substance contains 6.022× 10^23 particles (molecules, atoms or ions).
= (32.0 g/1)(1 mol CH₄O/32.04 g CH₄O)(6.02x10²³/1 mol CH₄O)
= 6.02 X 10²³ molecules of CH₄O
Hence, a sample of will contain number of molecules 6.02 X 10²³ molecules.
Learn more about the Moles calculation with the help of the given link:
brainly.com/question/21085277
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Answer:
mass of Cu = 26,34615g
Explanation:
The equation for reaction is
Cu + 2 AgNO3 ---> Cu(NO3)2 + 2 Ag
1mol 2mol 1mol 2mol
mass of Ag that reacted = 89.5g
molar mass of Ag =107.86h/mol
number of moles = mass/molar mass
number of moles of Ag = 89.5/107.86
=0.8298moles
from the stoichoimetric relation;
1mol of Cu reacted to produce 2moles of Ag;
hence 0.8298moles of Ag produce would require; 0.8298/2 moles of Cu to react
=0.4149moles of Cu
mass of Cu = moles of Cu * molar mass of Cu
molar mass(Cu) = 63.5g
mass of Cu = 0.4149 * 63.5
mass of Cu = 26,34615g
I pretty sure it true I’m not completely sure
Answer:
The concentration of the CaBr2 solution is 96 µmol/L
Explanation:
<u>Step 1:</u> Data given
Moles of Calciumbromide (CaBr2) = 4.81 µmol
Volume of the flask = 50.0 mL = 0.05 L
<u>Step 2:</u> Calculate the concentration of Calciumbromide
Concentration CaBr2 = moles CaBr2 / volume
Concentration CaBr2 = 4.81 µmol / 0.05 L
Concentration CaBr2 = 96.2 µmol /L = 96.2 µM
The concentration of the CaBr2 solution is 96 µmol/L
Answer:
Explanation:
Hello,
In this case, since we have grams of iron (III) oxide whose molar mass is 159.69 g/mol are able to compute the produced grams of iron by using its atomic mass that is 55.845 g/mol and their 2:4 molar ratio in the chemical reaction:
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