<span>0.032000000 cg
Rounding off ........>
0.0320 cg
The last zero means the measurement is accurate to that digit
</span>
As per the given chemical formula- Na2CO3.10H2O, one mole of the chemical compound contains 13 moles of oxygen atoms. Hence
Number of moles of oxygen atoms in one mole of Na2CO3.10H2O = 13
number of moles of oxygen atoms in 0.2 moles of Na2CO3.10H2O = 13 X 0.2 = 2.6
Now, one mole of a substance contains 6.022 X 10^23 particles of the substance. Thus
number of atoms of oxygen in one mole of oxygen atom = 6.022 X 10^23
number of moles of oxygen atoms in 2.6 moles of oxygen atoms = 2.6 X 6.022 X 10^23 = 15.657 X 10^23
= 1.566 X 10^24
Thus, there are 1.566 X 10^24 atoms of oxygen in 0.2 moles of Na2CO3.10H2O.
4.7
Answer:
202 L
Explanation:
Step 1: Write the balanced equation
C₆H₁₂O₆ + 6 O₂(g) ⇒ 6 CO₂(g) + 6 H₂O(l)
Step 2: Calculate the moles corresponding to 270 g of C₆H₁₂O₆
The molar mass of C₆H₁₂O₆ is 180.16 g/mol.
270 g × 1 mol/180.16 g = 1.50 mol
Step 3: Calculate the moles of CO₂ generated from 1.50 moles of glucose
The molar ratio of C₆H₁₂O₆ to CO₂ is 1:6. The moles of CO₂ formed are 6/1 × 1.50 mol = 9.00 mol
Step 4: Calculate the volume of 9.00 moles of CO₂ at STP
The volume of 1 mole of an ideal gas at STP is 22.4 L.
9.00 mol × 22.4 L/mol = 202 L
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