Answer:
Moles of water in 7.1×10²⁵ molecules are 118 mol.
Explanation:
Given data:
Number of molecules of water = 7.1×10²⁵ molecules
Moles of water in 7.1×10²⁵ molecules = ?
Solution;
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
1 mole = 6.022 × 10²³ molecules of water
7.1×10²⁵ molecules of water × 1 mol / 6.022 × 10²³ molecules of water
1.18×10² moles of water 0r 118 moles of water
Percentage Yield = (Actual Yield ÷ Theoretical Yield) × 100
The Actual Yield is given in the question as 21.2 g of NaCl. However, in order to find the theoretical yield, you have to write a balanced equation and use the mole ratio to calculate the mass of NaCl that would be produced.
Balanced Equation: CuCl + NaNO₃ → NaCl + CuNO₃
Moles of CuCl = Mass of CuCl ÷ Molar Mass of CuCl
= 31.0 g ÷ (63.5 + 35.5)g/mol
= 0.31 mol
the mole ratio of CuCl to NaCl is 1 : 1,
∴ if moles of CuCl = 0.31 mol,
then moles of NaCl = 0.31 mol
Now, Mass of NaCl = Moles of NaCl × Molar Mass of NaCl
= 0.31 mol × (23 + 35.5) g/mol
= 18.32 g
⇒ the THEORETICAL Yield of NaCl, in this case, is 18.32 g.
Now, since Percentage Yield = (Actual Yield ÷ Theoretical Yield) × 100
⇒ Percentage Yield of NaCl = (21.2g ÷ 18.32g) × 100
= 115.7 %
NOTE: Typically, the percentage yield of a reaction is less than 100%, however in a case where the mass of the substance is weighed with impurities, then that mass may be in excess of 100% as seen here.
Molar mass:
O2 = 31.99 g/mol
C8H18 = 144.22 g/mol
<span>2 C8H18(g) + 25 O2(g) = 16 CO2(g) + 18 H2O(g)
2 x 144.22 g --------------- 25 x 31.99 g
10.0 g ----------------------?? ( mass of O2)
10.0 x 25 x 31.99 / 2 x 144.22 =
7997.5 / 288.44 => 27.72 g of O2
hope this helps!
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Answer:
A. Chipping ice to flakes
Explanation:
You can't reverse the ice becoming flakes. They will stay like that until they melt
It is E. because the predators will have slightly more meals with their prey slightly increasing population.
