Answer : The correct option is, 13.7 mole
Solution : Given,
Moles of 
= 27.4 moles
The given balanced chemical reaction is,
From the balanced chemical reaction, we conclude that
As, 2 moles of react with 1 moles of 
So, 27.4 moles of react with 
moles of
Therefore, the number of moles of oxygen required are, 13.7 moles
Answer:
18.84 g of silver.
Explanation:
We'll begin by calculating the number atoms present in 5.59 g of sulphur. This can be obtained as follow:
From Avogadro's hypothesis,
1 mole of sulphur contains 6.02×10²³ atoms.
1 mole of sulphur = 32 g
Thus,
32 g of sulphur contains 6.02×10²³ atoms.
Therefore, 5.59 g of sulphur will contain = (5.59 × 6.02×10²³) / 32 = 1.05×10²³ atoms.
From the calculations made above, 5.59 g of sulphur contains 1.05×10²³ atoms.
Finally, we shall determine the mass of silver that contains 1.05×10²³ atoms.
This is illustrated below:
1 mole of silver = 6.02×10²³ atoms.
1 mole of silver = 108 g
108 g of silver contains 6.02×10²³ atoms.
Therefore, Xg of silver will contain 1.05×10²³ atoms i.e
Xg of silver = (108 × 1.05×10²³)/6.02×10²³
Xg of silver = 18.84 g
Thus, 18.84 g of silver contains the same number of atoms (i.e 1.05×10²³ atoms) as 5.59 g of sulfur
Answer:
C.
Explanation:
Scientific notation is a way of writing small and large numbers. It is a way of expressing a standard form into a scientific form.
NaCl consists of a salt, which is a nonmetal (Na) and Cl, which is a metal. When a nonmetal and a metal are combined, it makes the compound ionic.
0.66 M is the accurate molarity of the new solution of volume of 1200 ml.
Explanation:
Data given:
molarity of copper(II) sulphate, Mconc.= 2M
volume of 2M solution taken Vconc. = 400 ml
volume taken for dilution, Vdilute = 1200 ml
molarity of the diluted solution, Mdilute =?
We will use the formula for dilution as
Mconc Vconc = Mdilute x V dilute (conc is concentrated)
putting the values in the equation:
2 x 400 = Mdilute x 1200
Mdilute = 
Mdilute = 0.66 M
When the solution is diluted to the volume of 1200 ml its molarity changes to 0.66 M.
