94.6 g. You must use 94.6 g of 92.5 % H_2SO_4 to make 250 g of 35.0 % H_2SO_4.
We can use a version of the <em>dilution formula</em>
<em>m</em>_1<em>C</em>_1 = <em>m</em>_2<em>C</em>_2
where
<em>m</em> represents the mass and
<em>C</em> represents the percent concentrations
We can rearrange the formula to get
<em>m</em>_2= <em>m</em>_1 × (<em>C</em>_1/<em>C</em>_2)
<em>m</em>_1 = 250 g; <em>C</em>_1 = 35.0 %
<em>m</em>_2 = ?; _____<em>C</em>_2 = 92.5 %
∴ <em>m</em>_2 = 250 g × (35.0 %/92.5 %) = 94.6 g
The Earth rotates 365 times during each complete revolution.
HOPE THIS HELPS! ^_^
Answer:
54.5%
Explanation:
The percentage composition of oxygen in C₆H₈O₆ can be obtained as follow:
Molar mass of C₆H₈O₆ = (12×6) + (8×1) + (16×6)
= 72 + 8 + 96
= 176 g/mol
Next, there are 6 oxygen atoms in C₆H₈O₆. Therefore the mass of oxygen in C₆H₈O₆ is:
Mass of Oxygen = 16 × 6 = 96 g
Finally, we shall determine the percentage composition of oxygen in C₆H₈O₆ as follow:
Percentage of oxygen =
Mass of Oxygen/mass of C₆H₈O₆ × 100
Percentage of oxygen = 96 / 176 × 100
Percentage of oxygen = 54.5%
Thus, the percentage composition of oxygen in C₆H₈O₆ is 54.5%.
Molar mass CO₂ = 44.0 g/mol
44.0 g ----------------- 6.02x10²³ molecules
74.5 g ----------------- ??
74.5 x ( 6.02x10²³) / 44.0
= 1.019x10²⁴ molecules of CO2
Calculate the number <span>of atoms :
CO</span>₂<span> => 1 atom of Carbon , 2 atoms of Oxygen ( 1 + 2 = 3 atoms )
therefore:
</span>
1 molecule CO2 ---------------------- 3 atoms
1.019x10²⁴ CO2 --------------------- ?? atoms
3 x ( 1.019x10²⁴) / 1 =
= 3.057x10²⁴ atoms of CO₂
hope this helps!