Moles He = 7.83 x 10^24 / 6.02 x 10^23 =13.0
<span>mass He = 13.0 mol x 4.00 g/mol = 52.0 g</span>
Explanation:
Atomic radius decreases from left to right in a period.
Therefore Calcium would have a smaller atomic size.
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
Answer:
THE VOLUME OF THE NITROGEN GAS AT 2.5 MOLES , 1.75 ATM AND 475 K IS 55.64 L
Explanation:
Using the ideal gas equation
PV = nRT
P = 1.75 atm
n = 2.5 moles
T = 475 K
R = 0.082 L atm/mol K
V = unknown
Substituting the variables into the equation we have:
V = nRT / P
V = 2.5 * 0.082 * 475 / 1.75
V = 97.375 / 1.75
V = 55.64 L
The volume of the 2.5 moles of nitrogen gas exerted by 1.75 atm at 475 K is 55.64 L
