There are two isotopes of uranium abundant in nature
U235 and U238
As given that the sample has average molar mass of 237.482 amu
Let the amount of U235 in 100g sample = x
the amount of U238 in 100 g sample = 100-x
the average molar mass = [235(x) + 238 (100-x)] / 100
237.482 = [235(x) + 238 (100-x)] / 100
237.482 X 100 = [235(x) + 238 (100-x)]
23748.2 = 235 x + 23800 - 238x
51.8 = 3x
x = 17.27 g
So percentage of U-235 = 17.27 %
Answer:
um you dont need those 2 elements at the end
Answer:
The products are Al(NO₃)₃(aq) and H₂(g).
Al(s) + 3 HNO₃(aq) ⇒ Al(NO₃)₃(aq) + 1.5 H₂(g)
Explanation:
Let's consider the reaction between aluminum and nitric acid. This is a single replacement reaction, in which Al replaces H in HNO₃ to form aluminum nitrate. The unbalanced reaction is:
Al(s) + HNO₃(aq) ⇒ Al(NO₃)₃(aq) + H₂(g)
We start balancing N atoms by multiplying HNO₃ by 3.
Al(s) + 3 HNO₃(aq) ⇒ Al(NO₃)₃(aq) + H₂(g)
Finally, we get the balanced equation multiplying H₂ by 1.5.
Al(s) + 3 HNO₃(aq) ⇒ Al(NO₃)₃(aq) + 1.5 H₂(g)
This is a double replacement reaction; the ions switch twice.
