Just do 6 x 2.2
6 x 2.2 = 13.2
Answer : The average atomic mass of an element X is, 63.546 amu
Solution : Given,
Mass of isotope X-63 = 62.9296 amu
% abundance of isotope X-63 = 69.15% = 0.6915
Mass of isotope X-64 = 64.9278 amu
% abundance of isotope X-64 = 30.85% = 0.3085
Formula used for average atomic mass of an element X :
![\text{ Average atomic mass of an element X}=\sum[(62.9296\times0.6915)+(64.9278\times 0.3085)]](https://tex.z-dn.net/?f=%5Ctext%7B%20Average%20atomic%20mass%20of%20an%20element%20X%7D%3D%5Csum%5B%2862.9296%5Ctimes0.6915%29%2B%2864.9278%5Ctimes%200.3085%29%5D)
Therefore, the average atomic mass of an element X is, 63.546 amu
Looks like candy lol .....
Answer:
The chemical term in the equation for the precipitate of AgCl(s) is n=3.54*10^-3
Explanation:
the quantity of AgCl(s) in moles is:
n = 0.508g / 143.32 g/mol = 3.54*10^-3 mol
to verify it the mass of AgNO3 involved in the reaction should be
n AgNO3 required = n = 3.54*10^-3 mol
the mass of n involved should be higher than n AgNO3
n existing = V*N = 0.523 mol/L * 35*10^-3 L = 18.305*10^-3 mol
Answer:
0.581 L or 581 mL
Explanation:
As stated in the question, the combined gas law is (P1*V1/T1) = (P2*V2/T2)
Write down the amounts you are given.
V1 = 0.152 L (I was taught to always convert milliliters to liters)
P1 = 717 mmHg
T1 = 315 K
V2 = ?
P2 = 463 mmHg
T2 = 777 K
The variable that is being solved for is final volume. Fill in the combined gas law equation with the corresponding amounts and solve for V2.
(717 mmHg*0.152 L) / (315 K) = (463 mmHg*V2) / (777 K)
0.346 = (463*V2) / (777)
0.346*777 = (463*V2) / (777)*777
268.842 = 463*V2
268.842/463 = (463*V2)/463
V2 = 0.581
Pressure and volume are indirectly proportional. This checks out because the volume increased while pressure decreased. Volume and temperature are directly proportional. This checks out because both volume and temperature increased. This is a good way to check your answers. You can also solve each side of the combined gas law equation to see if they are both the same.
