Answer:
a) 128 grams
b) 81.25%
Explanation:
a) The mixture will be between the alloy and the pure copper, so the mass will be the sum of the mass of the compounds:
m = 50 + 78 = 128 g
b) In the alloy, 52% is copper, so to know the mass of copper in the alloy, we must multiply the total mass by 52% (which is 0.52)
0.52x50 = 26 g of copper
The total mass of copper in the mixture is 26 + 78 = 104 g
So, the percentual of copper is the mass of copper divided by the total mass
104/128 = 0.8125
Multiplying by 100%, to have the result in percentage
0.8125x100% = 81.25%
A) Nitrogen has an ATOMIC mass number of 14, but nitrogen gas consists of N₂ molecules, so the mass to use in this problem is 28 g/mol. Rates of effusion ∝ 1/√(mass), so
<span>√(mass unknown) /√28 = (rate N₂ effusion)/(rate unknown effusion) = 1.59 </span>
<span>∴ mass unknown = (1.59)²(28) = 70.78 g/mol </span>
<span>B) One possible gas that comes close for this mass is NF₃.</span>
Explanation:
There are ten d electrons in the outermost d subshell for Zn2+.
This has only only three sig figs because it is 546.000 the zeros are just place holders so thus meaning they are not significant. It is just another way to write 546 because in that number there is also 3 sig figs. They are just trying to throw you off. Just watch out because if the example is 540 then it is just two significant digits, and it were to be 540. then it be three cause that "." makes the zero significant. <span />