The heat that is needed to raise the temperature of 78.4 g of aluminium from 19.4 °c to 98.6°c is 5600.77 j
<u><em>calculation</em></u>
Heat(Q) = mass(M) x specific heat capacity (C) x change in temperature(ΔT)
where;
Q=?
M = 78. 4 g
C=0.902 j/g/c
ΔT=98.6°c -19.4°c =79.2°c
Q is therefore = 78.4 g x 0.902 j/g/c x 79.2°c =5600.77 j
Answer:
0.095 moles of Calcium is there in 5.74 x 1022 atoms of calcium.
Explanation:
- As we know, 6.023*10^23 atoms of an element is equal to its atomic weight.
And, 6.023*10^23 atoms of an element is also equal to 1 mole of the element.
We have,
- 6.023*10^23 atoms of element calcium equals to 1 mole of Calcium
- 5.74*10^22 atoms of element calcium equals to
(1/(6.023*10^.23)) * 5.74*10^22 moles of calcium
Therefore,
- 5.74 x 1022 atoms of calcium= 0.095 moles of calcium.
The most suitable answer is C becuase they would gain two elctrons to atain that stable OCTET thus becoming a anion with a charge of -2 and by virtue oxidation states of -2. There is however an exception with oxygen in two cases. But I still remain that the best answer would be C
