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>
1) Ca-37, with a half-life of 181.1(10) ms.
Answer:

Explanation:
Given that:

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

Multiplying (2) with equation (4) ; we have:

From equation (1) ; multiplying (-1) with equation (1); we have:

From equation (2); multiplying (3) with equation (2); we have:

Now; Adding up equation (5), (6) & (7) ; we get:



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(According to Hess Law)


Thw answer is PHj78 JJ CP30 R2D2
1 mole of carbon dioxide contains a mass of 44 g, out of which 12 g are carbon.
Hence, in this case the mass of carbon in 8.46 g of CO2:
(12/44) × 8.46 = 2.3073 g
1 mole of water contains 18 g, out of which 2 g is hydrogen;
Therefore, 2.6 g of water contains;
(2/18) × 2.6 = 0.2889 g of hydrogen.
Therefore, with the amount of carbon and hydrogen from the hydrocarbon we can calculate the empirical formula.
We first calculate the number of moles of each,
Carbon = 2.3073/12 = 0.1923 moles
Hydrogen = 0.2889/1 = 0.2889 moles
Then, we calculate the ratio of Carbon to hydrogen by dividing with the smallest number value;
Carbon : Hydrogen
0.1923/0.1923 : 0.2889/0.1923
1 : 1.5
(1 : 1.5) 2
= 2 : 3
Hence, the empirical formula of the hydrocarbon is C2H3