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
Answer:
The molar mass of 
is 96.8 g/mol
Explanation:
The given molecular formula -
Individual molar masses of each element in the compound is as follows.
Molar mass of nitrogen - 14.01 g/mol
Molar mass of of hydrogen = 1.008g/mol
Molar mass of carbon = 12.01 g/mol
Molar mass of oxygen =16.00 g/mol
Molar mass of is
![2\times[1(14.01)+4(1.008)]+1(12.01)+3(16.00)= 96.8g/mol](https://tex.z-dn.net/?f=2%5Ctimes%5B1%2814.01%29%2B4%281.008%29%5D%2B1%2812.01%29%2B3%2816.00%29%3D%2096.8g%2Fmol)
Therefore,The molar mass of is 96.8 g/mol
Answer:
3.052 × 10^24 particles
Explanation:
To get the number of particles (nA) in a substance, we multiply the number of moles of the substance by Avogadro's number (6.02 × 10^23)
The mass of Li2O given in this question is as follows: 151grams.
To convert this mass value to moles, we use;
moles = mass/molar mass
Molar mass of Li2O = 6.9(2) + 16
= 13.8 + 16
= 29.8g/mol
Mole = 151/29.8g
mole = 5.07moles
number of particles (nA) of Li2O = 5.07 × 6.02 × 10^23
= 30.52 × 10^23
= 3.052 × 10^24 particles.
Answer:
it is always necessary to use the roman numeral as the assigned charge of the metal.
Explanation:
This is so that one would know which Transition metal is being used. For example copper (II) would be Cu²+
Genetic engineering also called transformation
