C3H8 + 5O2 ------> 3CO2 + 4H2O
from reaction 1 mol 5 mol
given 1.82 mol x mol
x=(1.82*5)/1 = 9.10 mol CO2
Answer:
Elements in the same period have same number of electronic shell and electron is increased by one in every coming element with in same electronic shell.
Explanation:
Consider the second period of periodic table. This period consist of eight elements lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine and neon.
Electronic configuration of lithium:
Li₃ = [He] 2s¹
Electronic configuration of beryllium:
Be₄ = [He] 2s²
Electronic configuration of boron:
B₅ = [He] 2s² 2p¹
Electronic configuration of carbon:
C₆ = [He] 2s² 2p²
Electronic configuration of nitrogen:
N₇ = [He] 2s² 2p³
Electronic configuration of oxygen:
O₈ = [He] 2s² 2p⁴
Electronic configuration of fluorine:
F₉ = [He] 2s² 2p⁵
Electronic configuration of neon:
Ne₁₀ = [He] 2s² 2p⁶
All these elements present in same period having same electronic shell and number of electron increased by 1.
Answer:
Mass = 381.28 g
Explanation:
Given data:
Number of moles of HNO₃ = 16 mol
Mass of Cu needed to react with 16 mol of HNO₃ = ?
Solution:
Chemical equation:
3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 4H₂O + 2NO
Now we will compare the moles of Cu with HNO₃ from balance chemical equation.
HNO₃ : Cu
8 : 3
16 : 3/8×16 = 6
Mass of Cu needed:
Mass = number of moles × molar mass
Mass = 6 mol × 63.546 g/mol
Mass = 381.28 g
Covalent bonds form when atoms share electrons. This sharing allows each atom to achieve its octet of electrons and greater stability. Methane, CH 4<span>, the simplest organic compound, contains covalent bonds. Carbon has four valence electrons, while hydrogen has one valence electron. By sharing these outer‐shell electrons, carbon and hydrogen complete their valence shells and become more stable. The duet of electrons on the hydrogen is isoelectronic with helium and forms a complete shell.</span>
<u>Answer:</u>
<em>20, 44, 62 </em>
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<u>Explanation:</u>
To find the number of atoms of each element, we multiply coefficient and subscript
For example 
contains
5 × 1 = 5 ,Ca atoms and
5 × 2 = 10, Cl atoms
If there is a bracket in the chemical formula
For example 
we multiply coefficient × subscript × number outside the bracket to find the number of atoms
(Please note: 3 is the coefficient, and if there is no number given then 1 will be the coefficient )
So
3 × 3 = 9 , Ca atoms
3 × 1 × 2 = 6, P atoms
3 × 4 × 2 = 24, O atoms are present.
So let us find the number of atoms of each element on the left side of the equation
Number of C atoms = 2 × 10 = 20
Number of H atoms = 2 × 22 = 44
Number of O atoms = 31 × 2 = 62
20, 44, 62 are the Answers.
