Answer:
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From the Graham's law of effusion;
R1/R2 = √MM2/√MM1
Molar mass of chlorine gas is 71
Therefore;
1.87= √ 71 /√mm1
= 1.87² = 71/mm1
mm1 = 71/1.87²
= 71/3.4969
= 20.3
Thus, the molar mass of the other gas is 20.3 , and i think the gas is neon
Answer:
2.04 x 10²⁴ molecules
Explanation:
Given parameters:
Mass of Be(OH)₂ = 145.5g
To calculate the number of molecules in this mass of Be(OH)₂ we follow the following steps:
>> Calculate the number of moles first using the formula below:
Number of moles = mass/molarmass
Since we have been given the mass, let us derive the molar mass of Be(OH)₂
Atomic mass of Be = 9g
O = 16g
H = 1g
Molar Mass = 9 + 2(16 + 1)
= 9 + 34
= 43g/mol
Number of moles = 145.5/43 = 3.38mol
>>> We know that a mole is the amount of substance that contains Avogadro’s number of particles. The particles can be atoms, molecules, particles etc. Therefore we use the expression below to determine the number of molecules in 3.38mol of Be(OH)₂:
Number of
molecules= number of moles x 6.02 x 10²³
Number of molecules= 3.38 x 6.02 x 10²³
= 20.37 x 10²³ molecules
= 2.04 x 10²⁴ molecules
Answer:
1. bond in the molecule on the right
Explanation:
CH3CH2-OH
The compound above is an alcohol due to the presence of the OH bond. The wave number of the C - O bond is given as; 1050-1150 cm^-1.
CH3CH__O
The compound above is an aldehyde due to the presence of the CHO bond. The wave number of the C = O bond is given as; 1740-1720 cm^-1
Comparing both bonds, the C = O bond absorbs at a higher wave number.
- Dalton reported that seven pounds of Oxygen reacted with one pound of hydrogen to form water, but accurate modern experiment gives eight pounds to one.
- Lavoisier reported that water contains Oxygen 5.6 times more than hydrogen by weight.
- Dalton assumed that water contains 1 atom of oxygen and 1 atom of hydrogen and concluded that the relative weight of O must be 5.6 times as large as the H atom.
- By determining the relative mass we will be able to determine for example that one element has twice the mass of a second element.
