Answer:
There will be produced 1.71 moles of B which contain 1.03×10²⁴ molecules
Explanation:
The example reaction is:
2A → 3B
2 moles of A produce 3 moles of B
If we have the mass of A, we convert it to moles and then, we make the rule of three: 29.2 g / 25.6g/mol = 1.14 moles
Therefore 2 moles of A produce 3 moles of B
1.14 moles of A will produce (1.14 . 3) / 2 = 1.71 moles of B are produced
Now we can determine, the number of molecules
1 mol has NA molecules (6.02×10²³)
1.71 moles have (1.71 . NA) = 1.03×10²⁴ molecules
Answer:
Elements, in turn, are pure substances—such as nickel, hydrogen, and helium—that make up all kinds of matter.
Hope This Helps! Have A Nice Day!!
Answer:
<h3>The answer is option B</h3>
Explanation:
To calculate the number of atoms we must first calculate the number of moles
Molar mass = mass / number of moles
number of moles = mass / Molar mass
Molar mass (K) = 39.10mole
mass = 2.10g
number of moles = 2.10/ 39.10
= 0.0537mol
After that we use the formula
N = n × L
where n is the number of moles
N is the number of entities
L is the Avogadro's constant which is
6.02 × 10^23 entities
Number of K atoms is
N = 0.0537 × 6.02 × 10^13
<h3>N = 3.23×10^22 atoms of K</h3>
Hope this helps you.
<span>Answer:
A 0.04403 g sample of gas occupies 10.0-mL at 289.0 K and 1.10 atm. Upon further analysis, the compound is found to be 25.305% C and 74.695% Cl. What is the molecular formula of the compound?
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Seems like I did a problem very similar to this--this must be the "B" test. But the halogen was different.
25.305% C/12 = 2.108
74.695% Cl/35.5 = 2.104
So the empirical formula would be CH. However, there are many compounds which fit this bill, so we have to use the gas data. (And I made, in the previous problem, the simplifying assumption that 289C and 1.10 atm would offset each other, so I'll do that, too.)
0.044 grams/10 ml = x/22.4 liters
0.044g/0.010 liters = x/22.4 liters
22.4 liters/0.010 liters = 2240 (ratio)
2240 x .044 = 98.56 (actual atomic weight)
CCl = 35.5+12 or 47.5, so two of those is 95 grams/mole.
This is sufficiient to distinguish C2CL2, (dichloroacetylene)
from C6CL6 (hexachlorobenzene) which would
mass 3 times as much.</span>
