Answer:
1.8x10¹⁷ molecules of CO are in each breath we take
Explanation:
Parts per million, ppm, is an unit of concentration in chemistry used for very diluted solutions.
<em>A 9ppm of X in a solution means in 1 million of molecules (1x10⁶) you have only 9 molecules of X.</em>
In a breath we have 2x10²² molecules and 9 ppm are CO. Thus, CO molecules in each breath are:
2x10²² molecules × (9 molecules CO / 1x10⁶ molecules) =
<h3>1.8x10¹⁷ molecules of CO are in each breath we take</h3>
Explanation:
As molarity is the number of moles placed in a liter of solution. Therefore, no. of mole = Molarity × volume of solution in liter
Hence, moles of 
will be calculated as follows.
No. of mole of = 
= 0.0168 mole
According to the given reaction, 2 mole of react with 4 mole of KI.
Therefore, for 0.0168 mole amount of required will be as follows.
= 
= 0.0337 mole of KI
Hence, volume of KI required will be calculated as follows.
Volume = 
Volume of KI = 
= 0.1614 liter
= 161.4 ml (as 1 L = 1000 mL)
Thus, we can conclude that 161.4 ml volume of a 0.2089 M KI is required for the given situation.
This question is quite vague, as the initial concentration of ethanol is not provided. However, from experience I can tell you that most laboratory work is done with 98% ethanol, and not absolute ethanol (100%). So in order to calculate the final concentration, we need to take the given values, which includes the initial concentration (98%), the initial volume (50.0mL) and the final volume (100.0mL). We apply the following equation to calculate the final concentration:
C1V1 = C2V2
C1 = Initial concentration
C2 = Final concentration
V1 = Initial volume
V2 = Final volume
(98%)(50.0mL) = (C2)(100.0mL)
Therefore, the final concentration (C2) = 49%
Answer:
Explanation:
How best to demonstrate the equivalence of hydrogens in
H
2
C
=
C
H
2
?
1
H
NM
R spectroscopy would be the best way to show this. The 4 hydrogens give the one signal.
Explanation:
Answer:
attached below
Explanation:
write the chemical equations for the test and confirmation test
i.e.
i)Chemical equation for main copper test
ii)Chemical equation for copper confirmation test
iii)Chemical equation for main bismuth test
iv) Chemical equation for bismuth confirmation test
attached below are the chemical equations for the test and confirmation test
