45.6 mL of 0.0600 M of is added to 25.0 mL of HCl of unknown concentration then , the concentration of HCl is 0.032 M .
Calculation ,
Formula used : = ...( 1 )
Where = concentration of = 0.0600 M ( given )
= concentration of HCl = ? ( to be find )
= volume of = 25.0 mL ( given )
= volume of HCl = 45.6 mL ( given )
Now , put the value of all volume and concentration in equation ( 1 ) we get .
0.0600 M × 25.0 mL = × 45.6 mL
= 0.0600 M × 25.0 mL / 45.6 mL = 0.032 M
Therefore , concentration of unknown concentration of HCl is 0.032 M .
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Answer:
d. 3 signals: a singlet, a doublet, and a septet
Explanation:
In this case, we can start with the structure of
. When we draw the molecule we will obtain <u>2-methoxypropane</u> (see figure 1).
In 2-methoxypropane we will have three signals. The signal for the groups in the left, the and the in the right. Lets analyse each one:
-) in the right
In this carbon, we dont have any hydrogen as neighbors. Therfore we will have <u>singlet</u> signal in this carbon.
-)
In this case, we have 6 hydrogen neighbors ( the two methyl groups in the left). So, if we follow the <u>n + 1 rule</u> (where n is the amount of hydrogen neighbors):
For this carbon we will have a <u>septet</u>.
-) in the left
In this case we have only 1 hydrogen neighbor (the hydrogen in
). So, if we use the n+1 rule we will have:
We will have a doublet
With all this in mind the answer would be:
<u>d. 3 signals: a singlet, a doublet, and a septet
</u>
<u />
See figure 2 to further explanations
10 hectometers
One hectometer is 100 meters in length already, so 10 is 1000
Answer:
Explanation:
Hello,
In this case, the undergoing chemical reaction is shown below:
Thus, we obtain 0.0471 g of lead (II) sulfate, the iodide ions in the original solution are computed below by stoichiometry, taking into account that into 1 mole of lead (II) iodide there are 2 moles of iodide ions:
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We are given the chemical reaction and the amount of fuel used for the process. We use these data together with the molar masses to obtain what is asked. We do as as follows:
100 g C8H18 ( 1 / 114.33) ( 16 / 2 ) = 6.9973 mol CO2 produced