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Darina [25.2K]
2 years ago
15

A student measures the volume of a solution to be 0.01370 l. How many significant digits are in this measurement?

Chemistry
1 answer:
Mkey [24]2 years ago
7 0

A student measures the volume of a solution to be 0.01370 have 5 significant digits in this measurement.

<h3>What are significant digits?</h3>

The significant digits are the minimum number from zero to nine for reporting any measurement where the digits are uncertain.

The significant digits starting from zero are not significant digits, decimal is not a significant digit, and ending zero after the decimal are significant digits.

Therefore, the student measures the volume of a solution to be 0.01370 5 significant digits are in this measurement.

Learn more about significant digits, here:

brainly.com/question/1658998

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Which is the largest?<br> O 43.32 mL<br> 0 0,0056 L<br> O 0.000076 KL
mel-nik [20]

Answer:

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3 0
3 years ago
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Suppose the half-life is 9.0 s for a first order reaction and the reactant concentration is 0.0741 M 50.7 s after the reaction s
bazaltina [42]

<u>Answer:</u> The time taken by the reaction is 84.5 seconds

<u>Explanation:</u>

The equation used to calculate half life for first order kinetics:

k=\frac{0.693}{t_{1/2}}

where,

t_{1/2} = half-life of the reaction = 9.0 s

k = rate constant = ?

Putting values in above equation, we get:

k=\frac{0.693}{9}=0.077s^{-1}

Rate law expression for first order kinetics is given by the equation:

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}     ......(1)

where,

k = rate constant  = 0.077s^{-1}

t = time taken for decay process = 50.7 sec

[A_o] = initial amount of the reactant = ?

[A] = amount left after decay process =  0.0741 M

Putting values in equation 1, we get:

0.077=\frac{2.303}{50.7}\log\frac{[A_o]}{0.0741}

[A_o]=3.67M

Now, calculating the time taken by using equation 1:

[A]=0.0055M

k=0.077s^{-1}

[A_o]=3.67M

Putting values in equation 1, we get:

0.077=\frac{2.303}{t}\log\frac{3.67}{0.0055}\\\\t=84.5s

Hence, the time taken by the reaction is 84.5 seconds

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3 years ago
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3 years ago
How can you determine which bond in a structure is more polar without using an electronegativity table?
UkoKoshka [18]
To know this you pretty much do have to kind of memorize a few electronegativities. I don't recall ever getting a table of electronegativities on an exam.
From the structure, you have:

I remember the following electronegativities most because they are fairly patterned:
EN
H
=
2.1
EN
C
=
2.5
EN
N
=
3.0
EN
O
=
3.5
EN
F
=
4.0
EN
Cl
=
3.5
Notice how carbon through fluorine go in increments of
~
0.5
. I believe Pauling made it that way when he determined electronegativities in the '30s.
Δ
EN
C
−
Cl
=
1.0
Δ
EN
C
−
H
=
0.4
Δ
EN
C
−
C
=
0.0
Δ
EN
C
−
O
=
1.0
Δ
EN
O
−
H
=
1.4
So naturally, with the greatest electronegativity difference of
4.0
−
2.5
=
1.5
, the
C
−
F
bond is most polar, i.e. that bond's electron distribution is the most drawn towards the more electronegative compound as compared to the rest.
When the electron distribution is polarized and drawn towards a more electronegative atom, the less electronegative atom has to move inwards because its nucleus was previously favorably attracted to the electrons from the other atom.
That means generally, the greater the electronegativity difference between two atoms is, the shorter you can expect the bond to be, insofar as the electronegative atom is the same size as another comparable electronegative atom.
However, examining actual data, we would see that on average, in conditions without other bond polarizations occuring:
r
C
−
Cl
≈
177 pm
r
C
−
C
≈
154 pm
r
C
−
O
≈
143 pm
r
C
−
F
≈
135 pm
r
C
−
H
≈
109 pm
r
O
−
H
≈
96 pm
So it is not necessarily the least electronegativity difference that gives the longest bond.
Therefore, you cannot simply consider electronegativity. Examining the radii of the atoms, you should notice that chlorine is the biggest atom in the compound.
r
Cl
≈
79 pm
r
C
≈
70 pm
r
H
≈
53 pm
r
O
≈
60 pm
So assuming the answer is truly
C
−
C
, what would have to hold true is that:
The
C
−
F
bond polarization makes the carbon more electropositive (which is true).
The now more electropositive carbon wishes to attract bonding pairs from chlorine closer, thereby shortening the
C
−
Cl
bond, and potentially the
C
−
H
bond (which is probably true).
The shortening of the
C
−
Cl
bond is somehow enough to be shorter than the
C
−
C
bond (this is debatable).
5 0
3 years ago
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