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3 years ago

Which molecules would most likely cause a liquid to have the lowest viscosity?

Chemistry

1 answer:

IRINA_888 [86]3 years ago

8 0

That molecules would be small, no polar molecules

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Siapa penemu teori atom pertama?

Zielflug [23.3K]

Options please i think its John Dalton

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3 years ago

Suppose that some FeSCN2+ is added to the above solution to shift the equilibrium. When equilibrium is re-established, the follo

kiruha [24]

Explanation:

The given reaction equation will be as follows.

$[FeSCN^{2+}] \rightleftharpoons [Fe^{3+}] + [SCN^{-}]$

Let is assume that at equilibrium the concentrations of given species are as follows.

$[Fe^{3+}] = 8.17 \times 10^{-3}$ M

$[SCN^{-}] = 8.60 \times 10^{-3}$ M

$[FeSCN^{2+}] = 6.25 \times 10^{-2}$ M

Now, first calculate the value of $K_{eq}$ as follows.

$K_{eq} = \frac{[Fe^{3+}][SCN^{-}]}{[FeSCN^{2+}]}$

= $\frac{8.17 \times 10^{-3} \times 8.60 \times 10^{-3}}{6.25 \times 10^{-2}}$

= $11.24 \times 10^{-4}$

Now, according to the concentration values at the re-established equilibrium the value for $[FeSCN^{2+}]$ will be calculated as follows.

$K_{eq} = \frac{[Fe^{3+}][SCN^{-}]}{[FeSCN^{2+}]}$

$11.24 \times 10^{-4} = \frac{8.12 \times 10^{-3} \times 7.84 \times 10^{-3}}{[FeSCN^{2+}]}$

$[FeSCN^{2+}] = 5.66 \times 10^{-2}$ M

Thus, we can conclude that the concentration of $[FeSCN^{2+}]$ in the new equilibrium mixture is $5.66 \times 10^{-2}$ M.

7 0

3 years ago

Fill in the blank to complete each statement.

son4ous [18]

Answer:

5.physical change

6.chemical change

7.physical change

8.conservation of mass

9.thermal energy

10.physical change

I honeslty dont know if this is right

explanation:

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3 years ago

I will give brainliest to correct answer - no links

WINSTONCH [101]

Answer:

C - 23cm

Explanation:

6 0

3 years ago

Read 2 more answers

Suppose that you are performing a titration on a monoprotic acid. Titration of this monoprotic acid required a volume of 0.1L of

Alex73 [517]

Answer: A) More base is likely required to reach the endpoint for the diprotic acid than for the monoprotic acid under these conditions

Explanation:

The monoprotic acid (HA) has a valency of 1 and diprotic acid $(H_2A)$ has a valency of 2.

As the concentration and volume of the diprotic acid and the monoprotic acids are equal.

The neutralization reaction for monoprotic acid is:

$HA+BOH\rightarrow BA+H_2O$

The neutralization reaction for diprotic acid is:

$H_2A+2BOH\rightarrow B_2A+2H_2O$

Thus more number of moles of base are required for neutralization of diprotic acid and thus the volume required will be more as concentration and volume of the diprotic acid and the monoprotic acids are equal.

6 0

3 years ago