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Lerok [7]
3 years ago
14

Round off each measurement to the number of significant figures shown in parentheses.

Chemistry
1 answer:
Wewaii [24]3 years ago
6 0

We have to round off each measurement to the number of significant figures.

The correct answers are: 265.9, 0.00045 and 65 respectively.

265.991 when expressed in 4 significant figure it rounds off as 265.9 because  265.9 represents the value 265.991.

0.0004513 when expressed in 2 significant figure it rounds off as 0.00045 because zero before and after decimal are not significant.

6593 when expressed in 2 significant figure it rounds off as 65.


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If volumes are additive and 253 mL of 0.19 M potassium bromide is mixed with 441 mL of a potassium dichromate solution to give a
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Answer:

The concentration of the Potassium Dichromate solution is 0.611 M

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First of all, we need to understand that in the final solution we'll have potassium ions coming from KBr and also K2Cr2O7, so we state the dissociation equations of both compounds:

KBr (aq) → K+ (aq) + Br- (aq)

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According to these balanced equations when 1 mole of KBr dissociates, it generates 1 mole of potassium ions. Following the same thought, when 1 mole of K2Cr2O7 dissociates, we obtain 2 moles of potassium ions instead.

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1000 mL solution ----- 0.19 moles of KBr

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Now, we are told that the final concentration of K+ is 0.846 M. This means we have 0.846 moles of K+ in 1000 mL solution. Considering that volumes are additive, we calculate the amount of K+ moles we have in the final volume solution (441 mL + 253 mL = 694 mL):

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0.587124 final K+ moles - 0.04807 K+ moles from KBr = 0.539054 K+ moles from K2Cr2O7

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2 K+ moles ----- 1 K2Cr2O7 mole

0.539054 K+ moles ---- x = 0.269527 K2Cr2O7 moles

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441 mL ----- 0.269527 K2Cr2O7 moles

1000 mL ----- x = 0.6112 K2Cr2O7 moles = 0.6112 M

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