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fiasKO [112]
3 years ago
5

An aqueous KNO3 solution is made using 76.6 g of KNO3 diluted to a total solution volume of 1.84 L. (Assume a density of 1.05 g/

mL for the solution.) You may want to reference (Pages 552 - 557) Section 13.5 while completing this problem. Part A Calculate the molarity of the solution.
Chemistry
1 answer:
defon3 years ago
7 0

Answer:

The answer to your question is Molarity = 0.41

Explanation:

Data

mass of KNO₃ = 76.6 g

volume = 1.84 l

density = 1.05 g/ml

Process

1.- Calculate the molecular mass of KNO₃

molecular mass = 39 + 14 + (16 x 3) = 101 g

2.- Calculate the number of moles

                      101 g of KNO₃  --------------- 1 mol

                       76.6 g of KNO₃ ------------  x

                        x = (76.6 x 1) / 101

                        x = 0.76 moles

3.- Calculate molarity

Molarity = \frac{number of moles}{volume}

Substitution

Molarity = \frac{0.76}{1.84}

Result

Molarity = 0.41

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IgorLugansk [536]

Answer:

Yes, the given instance is an example of a trade-off.

Explanation:

  • The trade-off seems to be a strategy to minimize or surrender one or more favorable results in consideration for increasing either receiving other favorable results to optimize cumulative benefit or efficacy within such conditions.
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3 years ago
1. Three
Bess [88]

Answer:

Explination:

Given Data:

                  Trail 1          Trial 2         Trial 3

Student A 448.0 cm 485.6 cm 463.4 cm

Student B 450.5 cm 441.3 cm         446.8 cm

Student C 422.6 cm 445.2 cm 432.7 cm

Accepted Value = 435.0 cm

Required:

A: Accurate measurement =?

B: Reason for the answer =?

C: Precise measurement =?

D: Reason for the answer =?

Solution:

Student A:

Trail 1: 435.0 cm – 448.0 cm = (13.0 cm greater than accepted value)

Trail 2: 435.0 cm – 485.6 cm = (50.6 cm greater than accepted value)

Trial 3: 435.0 cm – 463.4 cm = (28.4 cm greater than accepted value)

Student B:

Trail 1: 435.0 cm – 450.5 cm = (14.5 cm greater than accepted value)

Trail 2: 435.0 cm – 441.3 cm = (6.3 cm greater than accepted value)

Trial 3: 435.0 cm – 446.8 cm = (11.8 cm greater than accepted value)

Student C:

Trial 1: 435.0 cm – 422.6 cm = (12.4 cm less than accepted value)

Trial 2: 435.0 cm – 445.2 cm = (10.2 cm greater than accepted value)

Trial 3: 435.0 cm – 432.7 cm = (1.3 cm less than accepted value)

A: The 3rd trial of the student C is accurate measurement = 432.7 cm.

B: The 3rd value of students’ C measurement is accurate because it is quite near to the accepted value, i.e.  435.0 cm.

As we know that Accuracy refers to the closeness of a measured value to a standard or known value.

This value has only the difference of 1.3 cm.

435.0 cm – 432.7 cm = (1.3 cm less than accepted value)

All the other have larger difference which is as above.

___________________

Student A:

1. 448.0 cm – 485.6 cm = (37.6 cm far)

2. 485.6 cm – 463.4cm = (22.2 cm far)

Student B:

1. 450.5 cm – 441.3 cm = (9.2 cm far)

2. 441.3 cm – 446.8 cm = (5.5 cm far)

Student C:

1. 422.6 cm – 445.2 cm = (22.6 cm far)

2. 445.2 cm – 432.7cm = (12.5 cm far)

So,

C: The values of Student B are more precise.

D: As we know that Precision refers to the closeness of two or more measurements to each other.

The measurements of student C are more close to each other. The values are only 9.2 cm and 5.5 cm far from each other.

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Explanation:

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B. Heating up the reaction will increase the entropy of a reaction.

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S = ΔH/T

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