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LiRa [457]
3 years ago
14

What is the volume of an 2.3 solution with 212 grams of calcium chloride dissolved in it

Chemistry
1 answer:
Travka [436]3 years ago
5 0

830 mL. A 2.3 mol/L solution of CaCl2 has a volume of 830 mL

I am guessing that the concentration of your solution is 2.3 mol/L.

a) Moles of CaCl2

MM of CaCl2 = 110.98 g/mol

Moles of CaCl2 = 212 g CaCl2 x (1 mol CaCl2/110.98 g CaCl2)

= 1.910 mol CaCl2

b) Volume of solution

V = 1.910 mol CaCl2 x (1 L solution/2.3 mol CaCl2) = 0.83 L solution

= 830 mL solution


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Which of the following statements is correct concerning the reaction 2 A + B → 2 C + 2 D?
Eddi Din [679]

Answer:

2 is the correct answer maybe

5 0
3 years ago
Use the data set to answer the question.
andrezito [222]

Answer:

It is both accurate and precise.

Explanation:

Precision and accuracy are two different terms used to describe data or measurements. Accuracy refers to how close a set of measurements/experimental values is to an accepted or correct value while Precision refers to how close a series of experimental values are to one another.

In the given set of data in the question below, the Correct Value is 59.2 while the experimental values are as follows;

Trial 1: 58.7

Trial 2: 59.3

Trial 3: 60.0

Trial 4: 58.9

Trial 5: 59.2

Based on comparison, it can be observed that these experimental values are close to the correct value (59.2). Hence, they are said to be ACCURATE. Also, the experimental values are close to one another, hence, they are said to be PRECISE.

Therefore, the data set is both accurate and precise.

7 0
3 years ago
The two naturally occuring isotopes of antimony are 121Sb (57.21%) and 123Sb (42.79%), with isotopic masses of 120.904 and 122.9
emmasim [6.3K]

Answer:

The average atomic weight = 121.7598 amu

Explanation:

The average atomic weight of natural occurring antimony can be calculated as follows :

To calculate the average atomic mass the percentage abundance must be converted to decimal.

121 Sb has a percentage abundance of 57.21%, the decimal format will be

57.21/100 = 0.5721 . The value is the fractional abundance of 121 Sb .

123 Sb has a percentage abundance of 42.79%, the decimal format will be

42.79/100 = 0.4279. The value is the fractional abundance of 123 Sb .

Next step is multiplying the fractional abundance to it masses

121 Sb = 0.5721 × 120.904 = 69.169178400

123 Sb = 0.4279 × 122.904 = 52.590621600

The final step is adding the value to get the average atomic weight.

69.169178400 + 52.590621600 = 121.7598 amu

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B  C   E

Explanation:

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8 0
2 years ago
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6 0
3 years ago
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