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8_murik_8 [283]
3 years ago
11

What is the mass percent of calcium chloride if 57 g of CaCl2 is dissolved in 334 g of water?

Chemistry
1 answer:
Zarrin [17]3 years ago
6 0

Answer:

Solubility in water Anhydrous: 74.5 g/100 mL (20 °C) Hexahydrate: 49.4 g/100 mL (−25 °C) 59.5 g/100 mL (0 °C) 65 g/100 mL (10 °C) 81.1 g/100 mL (25 °C) 102.2 g/100 mL (30.2 °C) α-Tetrahydrate: 90.8 g/100 mL (20 °C) 114.4 g/100 mL (40 °C) Dihydrate: 134.5 g/100 mL (60 °C) 152.4 g/100 mL (100 °C)

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Which is NOT a control in the following experiment?: An experiment was performed to determine how the amount of coffee grounds c
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Activity 2 Directions: Select among the choices thbest type of material to be used in making the objects at the left and explain
joja [24]

Answer:

A Walk in the City

Make a list of the number of cars, jeeps, tricycle, and even trucks you've seen on your way to school. On a Decision Making Chart, answer the question, "Does the volume of traffic affect the air quality in my local community?" Write your reasons for saying YES on the Reasons for column, and the reasons for saying NO on the Reasons Against column. At the bottom of a chart, make a position by writing your decision on the same question.

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3 0
2 years ago
2) Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800 mole of CO and 2.40 mole of H2
nadezda [96]

Answer:

Kc = 3.90

Explanation:

CO reacts with H_2 to form CH_4 and H_2O. balanced reaction is:

CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g)  +  H_2O(g)

No. of moles of CO = 0.800 mol

No. of moles of H_2 = 2.40 mol

Volume = 8.00 L

Concentration = \frac{Moles}{Volume\ in\ L}

Concentration of CO = \frac{0.800}{8.00} = 0.100\ mol/L

Concentration of H_2 = \frac{2.40}{8.00} = 0.300\ mol/L

                 CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g)  +  H_2O(g)

Initial            0.100      0.300             0   0

equi.            0.100 -x    0.300 - 3x     x    x

It is given that,

at equilibrium H_2O (x) = 0.309/8.00 = 0.0386 M

So, at equilibrium CO = 0.100 - 0.0386 = 0.0614 M

At equilibrium H_2 = 0.300 - 0.0386 × 3 = 0.184 M

At equilibrium CH_4 = 0.0386 M

Kc=\frac{[H_2O][CH_4]}{[CO][H_2]^3}

Kc=\frac{0.0386 \times 0.0386}{(0.184)^3 \times 0.0614} =3.90

8 0
3 years ago
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