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Lelechka [254]
3 years ago
7

If 50.0 g of water saturated with potassium chloride at 80.0°C is slowly evaporator to dryness, how many grams of the dry salt w

ill be recovered

Chemistry
1 answer:
balu736 [363]3 years ago
5 0

Answer:

25.0 g.

Explanation:

  • From the solubility curve that is shown in the attached image, the solubility of KCl per 100.0 g of water is about 50.0 g.
  • <em>So, a saturated solution of KCl in 50.0 g of water will contain about 25.0 g.</em>
  • <em>Thus, the grams of the dry salt that will be recovered is 25.0 g.</em>

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ZanzabumX [31]

You can detect salt in water without tasting by measuring the density of the water. Place a glass of spring water and a glass of the suspected salt water on a balance scale and the heavier one contains salt. Other ways to test for salt in water is to put a drop of water on the end of a nail and place in a gas flame. If the water contains salt, the flame will turn a yellow/orange color.

7 0
3 years ago
A single atom of an element has 21 neutrons, 20 electrons, and 20 protons. Which element is it?
sukhopar [10]

Answer:

ca

Explanation:

3 0
3 years ago
What is the mass of 0.100 mole of neon? (Watch sf’s)
dimaraw [331]

Answer:

The answer to your question is letter D. 2.02 g

Explanation:

Data

moles of Ne = 0.100

atomic mass of Neon = 20.18 g

Process

1.- Use proportions to find the answer

                   20.18 g of Ne ------------------  1 mol of Ne

                        x                 ------------------  0.1 moles

                        x = (0.1 x 20.18)/1

                        x = 2.018

2.- Consider the significant figures

      0.100 has three significant figures so the answer must be  2.02 g

4 0
3 years ago
Read 2 more answers
A blacksmith heated an iron bar to 1445 °C. The blacksmith then tempered the metal by dropping it into 42,800 mL of
Wittaler [7]

Answer:

6626 g

Explanation:

Given that:

Density of water = 1.00 g/ml, volume of water = 42800 ml.

Since density = mass/ volume

mass of water = volume of water * density of water = 42800 ml * 1 g/ml = 42800 g

Initial temperature of water = 22°C and final temperature of water = 45°C.

specific heat capacity for water = 4.184 J/g°C

ΔT water = 45 - 22 = 23°C

For iron:

mass = m,  

specific heat capacity for iron  = 0.444 J/g°C

Initial temperature of iron = 1445°C and final temperature of water = 45°C.

ΔT iron = 45 - 1445 = -1400°C

Quantity of heat (Q) to raised the temperature of a body is given as:

Q = mCΔT

The quantity of heat required to raise the temperature of water is equal to the temperature loss by the iron.

Q water (gain) + Q iron (loss) = 0

Q water = - Q iron

42800 g ×  4.184 J/g°C × 23°C = -m × 0.444 J/g°C × -1400°C

m = 4118729.6/621.6

m = 6626 g

8 0
3 years ago
What is the molarity of a kf(Aq) solution containing 116 g of kf in 1.00 l of solution
Oduvanchick [21]

Answer:

2.00 M

Explanation:

K=39 F=19

39+19=58

116/58= 2 mol/ 1 L = 2 M

8 0
3 years ago
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