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ira [324]
3 years ago
8

Increasing temperature increases solubility so that 480 grams (g) of sugar makes a saturated solution in 100 milliliters (mL) of

boiling water. If you add 240 g of sugar to 50 mL of boiling water, which will the solution be?
resaturated
supersaturated
saturated
Chemistry
1 answer:
valkas [14]3 years ago
8 0
Solubility at 100 °C = 480 g / 100 mL of solution =: saturated solution

The proportion 240 g / 50 mL  is equal to the saturated ratio 480 g / 100 mL

Then, 240 g of sugar in 50 mL of boiling water will make a saturated solution.

Answer: saturated


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3 years ago
A 1.87 L aqueous solution of KOH contains 155 g of KOH . The solution has a density of 1.29 g/mL . Calculate the molarity ( M ),
Semenov [28]

Answer:        

[KOH] : 1.47 M

[KOH] : 1.22 m

[KOH]: 6.42 % mass percent.      

Explanation:

First of all we must determine the volume of solution. We have to work with the density

Density = mass / volume

1.29 g/ml = mass / 1870 ml

1.29 g/ml . 1870 ml = 2412.3

Now we must convert the mass to moles

155g / 56.1 g/ mol = 2.76 moles

Now we can calculate molarity

2.76 mol / 1.87 L = 1.47 M

To calculate molality we have to find out the mass of solvent

mass solute + mass solvent = mass solution

155 g + mass solvent = 2412.3 g

2412.3g - 155g = 2257.3g

We have to convert the 2257.3 g to kg

2257.3 g = 2.25 kg

molality = 2.76 moles / 2.25 kg = 1.22 m

To find out the % mass percentation, we have to calculate the mass of solute in 100 g of solution.

In 2412.3 g of solution we have 155 g of KOH

In 100 g of solution, we would have (100 . 155) / 2412.3 = 6.42 %mass percent.

7 0
3 years ago
The half life for the decay of carbon-14 is 5.73 x 10 years. Suppose the activity due to the radioactive decay of the carbon-14
Elena-2011 [213]

Answer:

Age of the atifact is 4.2\times 10^{2} years

Explanation:

  • For first -order radioactive decay- A_{t}=A_{0}(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}
  • A_{t} represents activity of radioactive nuclide after t time, A_{0} represents initial activity of radioactive nuclide and t_{\frac{1}{2}} represents half-life
  • Here, A_{t}=19Bq, A_{0}=20Bq and t_{\frac{1}{2}}=5.73\times 10^{3}years

Plug-in all the given values in the above equation-

19=20\times (\frac{1}{2})^{\frac{t}{5.73\times 10^{3}}}

or, t=4.2\times 10^{2}

So, age of the atifact is 4.2\times 10^{2} years

6 0
3 years ago
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