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Furkat [3]
3 years ago
9

A researcher wants to test the solubility (property of being dissolved) of salt in water as the temperature of the water increas

es.
Design an experiment to test his hypothesis that as the water temperature increases the solubility of the salt in water also increases. Include the following in your experimental design: experimental setup, procedure for data collection methods and criteria for evaluating the hypothesis.



Materials: salt, water, beakers, heat source, thermometer, balance
Chemistry
1 answer:
aev [14]3 years ago
6 0
Have about 5 beakers all with different temperatures of water. Put in a teaspoon of salt at a time and when it stops dissolving stop adding and record how much salt it took. It should be more salt as the temperature rises. The independent variable is the waters temperature. The dependent variable is how much salt is used. Make sure that there is the same amount of water in each beaker. Or else it won’t work.
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Which property of light does this diagram show?
lidiya [134]
Full question need, post more
6 0
2 years ago
What is 316 Kelvin in Celsius ?
ss7ja [257]

Answer:

42.85 Celsius

5 0
3 years ago
To what volume would you need to dilute 200 mL of a 5.85M solution of Ca(OH)2 to make it a 1.95M solution?
Blizzard [7]

Answer: 600 mL

Explanation:

Given that;

M₁ = 5.85 m

M₂ = 1.95 m

V₁ = 200 mL

V₂ = ?

Now from the dilution law;

M₁V₁ = M₂V₂

so we substitute

5.85 × 200 = 1.95 × V₂

1170 = 1.95V₂

V₂ = 1170 / 1.95

V₂ = 600 mL

Therefore final volume is 600 mL

8 0
3 years ago
Which of the following would have the largest pKa?
garri49 [273]

Answer:

CH3CH2NH3+/CH3CH2NH2 would have the largest pKa

Explanation:

To answer this question we must know Kb of CH3CH2NH2 is 5.6x10⁻⁴, and for C6H5NH2 is 4.0x10⁻¹⁰. And the CH3CH2NH3+ and C6H5NH3+ are related with these substances because are their conjugate base. That means:

pKa of  CH3CH2NH3+ =  CH3CH2NH2;  C6H5NH3+ =  C6H5NH2

Also, Kw / Kb = Ka

Thus:

pKa of CH3CH2NH3+/CH3CH2NH2 is:

Kw / kb = Ka = 1.79x10⁻¹¹

-log Ka = pKa

pKa = 10.75

pKa of C6H5NH3+/ C6H5NH2 is:

Kw / kb = Ka = 2.5x10⁻⁵

-log Ka = pKa

pKa = 4.6

That means CH3CH2NH3+/CH3CH2NH2 would have the largest pKa

5 0
3 years ago
A sucrose solution is prepared to a final concentration of 0.250 M . Convert this value into terms of g/L, molality, and mass %.
MArishka [77]

Answer:

A. 85.6 g

= 0.0856 kg.

B. 0.00027 mol/g

= 0.27 mol/kg.

C. 8.39 %

Explanation:

Given:

Molar concentration = 0.25 M

Molar weight of sucrose = 342.296 g/mol

Density of solution = 1.02 g/mL

Mass of water = 934.4 g.

Density in g/l = 1.020 g/ml * 1000ml/1 l

= 1020 g/l

Mass of solution in 1 l of solution = 1020 g

Mass of solution = mass of solvent + mass of solute

Mass of sucrose = 1020 - 934.4

= 85.6 g of sucrose in 1 l of solution.

A.

Density of sucrose = mass/volume

= molar mass/molar concentration

= 342.296 * 0.25

= 85.6 g/l

Number of moles = mass/molar mass

= 85.6/342.296

= 0.25 mol

B.

Molality = number of moles of solute/mass of solvent

= 0.25/934.4

= 0.00027 mol/g

C.

% mass of sucrose = mass of sucrose/total mass of solution * 100

= 85.6/1020 * 100

= 8.39 %

6 0
3 years ago
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