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slega [8]
3 years ago
7

You are given mixture made of 290 grams of water and 14.2 grams of salt. Determine the % by mass of salt in the salt solution.

Chemistry
1 answer:
marishachu [46]3 years ago
3 0

Answer:

Solution is 4.67% by mass of salt

Explanation:

% by mass is the concentration that defines the mass of solute in 100g of solution.

In this case we have to find out the mass of solution with the data given:

Mass of solution = Mass of solute + Mass of solvent

Solute:  Salt → 14.2 g

Solvent: Water → 290 g

Solution's mass = 14.2 g + 290g = 304.2 g

% by mass = (mass of solute / mass of solution) . 100

(14.2 g / 304.2g) . 100 = 4.67 %

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motikmotik

The simplest and most helpful visual aid for this experiment would be if the student sets rulers of some sort or maybe pieces of wood that are properly marked with certain measuring unite. If they are set right next to each plant, the student will be able to monitor constantly how are the plants growing, and the smaller the measuring unites the better, as the student will be able to monitor the situation in higher detail.

Over time, the plants will most surely develop differently, so the student will have it easier to notice the changes. Considering that it is an experiment in question, the student should use proper measurements and to be able to have detailed numbers for it.

5 0
3 years ago
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MAXImum [283]

Answer:

the vapor pressure of this solution would increase if some of the water were allowed to evaporate

4 0
3 years ago
Which of the following elements would you expect to be ductile and
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Answer:

ductile - germanium

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8 0
3 years ago
Phosphoric acid has 3 pka values, which are 2.1, 6.9, and 12.4. draw the protonated form of phosphoric acid associated with the
Margaret [11]
As mentioned above, phosphoric acid has 3 pKa values, and after 3 ionization it gives 3 types of ions at different pKa values:

H₃PO₄(aq) + H₂O(l) ⇌ H₃O⁺(aq) + H₂PO₄⁻ (aq)         pKₐ₁ 
<span>

</span>H₂PO₄⁻(aq) + H₂O(l) ⇌ H₃O⁺(aq) + HPO₄²⁻ (aq)       pKₐ₂


HPO₄²⁻(aq) + H₂O(l) ⇌ H₃O⁺(aq) + PO₄³⁻ (aq)          pKₐ₃ 


At the highest pKa value (12.4) of phosphoric acid, the last OH group will lose its hydrogen. On the picture I attached, it is shown required protonated form of phosphoric acid before reaction whose pKa value is 12.4.


6 0
3 years ago
Calculate the solubility of benzene in water at 25 c in ppm. the required henry's law constant is 5.6 bar/mol/kg and benzene's s
kap26 [50]

The relationship between pressure and solubility of the gas is given by Henry's law as:

S_g = kP_g

where,

S_g is the solubility of the gas.

k is proportionality constant i.e. Henry's constant.

P_g is pressure of the gas.

k = 5.6 bar/mol/kg (given)

P_g = 0.13 bar (given)

Substituting the values,

S_g = 5.6 bar/mol/kg\times 0.13 bar = 0.728 mole/kg

To convert mole/kg to g/kg:

Molar mass of benzene, C_6H_6 = 6\times 12+6\times 1 = 78 g/mol

0.728\times 78 = 56.784 g/kg

Now for converting into ppm:

Since, 1 ppm = 0.001 g/kg

So, 56.784\times 1000 = 56784 ppm.

Hence, the solubility of benzene in water at 25^{o} C in ppm is 56784 ppm.


7 0
3 years ago
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