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2 years ago

11. A sample contains 25% water and weighs 201 grams. Determine the grams of water in the

Chemistry

1 answer:

77julia77 [94]2 years ago

5 0

Answer:

$m_{water}=50.25g$

Explanation:

Hello,

In this case, considering that the by-mass percent of water is:

$\% m/m=\frac{m_{water}}{m_{sample}}*100\%$

Given such percent and the mass of the sample, we can find the mass of water in grams in the sample by solving for it as shown below:

$m_{water}=\frac{\%m/m*m_{sample}}{100\%}\\ \\m_{water}=\frac{25\%*201g}{100\%}\\ \\m_{water}=50.25g$

Best regards.

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Analyze: The first shell can hold a maximum of two electrons. How does this explain the valence of hydrogen

chubhunter [2.5K]

Answer:

See explanation

Explanation:

Hydrogen has a valency of +1 or -1. Its electronic configuration is 1s1.

The 1s sub-level (first shell) is known to hold two electrons. This means that hydrogen may either loose this one electron in the 1s level to yield H^+ or accept another electron into this 1s level to form H^- (the hydride ion).

The formation of the hydride ion completes the 1s orbital.

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2 years ago

For the Bradford assay, the instructor will make a Bradford reagent dye by mixing 50 ml of 95% v/v ethanol with 100 mg of Coomas

jolli1 [7]

Answer:

4,25% v/v H3PO4

Explanation:

The concentration of phosphoric acid (H3PO4) is expressed as a volume / volume percentage, which means:

%v/v H3PO4 = (mL of pure H3PO4/mL of solution)*100%

In other words, <u>we are only interested in the final volume of the solution to which the phosphoric acid was diluted, regardless of its composition</u>. Which in this case is 1 L (1000 mL).

We can then apply the following equation, commonly used to calculate the initial or final concentration (or volume) of a substance when it is diluted:

Ci*Vi=Cf*Vf

<u>Where</u>:

Ci, is the initial concentration of the substance.

Vi, the initial volume of the substance

Cf, the final concentration reached after dilution

Vf, the final volume of the solution at which the substance was diluted

In this case, the incognite would be the final concentration of H3PO4 reached after dilution, that is, Cf. Therefore, we proceed to clear Cf from the previous equation and replace our data:

Cf = (Ci*Vi)/Vf = (85% v/v * 50 mL)/1000 mL = 4,25 % v/v

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2 years ago

Describe a single measurement that you could make that shows that a kettle is less than 100% efficient (providing evidence that

alexgriva [62]

Answer:

As you haven't explained what measurements you took before solving this problem, I will explain the general procedure to evaluate the efficiency of a kettle. I hope it helps you. I´ll send an attachement file with the full answer, since I couldn't write it here.

I assume that the material that is going to be heated in the kettle is water.

1- You have to boil water in it and take the time it takes to its boiling point (in seconds).

2- You have to evaluate the amount of energy the water absorbed Q with the efficiency formula which I explain in the attachement file.

3- Divide Q by the time it took to bring the water to boiling so you can have the power it consumed.

4- You divide the last value you obtained by the Kettles's power rating.

5- Multiply the last value by 100 to obtain a percentage value of efficiency.

Explanation:

Efficiency is the ration of a machine's useful work, in this case how much energy the water absorbed to get to its boiling point divided by the time it took to get to this point, and the total energy expended, in this case the kettles's power rating.

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2 years ago

Be sure to answer all parts. For each reaction, find the value of ΔSo. Report the value with the appropriate sign. (a) 3 NO2(g)

Liono4ka [1.6K]

Answer:

a. -268.13 J/K

b. -279.95 J/K

c. + 972.59 J/K

Explanation:

The value of the change in entropy (ΔS°) can be calculated by:

ΔS° = ∑n*S° products - ∑n*S° reactants, where n is the stoichiometric number of moles.

The values of S° for each substance can be found on a thermodynamic table.

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S°, NO2(g) = 240.06 J/mol.K

S°, H2O(l) = 69.91 J/mol.K

S°, HNO3(l) = 155.60 J/mol.K

S°, NO(g) = 210.76 J/mol.K

ΔS° = (210.76 + 2*155.60) - (3*240.06 + 69.91)

ΔS° = -268.13 J/K

b. N2(g) + 3F2(g) → 2NF3(g)

S°, N2(g) = 191.61 J/mol.K

S°, F2(g) = 202.78 J/mol.K

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ΔS° = (2*260.0 ) - (191.61 + 3*202.78)

ΔS° = -279.95 J/K

c. C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(g)

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S°, H2O(g) = 188.83 J/mol.K

ΔS° = (6*213.74 + 6*188.83) - (212 + 6*205.138)

ΔS° = +972.59 J/K

3 0

2 years ago

A laboratory manual gave precise instructions for carrying out the reaction described by the chemical equation, C2H6O + O2 C2H4O

Alexxandr [17]

C2H6O + O2 ---> C2H4O2 + H2O

using the molar masses:-
24+ 6 + 16 g of C2H6O produces 24 + 4 + 32 g C2H4O2 (theoretical)
46 g produces 60g

60 g C2H4O2 is produced from 46g C2H6O
1g . .................................46/60 g
700g ................................. (46/60) * 700 Theoretically

But as the yield is only 7.5%

the required amount is ((46/60) * 700 ) / 0.075 = 7155.56 g

= 7.156 kg to nearest gram. Answer

8 0

2 years ago