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elena55 [62]
3 years ago
11

A4 g sugar cube (Sucrose : C 12 H 22 O 11 ) is dissolved in a 350 ml teacup of 80 C water. What is the percent composition by ma

ss of the sugar solution ? Density of water at 80 degrees * C = 0.975g / m * l Select one :- 1.96\% . 1.63% 1.36 % d% e 1.16 \%​
Chemistry
1 answer:
lana66690 [7]3 years ago
7 0

Answer:

%Sgr = 1% (1 sig.fig.)

Explanation:

mass water = 350ml x 0.975g/ml = 341.25g

mass sugar added = 4g

solution mass = 341.25g + 4g = 345.25g

%sugar = (4g/345.25g)·100% = 1.1586% ≅ 1% (1 sig.fig)

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Calculate the value deltaG°​
atroni [7]

Answer:

ΔG=ΔG0+RTlnQ where Q is the ratio of concentrations (or activities) of the products divided by the reactants. Under standard conditions Q=1 and ΔG=ΔG0 . Under equilibrium conditions, Q=K and ΔG=0 so ΔG0=−RTlnK . Then calculate the ΔH and ΔS for the reaction and the rest of the procedure is unchanged.

Explanation:

4 0
3 years ago
A ___________________________ is made of two or more different substances.
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Mixture/ compound
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6 0
3 years ago
A certain radioactive nuclide has a half life of 1.00 hour(s). Calculate the rate constant for this nuclide. s-1 Calculate the d
Karo-lina-s [1.5K]

Answer:

k= 1.925×10^-4 s^-1

1.2 ×10^20 atoms/s

Explanation:

From the information provided;

t1/2=Half life= 1.00 hour or 3600 seconds

Then;

t1/2= 0.693/k

Where k= rate constant

k= 0.693/t1/2 = 0.693/3600

k= 1.925×10^-4 s^-1

Since 1 mole of the nuclide contains 6.02×10^23 atoms

Rate of decay= rate constant × number of atoms

Rate of decay = 1.925×10^-4 s^-1 ×6.02×10^23 atoms

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8 0
3 years ago
If you made 6 moles of NO2 How many grams of N2 did you use N2+2O2> 2NO2​
denis23 [38]

Answer: 1:2

Explanation:

Believe me its correct.

7 0
3 years ago
A sample of helium gas occupies a volume of 152.0 mL at a pressure of 717.0 mm Hg and a temperature of 315.0 K. What will the vo
kozerog [31]

Answer:

0.581 L  or  581 mL

Explanation:

As stated in the question, the combined gas law is (P1*V1/T1) = (P2*V2/T2)

Write down the amounts you are given.

V1 = 0.152 L (I was taught to always convert milliliters to liters)

P1 = 717 mmHg

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V2 = ?

P2 = 463 mmHg

T2 = 777 K

The variable that is being solved for is final volume. Fill in the combined gas law equation with the corresponding amounts and solve for V2.

(717 mmHg*0.152 L) / (315 K) = (463 mmHg*V2) / (777 K)

0.346 = (463*V2) / (777)

0.346*777 = (463*V2) / (777)*777

268.842 = 463*V2

268.842/463 = (463*V2)/463

V2 = 0.581

Pressure and volume are indirectly proportional. This checks out because the volume increased while pressure decreased. Volume and temperature are directly proportional. This checks out because both volume and temperature increased. This is a good way to check your answers. You can also solve each side of the combined gas law equation to see if they are both the same.

6 0
3 years ago
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