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artcher [175]
3 years ago
14

What is the mass of 4.5 moles of sodium fluoride?

Chemistry
1 answer:
Thepotemich [5.8K]3 years ago
3 0
Mass of sodium is 23 and mass of fluirine is 19 there mass of NaF is 42
42 g = 1 mole therefor 4.5 moles will have
4.5 × 42 = 189 g
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A 10 gram sample of H20 is sealed in a 1350 ml flask at 27°C. Given the fact that water has a vapor pressure of 26.7 mmHg at thi
Aleksandr-060686 [28]

Answer:

9.9652g of water

Explanation:

The establishment of the liquid-vapor equilibrium occurs when the vapour of water is equal to vapour pressurem 26.7 mmHg. Using gas law it is possible to know how many moles exert that pressure, thus:

n = PV / RT

Where P is pressure 26,7 mmHg (0.0351atm), V is volume (1.350L), R is gas constant (0.082 atmL/molK) and T is temperature (27°C + 273,15 = 300.15K)

Replacing:

n = 0.0351atm×1.350L / 0.082atmL/molK×300.15K

n = 1.93x10⁻³ moles of water are in gaseous phase. In grams:

1.93x10⁻³ moles × (18.01g / 1mol) = <u><em>0.0348g of water</em></u>

<u><em /></u>

As the initial mass of water was 10g, the mass of water that remains in liquid phase is:

10g - 0.0348g = <em>9.9652g of water</em>

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I hope it helps!

4 0
3 years ago
If the vapor's volume were to be incorrectly recorded as 125ml, how will this error affect the calculated molar mass of the unkn
pishuonlain [190]
Since you didn't give the actual volume (or any of the experimental values) I can only tell you how to do it. Do the calculation using the real (determined) volume of the flask. Then, re-do the calculation with v = 125ml. Take the two values and calculate % error; m = measured vol; g = guessed vol. 

<span>[mW (m) - mW (g)]/mW (m) x 100% </span>

<span>(they want % error so, if it is negative, just get rid of the sign) </span>
3 0
3 years ago
In a metabolic pathway, succinate dehydrogenase catalyzes the conversion of succinate to fumarate. the reaction is inhibited by
Pani-rosa [81]

Competitive inhibitor

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3 years ago
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Tanya [424]

Answer:

A. 2C + H₂ ⟶ CH₄  

Explanation:

A. 2C + H₂ ⟶ CH₄

UNBALANCED. 2C on the left and 1C on the right

B. 2Al₂O₃ ⟶ 4Al + 3O₂  

Balanced. Same number of each type of atom on each side.

C. 2H₂O₂ ⟶ 2H₂O + O₂  

Balanced. Same number of each type of atom on each side.

D. 2C₂H₆ + 7O₂ ⟶ 4CO₂ + 6H₂O

Balanced. Same number of each type of atom on each side.

8 0
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Alika [10]

Answer:

d

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