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Diano4ka-milaya [45]
3 years ago
11

What is the pressure of the gas if we have 3.50 moles of helium at -50.0°C

Chemistry
1 answer:
Marianna [84]3 years ago
7 0

Answer:

259.6kPa

Explanation:

because it is big

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What is the difference between docosahexaenoate (DHA) and 19,20-dihydroxydocosapentaenoate (19,20-DHDP)? Only DHA is a polyunsat
andrew11 [14]

Answer:

Explanation:

The correct answer is 19, 20 DHDP is more polar than DHA. This is as a result of the presence of two hydroxyl groups.

4 0
3 years ago
1- Alum used in cooking is potassium aluminum sulfate hydrate, KAl(SO4)2. XH2O. To find the value of X, you can heat the sample
Burka [1]

Answer:

THE VALUE OF X IS 7 AND THE FORMULA OF THE HYDRATED SALT IS KAl(SO4)2.7H20

Explanation:

1. write out the varibales given in thequestion:

Mass of the hydrated salt = 4.74 g

Mass of water lost = 2.16 g

Formula of the hydrated salt = KAl(SO4)2. XH20

2. calculate the molar mass of the salt and that of water of crystallization:

Molar mass of anhydrous salt = ( K = 39, Al = 27, S = 32, 0=16)

= ( 39 + 27 + 32*2 + 16 * 8

= (39 + 27 + 64 + 128)

= 258 g/mol

Molar mass of water = 18 g/mol

3. Use this expression to calculate X:

The expression,

XH20 / molar mass of anhydrous salt = Mass of water lost / Mass of hydrated salt.

X = molar mass of anhydrous salt * mass of water lost / mass of anhydrous salt * H20

where XH20 is the molar mass of water of crystallization, is used to calculate the value of X.

4. Solve for X:

So therefore:

X = 258 * 2.16 / 4.74 * 18

X = 557.28/ 85.32

X = 6.53

X is approximately 7.

The value of X is 7 and the formula pf the hydrated salt is KAl(SO4)2.7H20

7 0
3 years ago
The reaction for the formation of gaseous hydrogen fluoride (HF) from molecular hydrogen (H2) and fluorine (F2) has an equilibri
mestny [16]

Answer:

[H2]eq = 0.0129 M

[F2]eq = 1.0129 M

[HF]eq = 0.9871 M

Explanation:

  • H2(g) + F2(g) ↔ 2HF(g)

∴ Ke = [HF]² / [H2]*[F2] = 1.15 E2

experiment:

∴ n H2 = 3.00 mol

∴ n F2 = 6.00 mol

∴ V sln = 3.00 L

⇒ [H2]i = 3.00 mol / 3.00 L = 1 M

⇒ [F2]i = 6.00 mol / 3.00 L = 2 M

        [ ]i    change      [ ]eq

H2     1         1 - x         1 - x

F2     2        2 - x         2 - x

HF     -            x              x

⇒ K = (x)² / (1 - x)*(2 - x) = 1.15 E2

⇒ x² / (2 - 3x + x²) = 1.15 E2 = 115

⇒ x² = (2 - 3x + x²)(115)

⇒ x² = 230 - 345x + 115x²

⇒ 0 = 230 - 345x + 114x²

⇒ x = 0.9871

equilibrium:

⇒ [H2] = 1 - x = 1 - 0.9871 = 0.0129 M

⇒ [F2] = 2 - x = 2 - 0.9871 = 1.0129 M

⇒ [HF] = x = 0.9871 M

5 0
3 years ago
Iwsk3eidwujfdujeudcwix vvfedsxdwsx
Sedaia [141]

Answer: :)

Explanation:

7 0
2 years ago
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3. Determine the uses of the following materials:
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Answer:

D

Explanation:

D

3 0
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