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dem82 [27]
3 years ago
14

A 18.00 g g milk chocolate bar is found to contain 11.00 g g of sugar.1. How many milligrams of sugar does the milk chocolate ba

r contain?2. What will be the amount of sugar in milligrams if the size of the milk chocolate bar is reduced from 18.00 g to 4.000 g ? Express your answer in milligrams to four significant figures.
Chemistry
1 answer:
horrorfan [7]3 years ago
6 0

Answer:

1. 1.100x10⁴mg

2. 2444mg

Explanation:

1. The 18.00g milk chocolate bar contain 11.00g of sugar. In miligrams:

11.00g * (1000mg / 1g) = 1.100x10⁴mg

2. If a bar of 18.00g contain 11.00g of sugar, a bar of 4.000g will contain:}

4.000g bar * (11.00g / 18.00g)  = 2.444g of sugar.

In miligrams:

2.444g * (1000mg / 1g) = 2444mg

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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
How much heat is lost when 0.440 mol of steam condenses at 100 °C?
dusya [7]

Answer:

17,890 J

Explanation:

The amount of heat released by a gaseous substance when it condenses is given by the formula

Q=n\lambda_v

where

n is the number of moles of the substance

\lambda_v is the latent heat of vaporization

The formula can be applied if the substance is at its vaporization temperature.

In this problem, we have:

n = 0.440 mol is the number of moles of steam

\lambda_v=40,660 J/mol is the latent heat of vaporization of water

And the steam is already at 100C, so we can apply the formula:

Q=(0.440)(40660)=17,890 J

8 0
3 years ago
If the rate of decomposition of ammonia, NH3, at 1150 K is 2.10 x 10-6 mol/L/s, what is the
Alina [70]

Answer:

3.15 × 10⁻⁶ mol H₂/L.s

1.05 × 10⁻⁶ mol N₂/L.s

Explanation:

Step 1: Write the balanced equation

2 NH₃ ⇒ 3 H₂ + N₂

Step 2: Calculate the rate of production of H₂

The molar ratio of NH₃ to H₂ is 2:3. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of H₂ is:

2.10 × 10⁻⁶ mol NH₃/L.s × 3 mol H₂/2 mol NH₃ = 3.15 × 10⁻⁶ mol H₂/L.s

Step 3: Calculate the rate of production of N₂

The molar ratio of NH₃ to N₂ is 2:1. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of N₂ is:

2.10 × 10⁻⁶ mol NH₃/L.s × 1 mol N₂/2 mol NH₃ = 1.05 × 10⁻⁶ mol N₂/L.s

3 0
3 years ago
Why won’t sugar and Citric acid leave a sediment in salad dressing??
satela [25.4K]
Because it contains vinegar because it does not form layers when mixed with other liquids. Sugar or citric acid because they don't leave sediment.
7 0
3 years ago
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Select the correct answer. Which statement is true for a heating curve?
Anon25 [30]

Answer:

OB

Explanation:

tht is correct make me brainliest

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