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DerKrebs [107]
3 years ago
13

How many ounces are in a keg of beer?

Chemistry
1 answer:
valentinak56 [21]3 years ago
4 0
The answer is 1,984 ounces of beer.
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you have a solution of water and salt. if the total solution weighs 50 grams and the salt weighs 4 grams. what is the percent co
Natali5045456 [20]

Answer:

92%

Explanation:

I believe. Hope this was helpful.

8 0
3 years ago
Which aqueous solution has the highest boiling point at standard pressure?(1) 1.0 M KC1(aq) (3) 2.0 M KCl(aq)(2) 1.0 M CaC12(aq)
miss Akunina [59]

Answer:

(4) 2.0 M CaCl₂(aq).

Explanation:

  • Adding solute to water elevates the boiling point.
  • The elevation in boiling point (ΔTb) can be calculated using the relation:

<em>ΔTb = i.Kb.m,</em>

where, ΔTb is the elevation in boiling point.

i is the van 't Hoff factor.

  • van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.

Kb is the molal elevation constant of water.

m is the molality of the solution.

<u><em>(1) 1.0 M KCl(aq):</em></u>

i for KCl = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

suppose molarity = molality, m = 1.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (2)(Kb)(1.0 m) = 2(Kb).

<u><em>(2) 2.0 M KCl(aq):</em></u>

i for KCl = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

suppose molarity = molality, m = 2.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (2)(Kb)(2.0 m) = 4(Kb).

<u><em>(3) 1.0 M CaCl₂(aq):</em></u>

i for CaCl₂ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.

suppose molarity = molality, m = 1.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (3)(Kb)(1.0 m) = 3(Kb).

<u><em>(4) 2.0 M CaCl₂(aq):</em></u>

i for CaCl₂ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.

suppose molarity = molality, m = 2.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (3)(Kb)(2.0 m) = 6(Kb).

  • <em>So, the aqueous solution has the highest boiling point at standard pressure is: (4) 2.0 M CaCl₂(aq).</em>

<em></em>

6 0
3 years ago
Both HNO3(aq) and CH3COOH(aq) can be classified as *
lord [1]

Answer:

acids

Explanation:

HNO3 is a strong acid (Nitric Acid)

CH3COOH is a weak acid (Acetic Acid

6 0
3 years ago
Read 2 more answers
The molecular weight of a gas is ________ g/mol if 3.5 g of the gas occupies 2.1 l at stp
bija089 [108]
<span>Pre-1982 definition of STP: 37 g/mol Post-1982 definition of STP: 38 g/mol This problem is somewhat ambiguous because the definition of STP changed in 1982. Prior to 1982, the definition was 273.15 K at a pressure of 1 atmosphere (101325 Pascals). Since 1982, the definition is 273.15 K at a pressure of exactly 100000 Pascals). Because of those 2 different definitions, the volume of 1 mole of gas is either 22.414 Liters (pre 1982 definition), or 22.71098 liters (post 1982 definition). And finally, there's entirely too many text books out there that still use the 35 year obsolete definition. So let's solve this problem using both definitions and you need to pick the correct answer for the text book you're using. First, determine how many moles of gas you have. Just simply divide the volume you have by the molar volume. Pre-1982: 2.1 / 22.414 = 0.093691443 moles Post-1982: 2.1 / 22.71098 = 0.092466287 moles Now determine the molar mass. Simply divide the mass by the moles. So Pre-1982: 3.5 g / 0.093691443 moles = 37.35666667 g/mol Post-1982: 3.5 g / 0.092466287 moles = 37.85163333 g/mol Finally, round to 2 significant figures. So Pre-1982: 37 g/mol Post-1982: 38 g/mol</span>
5 0
4 years ago
Which definition best describes the term activation energy?
dalvyx [7]

Answer:

B

Explanation:

bshdiidididjdjjdididid

4 0
3 years ago
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