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o-na [289]
3 years ago
5

A sample of As weighs 73.2 grams. Will a sample of Kr that contains the same number of atoms weigh more or less than 73.2 grams?

(more, less): Calculate the mass of a sample of Kr that contains the same number of atoms. grams of Kr
Chemistry
1 answer:
My name is Ann [436]3 years ago
8 0

Answer:

This means a sample of 73.2 grams As atoms weighs less than the same amount of Kr atoms

Explanation:

Step 1: Data given

Mass of As = 73.2 grams

Molar mass As = 74.92 g/mol

Molar mass of Kr = 83.80 g/mol

Step 2: Calculate moles As

Moles As = Mass As / molar mass As

Moles As = 73.2 grams . 74.92 g/mol

Moles As = 0.977 moles

Step 3: Calculate As atoms

As atoms = moles As * number of Avogadro

As atoms = 0.977 moles * 6.02 * 10^23

As atoms = 5.88 *10^23 As atoms

Step 4: Calculate moles Kr

Moles Kr = Atoms Kr / number of Avogadro

Moles Kr = 5.88 * 10^23 Kr atoms / 6.02 *10^23

Moles Kr = 0.977 moles

Step 5: Calculate mass Kr

Mass Kr = moles Kr * molar mass Kr

Mass Kr = 0.977 moles * 83.80 g/mol

Mass Kr = 81.9 grams

This means a sample of 73.2 grams As atoms weighs less than the same amount of Kr atoms

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What is the mass of 5.00 moles of KaS04?
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Answer: D

Explanation:

I assume you meant \text{K}_{2}\text{SO}_{4}.

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  • The atomic mass of sulfur is 32.065 g/mol.
  • The atomic mass of oxygen is 15.9994 g/mol.

So, the formula mass of potassium sulfate is 2(39.0983)+32.065+4(15.9994)=174.2592 g/mol.

So, 5.00 moles have a mass of (5.00)(174.2592), which is about <u>870 g</u>

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2 years ago
A mole is the amount of a substance that contains as many particles as the number of atoms in 12 grams of what isotope?
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4 years ago
A 76.0-gram piece of metal at 96.0 °C is placed in 120.0 g of water in a calorimeter at 24.5 °C. The final temperature in the ca
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The specific heat capacity of the metal given the data from the question is 0.66 J/gºC

<h3>Data obtained from the question</h3>
  • Mass of metal (M) = 76 g
  • Temperature of metal (T) = 96 °C
  • Mass of water (Mᵥᵥ) = 120 g
  • Temperature of water (Tᵥᵥ) = 24.5 °C
  • Equilibrium temperature (Tₑ) = 31 °C
  • Specific heat capacity of the water (Cᵥᵥ) = 4.184 J/gºC
  • Specific heat capacity of metal (C) =?

<h3>How to determine the specific heat capacity of the metal</h3>

The specific heat capacity of the sample of the metal can be obtained as follow:

Heat loss = Heat gain

MC(M –Tₑ) = MᵥᵥCᵥᵥ(Tₑ – Tᵥᵥ)

76 × C × (96 – 31) = 120 × 4.184 × (31 – 24.5)

C × 4940 = 3263.52

Divide both side by 4940

C = 3263.52 / 4940

C = 0.66 J/gºC

Learn more about heat transfer:

brainly.com/question/6363778

#SPJ1

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