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gogolik [260]
3 years ago
11

What is the amount, in moles, of each elemental sample? a. 11.8 g Ar b. 3.55 g Zn c. 26.1 g Ta d. 0.211 g Li

Chemistry
1 answer:
spayn [35]3 years ago
8 0

Answer:

A. 0.295 mole

B. 0.055 mole

C.0.144 mole

D. 0.03 mole

Explanation:

To find the amount in moles, we simply use a mathematical relation that connects mass, atomic mass and number of moles.

Number of moles = mass/atomic mass

A. Atomic mass of Argon is 40

n = 11.8/40 = 0.295 mole

B. Atomic mass of zinc is 65

n = 3.55/65 = 0.055 mole

C. Atomic mass of Tantalum is 181

n = 26.1/181 = 0.144 mole

D. Atomic mass of lithium is 7

n = 0.211/7 = 0.03 mole

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8 0
3 years ago
Given the following equation: Cu + 2 AgNO3 ---> Cu(NO3)2 + 2
Temka [501]

Given the following equation; Cu + 2AgNO3 = Cu(NO3)2 + 2Ag, 48.97 grams of Cu are needed to react with 262g of AgNO3.

<h3>How to calculate mass of substances?</h3>

The mass of a substance can be calculated using the following steps:

Cu + 2AgNO3 = Cu(NO3)2 + 2Ag

1 mole of Cu react with 2 moles of AgNO3

  • Molar mass of AgNO3 = 169.87 g/mol
  • Molar mass of Cu = 63.5g/mol

moles of AgNO3 = 262g/169.87g/mol = 1.54mol

1.54 moles of AgNO3 will react with 0.77 moles of Cu.

mass of Cu = 0.77 × 63.5 = 48.97g

Therefore, given the following equation; Cu + 2AgNO3 = Cu(NO3)2 + 2Ag, 48.97 grams of Cu are needed to react with 262g of AgNO3.

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8 0
2 years ago
An atom of boron has an atomic number of 5 and an atomic mass of 11. the atom contains
kirill [66]
The atom contain 6 neutrons, if that's the question
8 0
3 years ago
To understand the relation between the strength of an acid or a base and its pKa and pKb values. The degree to which a weak acid
elena-14-01-66 [18.8K]

Answer:

pKa = 3.675

Explanation:

  • pKa = - Log Ka

∴ <em>C</em> X-281 = 0.079 M

∴ pH = 2.40

let X-281 a weak acid ( HA ):

∴ HA ↔ H+ + A-

⇒ Ka = [H+] * [A-] / [HA]

mass balance:

⇒<em> C</em> HA = 0.079 M = [HA] + [A-]

⇒ [HA] = 0.079 - [A-]

charge balance:

⇒ [H+] = [A-] + [OH-]... [OH-] is negligible; it comes from to water

⇒ [H+] = [A-]

∴ pH = - log [H+] = 2.40

⇒ [H+] = 3.981 E-3 M

replacing in Ka:

⇒ Ka = [H+]² / ( 0.079 - [H+] )

⇒ Ka = ( 3.981 E-3 )² / ( 0.079 - 3.981 E-3 )

⇒ Ka = 2.113 E-4

⇒ pKa = - Log ( 2.113 E-4 )

⇒ pKa = 3.675

8 0
3 years ago
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igor_vitrenko [27]

Answer:

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Explanation:

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4 0
2 years ago
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