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Ulleksa [173]
4 years ago
6

Determine the mass of carbon iv oxide ,produced on burning 104g of ethyne​

Chemistry
1 answer:
mars1129 [50]4 years ago
7 0

163 grams (3 sig. fig.).

<h3>Explanation</h3>
  • Formula of <em>carbon(IV) oxide</em> (a.k.a. carbon dioxide): \text{CO}_2.
  • Molar mass of \text{CO}_2: \underbrace{12.01}_{\text{C}} + 2\times\underbrace{16.00}_{\text{O}}=44.01\;\text{g}\cdot\text{mol}^{-1}.
  • Formula of ethyne: structural \text{H}-\text{C}\equiv\text{C}-\text{H} or molecular \text{C}_2\text{H}_2.
  • Molar mass of \text{C}_2\text{H}_2: 2\times\underbrace{12.01}_{\text{C}}+2 \times\underbrace{16.00}_{\text{O}} = 56.02\;\text{g}\cdot\text{mol}^{-1}.

All carbon atoms in that 104 grams of ethyne will end up in \text{CO}_2. Number of moles of molecules in 104 grams of ethyne:

n = \dfrac{m}{M} = \dfrac{104}{56.02} = 1.85648\;\text{mol}.

There are two carbon atoms in each ethyne molecule. Number of carbon atoms in that many ethyne molecules:

n(\text{C}) = 2\;n(\text{C}_2\text{H}_2) = 3.71296\;\text{mol}.

There are one carbon atom in each \text{CO}_2 molecule. In case oxygen is in excess, all those carbon atoms from that 104 grams of ethyne will make n(\text{CO}_2) = n(\text{C}) =3.71296\;\text{mol} of \text{CO}_2.

Mass of all those \text{CO}_2 molecules:

m = n\cdot M = 163\;\text{g}. (3 sig. fig. as in the mass of ethyne.)

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Cardenolides, with the chemical formula CH₁₈C₂₀H₁₅CHCO₂ have (D) 23 carbon atoms, 34 hydrogen atoms, and 2 oxygen atoms.

Milkweed contains a poison known as cardenolides. The chemical formula for cardenolides CH₁₈C₂₀H₁₅CHCO₂.

The subscripts in the formula represent the atomicities, that is the number of atoms of each element in each part of the formula.

We can calculate the total number of atoms of each element by adding its atomicities.

<h3>Carbon atoms</h3>

C =  1 + 20 + 1 + 1 = 23

<h3>Hydrogen atoms</h3>

H = 18 + 15 + 1 = 34

<h3>Oxygen atoms</h3>

O = 2

Cardenolides, with the chemical formula CH₁₈C₂₀H₁₅CHCO₂ have (D) 23 carbon atoms, 34 hydrogen atoms, and 2 oxygen atoms

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5 0
3 years ago
The density of aluminium is 2.7 g/cm3. Find the mass in grams of a bar of aluminum measuring 2.7 cm by 1.2 cm by 14.3 cm.
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Answer:125.0964g

Explanation:

Density = mass/volume

Density = 2.7 gcm-3

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Mass=?

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How much does the mass of 12.65-g sample of copper(II) nitrate hexahydrate decrease when heated?
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Sort these species into isoelectronic groups. It doesn\'t matter which group goes in which box, so long as the correct species a
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Answer : The isoelectronic groups are:

He,Be^{2+},Li^{+}

N^{3-},Mg^{2+}

S^{2-},Ti^{4+},K^{+}

Explanation :

Isoelectronic : It is defined as the compound or molecule having the same number of electrons and the same number of electronic structure.

  • The element is helium. The number of electrons are 2.
  • The element is beryllium. The number of electrons are 4. The number of electrons in Be^{2+} = 4 - 2 = 2
  • The element is lithium. The number of electrons are 3. The number of electrons in Li^{+} = 3 - 1 = 2
  • The element is nitrogen. The number of electrons are 7. The number of electrons in N^{3-} = 7 + 3 = 10
  • The element is neon. The number of electrons are 10.
  • The element is sulfur. The number of electrons are 16. The number of electrons in S^{2-} = 16 + 2 = 18
  • The element is magnesium. The number of electrons are 12. The number of electrons in Mg^{2+} = 12 - 2 = 10
  • The element is titanium. The number of electrons are 22. The number of electrons in Ti^{4+} = 22 - 4 = 18
  • The element is potassium. The number of electrons are 19. The number of electrons in K^{+} = 19 - 1 = 18

The isoelectronic groups are:

He,Be^{2+},Li^{+}

N^{3-},Mg^{2+}

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7 0
3 years ago
The half-life of radon gas is approximately four days. Four weeks after the introduction of radon into a sealed room, the fracti
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The fraction of the original amount remaining is closest to 1/128

<h3>Determination of the number of half-lives</h3>
  • Half-life (t½) = 4 days
  • Time (t) = 4 weeks = 4 × 7 = 28 days
  • Number of half-lives (n) =?

n = t / t½

n = 28 / 4

n = 7

<h3>How to determine the amount remaining </h3>
  • Original amount (N₀) = 100 g
  • Number of half-lives (n) = 7
  • Amount remaining (N)=?

N = N₀ / 2ⁿ

N = 100 / 2⁷

N = 0.78125 g

<h3>How to determine the fraction remaining </h3>
  • Original amount (N₀) = 100 g
  • Amount remaining (N)= 0.78125 g
  • Fraction remaining =?

Fraction remaining = N / N₀

Fraction remaining = 0.78125 / 100

Fraction remaining = 1/128

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