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dedylja [7]
2 years ago
14

BEBE

Chemistry
1 answer:
DIA [1.3K]2 years ago
3 0

Reaction 1 : yes

Reaction 2 : no

<h3>Further explanation</h3>

The metal activity series is expressed in voltaic series  

Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au  

The more to the left, the metal is more reactive (easily release electrons) and the stronger reducing agent  

The more to the right, the metal is less reactive (harder to release electrons) and the stronger oxidizing agent

So that the metal located on the left can push the metal on the right in the redox reaction  

From activity series of Halogen :

F₂>Cl₂>Br₂>I₂

F₂ is the strongest oxidizing agent

1. Reaction

Cl₂ + 2Rbl - 2RbCI+ I₂

Cl₂>I₂⇒reaction can occur⇒yes, reactions will take place.

2. Reaction

I₂ + NiBr₂ - NI₂ + Br₂

Br₂>I₂⇒Reaction can't occur⇒no, reaction will not take place

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Determine the mass of carbon iv oxide ,produced on burning 104g of ethyne​
mars1129 [50]

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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