Answer:
Explanation:
The symbol for TIN is Sn.
When Sn loses 2 electrons, it gets a double positive charge ( +2 ) and becomes 
.
It becomes a cation.
The name of Ion is Tin ( II ) Ion.
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Hope this helped!
<h3>~AH1807</h3>
<span>CH4 + 4 Cl2 → CCl4 + 4 HCl
(4.00 mol CH4) x (1/1) x (0.70) = 2.80 mol CCl4
(4.00 mol CH4) x (4/1) x (0.70) = 11.2 mol HCl
CCl4 + 2 HF → CCl2F2 + 2 HCl
(2.80 mol CCl4) x (2/1) x (0.70) = 3.92 mol HCl
11.2 mol + 3.92 mol = 15.1 mol HCl from both steps</span>
Answer
I think it might be B
Explanation
Answer:
The correct option here is the first option
Explanation:
Covalent bond is the bond that involves the sharing of electrons between the participating atoms. The electrons (in the outermost shells of the atoms) that are involved this sharing are called the "shared pair" while those electrons (in the outermost shells of the atoms) that are not involved in this sharing are called the "lone pair". Bonding eventually leads to each of the participating atoms achieving it's octet configuration.
Carbon will bind covalently with fluorine (to form carbon tetrafluoride) with each of the electrons on the outermost shell of the carbon been shared covalently with fluorine atoms (that also requires just one electron to achieve it's octet configuration). Thus, at the end, we would have one carbon atom being covalently linked to four flourine atoms.
Answer:
KBr is limiting reactant.
Explanation:
Given data:
Mass of KBr =4g
Mass of Cl₂ = 6 g
Limiting reactant = ?
Solution:
Chemical equation:
2KBr + Cl₂ → 2KCl + Br₂
Number of moles of KBr:
Number of moles = mass/molar mass
Number of moles = 4 g/ 119 gmol
Number of moles = 0.03 mol
Number of moles of Cl₂:
Number of moles = mass/molar mass
Number of moles = 6 g/ 70 gmol
Number of moles = 0.09 mol
Now we will compare the moles of reactant with product.
KBr : KCl
2 : 2
0.03 : 0.03
KBr : Br₂
2 : 1
0.03 : 1/2×0.03= 0.015
Cl₂ : KCl
1 : 2
0.09 : 2/1×0.09 = 0.18
Cl₂ : Br₂
1 : 1
0.09 : 0.09
Less number of moles of product are formed by the KBr thus it will act as limiting reactant while Cl₂ is present in excess.
