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VARVARA [1.3K]
2 years ago
8

Number of paired electrons in Li

Chemistry
2 answers:
inessss [21]2 years ago
8 0
In the lithium atom the two electrons in the 1s orbital are paired, and the electron in the 2s orbital is unpaired.
Hunter-Best [27]2 years ago
6 0
It has 3 electrons
2 are pair
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2-bromo-3,4-dimethylpentane is combined with t-butoxide. The product of this reaction is.
Andrej [43]

2-bromo-3,4-dimethylpentane is combined with t-butoxide. The product of this reaction is 3,4 dimethyl - 1- pentene.  

The reaction of 2-bromo-3,4-dimethylpentane is combined with t-butoxide forms 2 alkene in the elimination reaction due to steric hindrance.  The least stable alkene 3,4 dimethyl - 1- pentene is easy to make. the  t-butoxide is (CH₃)₃CO⁻. The reaction involves in this reaction is E2 elimination reaction. This reaction involves the one step reaction. The product will also form that is 3,4 dimethyl - 2 - pentene.  so the reaction involve Elimination reaction and the product due to steric hindrance is 3,4 dimethyl - 1- pentene

Thus, 2-bromo-3,4-dimethylpentane is combined with t-butoxide. The product of this reaction is 3,4 dimethyl - 1- pentene.  

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4 0
1 year ago
Will a precipitate form from Pb(NO3)2(aq) and HCl(aq)?
Rus_ich [418]
1. Write out the formula Pb(NO3)2 (aq) + 2HCl (aq) ----> PbCl2 + 2HNO3 2. Use solubility guidelines (gotta memorize 'em) for the products to see if a solid forms Nitrates are always soluble so 2HNO3 (aq) Chlorides (Cl) are always soluble except for when you mix them with copper, lead, mercury, or silver. Since you mixed it with lead (Pb) it is solid and forms a precipitate. PbCl2 (s)
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Im confused! Can someone please show me the step by step process??
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3 years ago
Calculate the molarity of a solution prepared by dissolving 11.9 g of beryllium chloride in a total volume of 292.2 mL of soluti
anyanavicka [17]

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8793

Explanation:

8 0
3 years ago
How many grams of PbBr2 will precipitate when excess CuBr2 solution is added to 77.0 mL of 0.595 M Pb(NO3)2 solution?Pb(NO3)2(aq
Alexxx [7]

Answer:

16.89g of PbBr2

Explanation:

First, let us calculate the number of mole of Pb(NO3)2. This is illustrated below:

Molarity of Pb(NO3)2 = 0.595M

Volume = 77mL = 77/1000 = 0.077L

Mole =?

Molarity = mole/Volume

Mole = Molarity x Volume

Mole of Pb(NO3)2 = 0.595x0.077

Mole of Pb(NO3)2 = 0.046mol

Convert 0.046mol of Pb(NO3)2 to grams as shown below:

Molar Mass of Pb(NO3)2 =

207 + 2[ 14 + (16x3)]

= 207 + 2[14 + 48]

= 207 + 2[62] = 207 +124 = 331g/mol

Mass of Pb(NO3)2 = number of mole x molar Mass = 0.046 x 331 = 15.23g

Molar Mass of PbBr2 = 207 + (2x80) = 207 + 160 = 367g/mol

Equation for the reaction is given below:

Pb(NO3)2 + CuBr2 —> PbBr2 + Cu(NO3)2

From the equation above,

331g of Pb(NO3)2 precipitated 367g of PbBr2

Therefore, 15.23g of Pb(NO3)2 will precipitate = (15.23x367)/331 = 16.89g of PbBr2

8 0
3 years ago
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