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Basile [38]
3 years ago
13

I need help with problem 103 all of the questions.

Chemistry
1 answer:
AfilCa [17]3 years ago
4 0
A. is decomposition so HCL = H2 + Cl2

not balanced cause hcl needs 2

2HCL = H2 + Cl2

balanced

b. Br2 + Al-i = AlBr3 + I2 single rep.

not balanced since br need 3 so watch carefully cause many changes needed


 3Br2 + Al-i = AlBr3 + I2 not right is unbalanced so make it 2

3Br2 + Al-i = 2AlBr3 + I2 now left Al is unbal. so make 2 there

3Br2 + 2Ali = 2AlBr3 + I2 

Balanced

C. Na + S = Na2S synthesis reaction is not bal. left Na needs 2

2Na + S = Na2S  balanced.

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3 0
3 years ago
Std 10th
kondaur [170]

Answer:

zinc and lead or copper and tin

Explanation:

these elements react both as an acid as well as a base

5 0
3 years ago
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shepuryov [24]

Answer:

Explanation:

All three lighter boron trihalides, BX3 (X = F, Cl, Br), form stable adducts with common Lewis bases. Their relative Lewis acidities can be evaluated in terms of the relative exothermicities of the adduct-forming reaction. Such measurements have revealed the following sequence for the Lewis acidity: BF3 < BCl3 < BBr3 (in other words, BBr3 is the strongest Lewis acid).

This trend is commonly attributed to the degree of π-bonding in the planar boron trihalide that would be lost upon pyramidalization (the conversion of the trigonal planar geometry to a tetrahedral one) of the BX3 molecule, which follows this trend: BF3 > BCl3 > BBr3 (that is, BBr3 is the most easily pyramidalized). The criteria for evaluating the relative strength of π-bonding are not clear, however. One suggestion is that the F atom is small compared to the larger Cl and Br atoms, and the lone pair electron in the 2pzorbital of F is readily and easily donated, and overlaps with the empty 2pz orbital of boron. As a result, the [latex]\pi[/latex] donation of F is greater than that of Cl or Br. In an alternative explanation, the low Lewis acidity for BF3 is attributed to the relative weakness of the bond in the adducts F3B-L.

3 0
2 years ago
Read 2 more answers
The three major types of faults are normal, reverse and syncline. true or false
Novosadov [1.4K]
The three major types of faults are Normal, Reverse and Strike-slip faults. 


Answer: FALSE
6 0
3 years ago
Read 2 more answers
At a certain temperature, the solubility of strontium arsenate, sr3(aso4)2, is 0.0480 g/l. what is the ksp of this salt at this
Nadusha1986 [10]
Given the solubility of strontium arsenate is 0.0480 g/l . we have to convert it into mol/L by dividing it over molar mass (540.7 g/mol)
Molar solubility = 0.0480 / 540.7 = 8.9 x 10⁻⁵ mol/L
Dissociation equation:
Sr₃(AsO₄)₂(s) → 3 Sr²⁺(aq) + 2 AsO₄³⁻(aq)
                             3 s                   2 s
Ksp = [Sr²⁺]³ [AsO₄³⁻]²
       = (3s)³ (2s)²
       = 108 s⁵
Ksp = 108 (8.9 x 10⁻⁵) = 5.95 x 10⁻¹⁹ 
8 0
3 years ago
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