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natita [175]
3 years ago
13

6. Identify each element as a metal, metalloid, or nonmetal

Chemistry
1 answer:
zhannawk [14.2K]3 years ago
7 0
A. Metal
B. Metalloid
C. Nonmetal
D. Metal
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The conjugate acid of a particular halide has a pka = 3. a student wants to use this halide in a substitution reaction thinking
Taya2010 [7]

Since the leaving ability of the halide ions increasees as the basicity of the halide decreases.

If the basicity of the halide decreases as the its conjugate acid is strong.

Since the pKa value of conjuage acid of haldie is 3, it is a weak acid. So, it halide is not a good leaving group.

Therefore the answer is No, because a good leaving group is the conjugate base of a strong acid. The halide not acidic enough to be a good leaving group.

5 0
3 years ago
Raising the concentration of hydronium ions, H3O+, in a solution will result in the pH level _________.
algol [13]

Answer:B) Going Down

Explanation:

The higher the concentration of hydro in ions in a solution, the more acidic it is, and the lower the pH is.

(I just answered the question on USA TP!)

3 0
3 years ago
Which statement about the alkali metals is correct?
icang [17]

Answer:

Sofia the firstttdbshsgshaksus

5 0
3 years ago
Which metals may be oxidized by H+ under standard-state conditions? Ag+(aq) + e– → Ag(s) E° = 0.80 V Cu2+(aq) + 2e– → Cu(s) E° =
Debora [2.8K]
Answer is: tin and zinc, because they standard potential as less than zero.
Tin and zinc are oxidized to tin and zinc cations (with +2 charge) and hydrogen anions are reduced to hydrogen molecules with neutral charge.
Zn → Zn²⁺ + 2e⁻; 2H⁺ + 2e⁻ → H₂.
<span>Oxidation is increase of oxidation number  and reduction is decrease of oxidation number.</span>
8 0
3 years ago
How many grams of sulfur trioxide are produced 18 mol O2 react with sufficient sulfur? Show all your work S8 +12O 2&gt; 8SO3
SOVA2 [1]

Answer:

m_{SO_3}=2.31x10^4gSO_3

Explanation:

Hello!

In this case, since the reaction between sulfur and oxygen is:

S_8 +\frac{1}{2} O_2 \rightarrow 8SO_3

Whereas there is a 1/2:8 mole ratio between oxygen and SO3, and we can compute the produced grams of product as shown below:

m_{SO_3}=18molO_2*\frac{8molSO_3}{1/2molO_2} *\frac{80.06gSO_3}{1molSO_3} \\\\m_{SO_3}=23,057.3gSO_3=2.31x10^4gSO_3

Best regards!

7 0
3 years ago
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