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Marrrta [24]
3 years ago
13

A metal (M) forms an oxide with the formula MO. If the oxide contains 7.17 % O by mass, what is the identity of the metal?

Chemistry
1 answer:
Nataly_w [17]3 years ago
7 0

The metal is lead, Pb.

One unit of the oxide contains one atom of O (16.00 u).

∴ Mass of oxide = 16.00 u O × (100 u MO/7.17 u O) = 223.15 u MO

Mass of M = mass of MO – mass of O = 223.15 u -16.00 u = 207.2 u

The only element with an atomic mass of 207.2 u is lead (Pb) and the formula of the oxide is PbO.

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Astatine belongs to which element group: nonmetal, halogen, noble gas?
Mkey [24]

Answer:

halogen

Explanation:

It belongs to Group 17 on the periodic table, which is a halogen

8 0
3 years ago
Enter the ions formed when (NH4)2S dissolves in water.
lisabon 2012 [21]

The ions formed are NH4(+) and S(2-)

The dissolution reaction of (NH4) 2S in water is as follows:


(NH4) 2S ==> 2 NH4 (+) + S (2-).



Ammonium sulfide is the ammonium salt of hydrogen sulfide. It has the formula (NH4) 2S and belongs to the sulfide family.


It is a relatively unstable compound (crystals decomposing at -18 ° C, but exists and is more stable in aqueous solution.) With a pKa exceeding 15, the hydrosulfide ion cannot be significantly deprotonated by ammonia. Thus, such solutions consist mainly of a mixture of ammonia and hydrosulphide of ammonium, it has a smell, close to that of hydrogen sulfide, and its aqueous solutions can be precisely by emitting H2S.

5 0
3 years ago
Read 2 more answers
Jim, Jill, Robert, and Kim each run paper chromatography on an unknown aqueous mixture. Jim gets a red band and a blue band, Jil
Romashka-Z-Leto [24]

Answer:

Robert

Explanation:

There is not more than one colour

4 0
2 years ago
There are four different starting molecules that one might use to synthesize the illustrated alkyl halide as the major product u
Mkey [24]

Answer:

Explanation:

An electrophilic addition reaction occurs when an electrophile attacks a substrate, with the end result being the inclusion of one or many comparatively straightforward molecules along with multiple bonds.

In the given question, the hydrogen bromide provides the electrophile while the bromide is the nucleophile. The mechanism proceeds with the attack of the electrophile on the carbon, followed by deprotonation. This process is continued with a formation of carbocation and the bromide(nucleophile) finally bonds to the carbocation to form a stable product.

The first diagram showcases the possible various starting molecules for the synthesis while the second diagram illustrates their mechanism.

6 0
2 years ago
When 6.0 mol Al react with 13 mol HCl, what is the limiting reactant, and how many moles of H2 can be formed? 2Al + 6HCl → 2AlCl
4vir4ik [10]

Well, we need to find the ratio of Al to the other reactant.


Al:HCl = 1:3


--> this means that for every 1 Al used, you have to use 3 HCl.



6*3 = 18 moles of HCl needed to fully react with 6 moles of Al. Since 13<18, HCL is the limiting reactant.



The ratio of HCl:AlCl = 3:1



13/3 = 4.3333...



The final answer is HCl is the limiting reactant with 4.3 moles of AlCl3 able to be produced.



Hope this helps!!! :)


6 0
2 years ago
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