Answer:
(D)
Explanation:
Acids are the species which furnish hydrogen ions in the solution or is capable of forming bonds with electron pair species as they are electron deficient species.
When an acid donates a proton, it changes into a base which is known as its conjugate base.
Bases are the species which furnish hydroxide ions in the solution or is capable of forming bonds with electron deficient species as they are electron rich species. When a base accepts a proton, it changes into a acid which is known as its conjugate acid.
The acid and the base which is only differ by absence or presence of the proton are known as acid conjugate base pair.
Thus,
(A)
does not have any conjugate base.
(B)
is a base and does not have any conjugate base.
(C) The conjugate base of the acid,
is
but not vice versa.
(D) The conjugate base of the acid,
is
(E)
is a acid conjugate base pair.
Hence, (D) is the answer.
This question is providing two ions, Sn⁴⁺ and F¹⁻ so that the compound they form is asked to be named. At the end, according to the IUPAC rules, the name will be B: Tin (IV) fluoride.
<h3>Nomenclature:</h3>
In chemistry, nomenclature is used to assign the proper name of a chemical compound, so that it can be recognized in any part of the world. In this particular case, since this is a binary salt with a cation that has two oxidation states, Sn⁴⁺ and Sn²⁺, we should assign the correct roman numeral, which is tin (VI).
On the other hand, since fluorine takes a negative charge when forming binary salts, as a 1-, its assigned name would be fluoride. In such a way, by combining them, we find the name as B: Tin (IV) fluoride.
Learn more about nomenclature: brainly.com/question/6496952
Answer:
333.3mL
Explanation:
Using the formula as follows:
C1V1 = C2V2
Where;
C1 = initial concentration (M)
C2 = final concentration (M)
V1 = initial volume (mL)
V2 = final volume (mL)
According to the information provided in this question,
C1 = 4.00M
C2 = 1.50M
V1 = 125mL
V2 = ?
Using C1V1 = C2V2
4 × 125 = 1.5 × V2
500 = 1.5V2
V2 = 500/1.5
V2 = 333.3mL
Therefore, the CuSO4 solution needs to be diluted to 333.3mL to make 1.50 M solution.
Answer: 1. ![NH_3=82.4\%](https://tex.z-dn.net/?f=NH_3%3D82.4%5C%25)
2. ![CO(NH_2)_2=46.7\%](https://tex.z-dn.net/?f=CO%28NH_2%29_2%3D46.7%5C%25)
3. .
4.
has highest nitrogen content.
Explanation:
To calculate the mass percent of element in a given compound, we use the formula:
1.
![\text{Mass percent of nitrogen}=\frac{\text{Mass of nitrogen}}{\text{Mass of} NH_3}=\frac{14}{17}\times 100=82.4\%](https://tex.z-dn.net/?f=%5Ctext%7BMass%20percent%20of%20nitrogen%7D%3D%5Cfrac%7B%5Ctext%7BMass%20of%20nitrogen%7D%7D%7B%5Ctext%7BMass%20of%7D%20NH_3%7D%3D%5Cfrac%7B14%7D%7B17%7D%5Ctimes%20100%3D82.4%5C%25)
2.
![\text{Mass percent of nitrogen}=\frac{\text{Mass of nitrogen}}{\text{Mass of} CO(NH_2)_2}=\frac{14\times 2}{12\times 1+16\times 1+2\times 14+4\times 1}\times 100=\frac{28}{60}\times 100=46.7\%](https://tex.z-dn.net/?f=%5Ctext%7BMass%20percent%20of%20nitrogen%7D%3D%5Cfrac%7B%5Ctext%7BMass%20of%20nitrogen%7D%7D%7B%5Ctext%7BMass%20of%7D%20CO%28NH_2%29_2%7D%3D%5Cfrac%7B14%5Ctimes%202%7D%7B12%5Ctimes%201%2B16%5Ctimes%201%2B2%5Ctimes%2014%2B4%5Ctimes%201%7D%5Ctimes%20100%3D%5Cfrac%7B28%7D%7B60%7D%5Ctimes%20100%3D46.7%5C%25)
3.
![\text{Mass percent of nitrogen}=\frac{\text{Mass of nitrogen}}{\text{Mass of} NH_4NO_3}=\frac{14\times 2}{14\times 2+16\times 3+4\times 1}\times 100=\frac{28}{80}\times 100=35\%](https://tex.z-dn.net/?f=%5Ctext%7BMass%20percent%20of%20nitrogen%7D%3D%5Cfrac%7B%5Ctext%7BMass%20of%20nitrogen%7D%7D%7B%5Ctext%7BMass%20of%7D%20NH_4NO_3%7D%3D%5Cfrac%7B14%5Ctimes%202%7D%7B14%5Ctimes%202%2B16%5Ctimes%203%2B4%5Ctimes%201%7D%5Ctimes%20100%3D%5Cfrac%7B28%7D%7B80%7D%5Ctimes%20100%3D35%5C%25%20)
4.
![\text{Mass percent of nitrogen}=\frac{\text{Mass of nitrogen}}{\text{Mass of}(NH_4)_2SO_4}=\frac{14\times 2}{14\times 2+16\times 4+32\times 1+8\times 1}\times 100=\frac{28}{132}\times 100=21.2\%](https://tex.z-dn.net/?f=%5Ctext%7BMass%20percent%20of%20nitrogen%7D%3D%5Cfrac%7B%5Ctext%7BMass%20of%20nitrogen%7D%7D%7B%5Ctext%7BMass%20of%7D%28NH_4%29_2SO_4%7D%3D%5Cfrac%7B14%5Ctimes%202%7D%7B14%5Ctimes%202%2B16%5Ctimes%204%2B32%5Ctimes%201%2B8%5Ctimes%201%7D%5Ctimes%20100%3D%5Cfrac%7B28%7D%7B132%7D%5Ctimes%20100%3D21.2%5C%25)
Thus
has highest nitrogen content of 82.4%.
Given :
A balanced chemical equation :
![Zn +Cl_2->ZnCl_2](https://tex.z-dn.net/?f=Zn%20%2BCl_2-%3EZnCl_2)
To Find :
How many grams of zinc are needed to produce 12 grams of zinc chloride.
Solution :
Moles of
,
![n=\dfrac{Given \ wt}{Molecular\ Mass}\\\\n =\dfrac{12}{136.30}\ mol\\\\n=0.088\ mol](https://tex.z-dn.net/?f=n%3D%5Cdfrac%7BGiven%20%5C%20wt%7D%7BMolecular%5C%20Mass%7D%5C%5C%5C%5Cn%20%3D%5Cdfrac%7B12%7D%7B136.30%7D%5C%20mol%5C%5C%5C%5Cn%3D0.088%5C%20mol)
Now, by balanced chemical equation we can say that 1 mol of Zn produce
1 mol of
.
So, 0.088 mol of Zn is required to produced 0.088 mol of
.
![Mass \ required = molecular \ mass \times moles\\\\m = 65.38 \times 0.088\\\\m=5.8 \ gm](https://tex.z-dn.net/?f=Mass%20%5C%20required%20%3D%20molecular%20%5C%20mass%20%5Ctimes%20moles%5C%5C%5C%5Cm%20%3D%2065.38%20%5Ctimes%200.088%5C%5C%5C%5Cm%3D5.8%20%5C%20gm)
Therefore, 5.8 grams of zinc is required.
Hence, this is the required solution.