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Vika [28.1K]
3 years ago
11

The element X forms ions of charge +2 and the element Y forms ions of charge -3. The most likely formula for a binary compound b

etween these two elements IS:
Chemistry
2 answers:
Iteru [2.4K]3 years ago
6 0

Answer:

X_3Y_2 is the formula.

Explanation:

Given:

Element X forms charge +2.Hence X is the cation and it's valency is 2.

Element Y forms charge -3.Hence Y is the anion and its valency is 3.

Method:

The number of cations = valency of anion

The number of anions = valency of cation.

Hence the correct formula will be X_3Y_2.

tester [92]3 years ago
5 0

Answer:

X3Y2

Explanation:

To represent the chemical formula of the binary compound, the valencies of both elements need to be exchanged.

Valency is the combining power of elements, ions or radicals. It is the value that shows how an element tend to combine with other elements.

In this particular question, the valence of the element X is +2, this means when it enters into chemical combination, it goes in with the power of positive 2. Same goes for the element Y, when it goes into chemical combination, it is expected that it goes in with the negative power of 3.

To completely write the chemical formula of the compound formed from the combination of both, we simply exchange the valency as said above. It must be noted that while the less electronegative element carry a positive or less negative charge , the more electronegative element will carry a negative or more negative charge.

We write the less electronegative element before the more electronegative element while writing the formula of the compound

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Which element will chemically combine with fluorine to combine salt?
Natali5045456 [20]

Answer:

When an atom of sodium and an atom of fluorine combine to form the salt, sodium fluoride, an ionic bond, is formed.

Explanation:

8 0
2 years ago
For each of the following units and concentration values, mention if they are parts per million (ppm), parts per billion (ppb) o
arsen [322]

Answer:

a) ppm

b) ppm

c) ppb

d) ppt

e) ppb

Explanation:

a) You know that 1000 g are 1 kg, and 1000 kg are 1 ton, so (1000)*(1000) g are 1 ton, so 1,000,000 grams are one ton.

b) 1000 mg are 1 g, and 1000 g are 1 liter, so 1,000,000 grams are one liter.

c) You know that 1000 ug are 1 mg, so with the b), we just need to multiply the answer by 1000, so 1,000,000,000 ug are 1 liter.

d) The same as c, 1000 ng are 1 mg. So we are talkinf of ppt.

e) 1000 mg are 1 g. And 1000 g are 1 kg, then 1000 kg are one ton. So 1,000,000,000 mg are one ton.

7 0
3 years ago
3.5g of a Certain compound X, known to be made of carbon, hydrogen, and perhaps oxygen, and to have a molecular molar mass of 15
shutvik [7]

Answer:

C₅H₁₀O₅

Explanation:

1. Calculate the mass of each element in 2.78 mg of X.

(a) Mass of C

\text{Mass of C} = \text{5.13 g CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{1.400 g C}

(b) Mass of H

\text{Mass of H} = \text{2.10 g H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g H$_{2}$O}} = \text{0.2349 g H}

(c) Mass of O

Mass of O = 3.5 - 1.400 - 0.2349 = 1.87 g

2. Calculate the moles of each element

\text{Moles of C = 1400  mg C}\times\dfrac{\text{1 mmol C}}{\text{12.01 mg C }} = \text{116.6 mmol C}\\\\\text{Moles of H = 234.9 mg H} \times \dfrac{\text{1 mmol H}}{\text{1.008 mg H}} = \text{233.1 mmol H}\\\\\text{Moles of O = 1870 mg O} \times \dfrac{\text{1 mmol O}}{\text{16.00 mg O}} = \text{116 mmol O}

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

\text{C: } \dfrac{116.6}{116.6}= 1\\\\\text{H: } \dfrac{233.1}{116.6} = 1.999\\\\\text{O: } \dfrac{116}{116.6} = 1.00

4. Round the ratios to the nearest integer

C:H:O = 1:2:1

5. Write the empirical formula

The empirical formula is CH₂O.

6. Calculate the molecular formula.

EF Mass = (12.01 + 2.016  + 16.00) u  = 30.03 u

The molecular formula is an integral multiple of the empirical formula.

MF = (EF)ₙ

n = \dfrac{\text{MF Mass}}{\text{EF Mass }} = \dfrac{\text{150 u}}{\text{30.03 u}} = 5.00  \approx 5

MF = (CH₂O)₅ = C₅H₁₀O₅

The molecular formula of X is C₅H₁₀O₅.

8 0
3 years ago
If 45.0 mL of a 0.0500 M HNO3, 10.0 mL of a 0.0500 M KSCN, and 30.0 mL of a 0.0500 M Fe(NO3)3 are combined, what is the initial
jeka94

Answer:

the  initial concentration of SCN- in the mixture is 0.00588 M

Explanation:

The computation of the initial concentration of the SCN^- in the mixture is as follows:

As we know that

KSCN \rightarrow K^ + SCN^-

As it is mentioned in the question that KSCN is present 10 mL of 0.05 M

So, the total milimoles of SCN^- is

= 10 × 0.05

= 0.5  m moles

The total volume in mixture is

= 45 + 10 + 30

= 85 mL

Now the initial concentration of the SCN^- is

= 0.5 ÷ 85

= 0.00588 M

hence, the  initial concentration of SCN- in the mixture is 0.00588 M

5 0
3 years ago
A solution may contain Ag+, Pb2+, and/or Hg22+. A white precipitate forms when 6 M HCl is added. The precipitate is partially so
omeli [17]

Answer:

All three are present

Explanation:

Addition of 6 M HCl would form precipitates of all the three cations, since the chlorides of these cations are insoluble: AgCl (s), PbCl_2 (s), Hg_2Cl_2 (s).

  • Firstly, the solid produced is partially soluble in hot water. Remember that out of all the three solids, lead(II) choride is the most soluble. It would easily completely dissolve in hot water. This is how we separate it from the remaining precipitate. Therefore, we know that we have lead(II) cations present, as the two remaining chlorides are insoluble even at high temperatures.
  • Secondly, addition of liquid ammonia would form a precipitate with silver: AgCl (s) + 2 NH_3 (aq) + H_2O (l)\rightarrow [Ag(NH_3)_2]OH (s) + HCl (aq); Silver hydroxide at higher temperatures decomposes into black silver oxide: 2 [Ag(NH_3)_2]OH (s)\rightarrow Ag_2O (s) + H_2O (l) + 4 NH_3 (g).
  • Thirdly, we also know we have Hg_2^{2+} in the mixture, since addition of potassium chromate produces a yellow precipitate: Hg_2^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow Hg_2CrO_4 (s). The latter precipitate is yellow.
3 0
3 years ago
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