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Firlakuza [10]
3 years ago
9

Write the formulas for the following compounds: (a) rubidium nitrite, (b) potassium sulfide, (c) sodium hydrogen sulfide, (d) ma

gnesium phosphate, (e) calcium hydrogen phosphate, (f) lead(ll) carbonate, (g) tin(ll) fluoride, (h)ammonium sulfate, (i) silver perchlorate, (j) boron trichloride.
Chemistry
1 answer:
choli [55]3 years ago
6 0

Answer:

All are having different valent cation and anion like mono,di and trivalent polyatomic ions .

A. RbNO3

B. K2S

C. NaHS

D. Mg3(PO4)2 formed by divalent Mg+2 and trivalent PO43-

E. CaHPO4

F. PbCO3 , lead is in Pb+2 form

G. SnF2

H. (NH4)2SO4

I. AgClO4

J. BCl3

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18 An important environmental consideration is the appropriate disposal of cleaning solvents. An environmental waste treatment c
Katyanochek1 [597]

Answer:

a) Percentage by mass of carbon: 18.3%

   Percentage by mass of hydrogen: 0.77%

b)  Percentage by mass of chlorine: 80.37%

c) Molecular formula: C_{2} H Cl_{3}

Explanation:

Firstly, the mass of carbon must be determined by using a conversion factor:

0.872g CO _{2} *\frac{12g C}{44g CO_{2} } = 0.238g CO_{2}

The same process is used to calculate the amount of hydrogen:

0.089g H_{2}O*\frac{2g H}{18g H_{2}O }  = 0.010g H

The percentage by mass of carbon and hydrogen are calculated as follows:

%C\frac{0.238g}{1.3g} *100%= 18.3%

%H\frac{0.010g}{1.3g} *100%=0.77%

From the precipation data it is possible obtain the amount of chlorine present in the compound:

1.75 AgCl*\frac{35.45g Cl}{143.45g AgCl}= 0.43g AgCl

Let's calculate the percentage by mass of chlorine:

%Cl=\frac{0.43g}{0.535g} * 100%= 80.37%

Assuming that we have 100g of the compound, it is possible to determine the number of moles of each element in the compound:

18.3g C*\frac{1mol C}{12g C} = 1.52mol C

0.77g H*\frac{1mol H}{1g H} = 0.77mol H

80.37gCl*\frac{1molCl}{35.45g Cl} = 2.27mol Cl

Dividing each of the quantities above by the smallest (0.77mol), the  subscripts in a tentative formula would be

C=\frac{1.52}{0.77} = 1.97 ≈ 2

H = \frac{0.77}{0.77} = 1

Cl =\frac{2.27}{0.77}=2.94≈3

The empirical formula for the compound is:

C_{2} H Cl_{3}

The mass of this empirical formula is:

mass of C + mass of H + mass of Cl= 24g +1+ 106.35 =131.35g

This mass matches with the molar mass, which means that the supscript in the molecular formula are the same of the empirical one.

5 0
3 years ago
as the elements period 3 are considered in order of increasing atomic number, the number of principal energy levels in each succ
Diano4ka-milaya [45]

Answer:

stay the same.

Explanation:  Period 3 consists of the full 1s, 2s, and 2p electron orbitals, plus the 3s and 3p valence orbitals, which are filled with a total of 8 more electrons as we move from left (Na) to the far right (Ar):

Na:  1s2 2s2 2p6 3s1

Ar:    s2 2s2 2p6 3s2 3p6

As we move from left to right, and ignoring the already-filled 1s, 2s, and 2p orbitals, the period three starting and ending elements have the following:

Na: 3s1

Ar: 3s2, 3p6

All the new electrons electrons filled the third energy level (3s and 3p).  So the energy level does not change, just the orbitals.

5 0
2 years ago
Question 4
vazorg [7]

Answer:

H(aq) + NO3 (aq) + HF(aq)

Explanation:

In the given mixture of HNO3 (Nitric Acid) and HF (hydrofluoric acid) in water  the major species present are H(aq) + NO3 (aq) + HF(aq).

On the reaction of  HNO3 (Nitric Acid) and HF (hydrofluoric acid) in water  , it will give a polar solution and will form a homogenous mixture.

Hence, the correct answer is "H(aq) + NO3 (aq) + HF(aq)".

4 0
2 years ago
A 2.4L balloon filled with helium at room temperature 25oC is put into liquid nitrogen
Stels [109]

Answer:

V₂ = 0.62 L

Explanation:

Given data:

Initial volume = 2.4 L

Initial temperature = 25°C

Final temperature = -196°C

Final volume = ?

Solution:

Initial temperature = 25°C (25+273 = 298 K)

Final temperature = -196°C ( -196+273 = 77 K)

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 2.4 L × 77 K / 298 k

V₂ = 184.8 L.K / 298 K

V₂ = 0.62 L

5 0
2 years ago
Weight and mass are essentially the same measurements.<br> True or false?
Serhud [2]

The answer to this question is False

7 0
3 years ago
Read 2 more answers
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