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DIA [1.3K]
3 years ago
10

An unknown solid is entirely soluble in water. On addition of dilute HCl, a precipitate forms. After the precipitate is filtered

off, the pH is adjusted to about 1 and H2S is bubbled in; a precipitate again forms. After filtering off this precipitate, the pH is adjusted to 8 and H2S is again added; no precipitate forms. No precipitate forms upon addition of (NH4)2HPO4. The remaining solution shows a yellow color in a flame test.
Based on these observations, which of the following compounds might be present, which are definitely present, and which are definitely absent.
CdS, Pb(NO3)2, HgO, ZnSO4, Cd(NO3)2, and Na2SO4
Chemistry
1 answer:
Karo-lina-s [1.5K]3 years ago
5 0

Answer:

Pb(NO3)2

Cd(NO3)2

Na2SO4

Explanation:

In the first part, addition of HCl leads to the formation of PbCl2 which is poorly soluble in water. This is the first precipitate that is filtered off.

When the pH is adjusted to 1 and H2S is bubbled in, CdS is formed. This is the second precipitate that is filtered off.

After this precipitate has been filtered off and the pH is adjusted to 8, addition of H2S and (NH4)2HPO4 does not lead to the formation of any other precipitate.

The yellow flame colour indicates the presence of Na^+ which must come from the presence of Na2SO4.

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A liquid mixture composed of 20% CH4, 30% C2H4, 35% C2H2, and 15% C2H20. What is the average molecular weight of the mixture? a)
blondinia [14]

<u>Answer:</u> The correct answer is Option c.

<u>Explanation:</u>

We are given:

Mass percentage of CH_4 = 20 %

So, mole fraction of CH_4 = 0.2

Mass percentage of C_2H_4 = 30 %

So, mole fraction of C_2H_4 = 0.3

Mass percentage of C_2H_2 = 35 %

So, mole fraction of C_2H_2 = 0.35

Mass percentage of C_2H_2O = 15 %

So, mole fraction of C_2H_2O = 0.15

We know that:

Molar mass of CH_4 = 16 g/mol

Molar mass of C_2H_4 = 28 g/mol

Molar mass of C_2H_2 = 26 g/mol

Molar mass of C_2H_2O = 48 g/mol

To calculate the average molecular mass of the mixture, we use the equation:

\text{Average molecular weight of mixture}=\frac{_{i=1}^n\sum{\chi_im_i}}{n_i}

where,

\chi_i = mole fractions of i-th species

m_i = molar masses of i-th species

n_i = number of observations

Putting values in above equation:

\text{Average molecular weight}=\frac{(\chi_{CH_4}\times M_{CH_4})+(\chi_{C_2H_4}\times M_{C_2H_4})+(\chi_{C_2H_2}\times M_{C_2H_2})+(\chi_{C_2H_2O}\times M_{C_2H_2O})}{4}

\text{Average molecular weight of mixture}=\frac{(0.20\times 16)+(0.30\times 28)+(0.35\times 26)+(0.15\times 42)}{4}\\\\\text{Average molecular weight of mixture}=6.75

Hence, the correct answer is Option c.

3 0
4 years ago
A 211 g sample of barium carbonate, baco3, reacts with a solution of nitric acid to give barium nitrate, carbon dioxide, and wat
bija089 [108]
Answer:
            Mass  =  47.04 g 

            Volume  =  23.94 L 

Solution:

The equation for given reaction is as follow,

                  BaCO₃  +  2 HNO₃     →     Ba(NO₃)₂  +  CO₂  +  H₂O

According to this equation,

            197.34 g (1 mole) BaCO₃ produces  =  44 g (1 mole) of CO₂
So,
                       211 g of BaCO₃ will produce  =  X g of CO₂

Solving for X,
                     X  =  (211 g × 44 g) ÷ 197.34 g

                     X  =  47.04 g of CO₂

As we know,

                  44 g (1 mole) CO₂ at STP occupies  =  22.4 L volume
So,
                                47.04 g of CO₂ will occupy  =  X L of Volume

Solving for X,
                     X  =  (47.04 g × 22.4 L) ÷ 44 g

                     X  =  23.94 L Volume
8 0
4 years ago
How many bonding and nonbonding electrons are on the carbon atom in CO2?
il63 [147K]

Answer:

I believe the answer is B

Explanation:

to be honest i'm not completely sure, sorry

4 0
3 years ago
Read 2 more answers
What is the difference between an atom and a compound?
puteri [66]
An atom cannot be broken down any smaller whereas a compound is made up of atoms and can be broken down into smaller pieces (the individual atoms that make it up)
Hope this helps!
6 0
4 years ago
Read 2 more answers
-HELP-<br>How many molecules are contained in 125 grams of oxygen gas (O2)?
strojnjashka [21]

Answer:

62.50

Explanation:

I divided the 125 in the result is 62.50 but it is considered a 62.

never mine:)

3 0
3 years ago
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