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3 years ago

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Describe use of H₂S an analytical reagent?

1 answer:

Gemiola [76]3 years ago

7 0

Answer:

The main use for hydrogen sulfide is in the production of sulfuric acid and elemental sulfur. ... H2S is used to prepare the inorganic sulfides you need to make those products. As a reagent and intermediate, hydrogen sulfide is beneficial because it can prepare other types of reduced sulfur compounds.

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There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with

grandymaker [24]

$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$ .

Assuming complete decomposition of both samples,

$m(\text{Hg}) = m(\text{residure})$
$m(\text{O}) = m(\text{loss})$

First compound:

$m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g$

$n = m/M$ ; $0.6498 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.048 \; g / 16 \; g \cdot mol^{-1}= 0.003 \; mol$
$n(\text{Hg atoms}) = 0.6018 \; g / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol$

Oxygen and mercury atoms seemingly exist in the first compound at a $1:1$ ratio; thus the empirical formula for this compound would be $\text{Hg} \text{O}$ where the subscript "1" is omitted.

Similarly, for the second compound

$m(\text{O}) = m(\text{loss}) = 0.016 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012 \; g$

$n = m/M$ ; $0.4172 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.016 \; g / 16 \; g \cdot mol^{-1}= 0.001 \; mol$
$n(\text{Hg atoms}) = 0.4012 \; g / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol$

$n(\text{Hg}) : n(\text{O}) \approx 2:1$ and therefore the empirical formula

$\text{Hg}_{2} \text{O}$ .

8 0

3 years ago

Nah I'm gonna fail this quiz help please

Eva8 [605]

Answer:

It has 2 eyes

Explanation:

Quantitive observation means that it is facts based of numbers, therefore you can count how many eyes it has, but not that it has red eyes

4 0

3 years ago

How much cooking liquid is needed when employing the pasta method?

Sonbull [250]

A l l of IT? idk im 7

8 0

3 years ago

Read 2 more answers

Which states of matter can form solutions?

IrinaVladis [17]

Answer:

c

Explanation:

spongebob cheese pants

8 0

3 years ago

Read 2 more answers

Determine the nuclear composition (number of protons and neutrons) of the following isotopes. (a) chromium-52 protons neutrons (

madreJ [45]

Answer: a) chromium-52 : protons = 24, neutrons = 28

selenium - 80 : protons = 34, neutrons = 46

molybdenum-98: protons = 42, neutrons = 56

xenon-132: protons = 54, neutrons = 78

ytterbium-174 : protons = 70, neutrons = 104

Explanation:

Atomic number : It is defined as the number of electrons or number of protons present in a neutral atom.

Thus, number of protons = atomic number

Mass number is the number of the entities present in the nucleus which is the equal to the sum of the number of protons and electrons.

Mass number = Number of protons + Number of neutrons

a) chromium-52 :

Atomic number of Chromium is 24 hence number of protons = 24

52= 24+ Number of neutrons

Number of neutrons = 28

b) selenium - 80 :

Atomic number of selenium is 34 hence number of protons = 34

80 = 34 + Number of neutrons

Number of neutrons = 46

c) molybdenum - 98 :

Atomic number of molybdenum is 42 is hence number of protons = 42

98 = 42 + Number of neutrons

Number of neutrons = 56

d) xenon-132:

Atomic number of xenon is 54 hence number of protons = 54

132 = 54 + Number of neutrons

Number of neutrons = 78

e) ytterbium-174:

Atomic number of ytterbium is 70 and hence number of protons = 70

174 = 70 + Number of neutrons

Number of neutrons = 104

6 0

3 years ago