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prisoha [69]
1 year ago
5

5. given the ratio of reactants, what is the possibility of obtaining di- and polychlorinated product? explain.

Chemistry
1 answer:
Vladimir79 [104]1 year ago
3 0

The ratio of reactants is chlorination of <u>2,3</u> dimethyl butane the possibility of obtaining do and the polychlorinated product is not seen.

When a mixture of methane and chlorine is exposed to ultraviolet light a substitution reaction occurs and the organic product is chloromethane. Because there are various hydrogen atoms that can be extracted in the first propagation step.

Abstraction of a hydrogen atom from the middle carbon of propane results in 2-chloropropane. In the presence of sunlight, methane reacts with chlorine to form chloromethane. The chlorination of methane is a free radical substitution reaction. Chlorine cannot turn into free radicals in the dark, so no reaction takes place. Therefore, the presence of sunlight is essential for the reaction to proceed.

Learn more about The reactants here:- brainly.com/question/6421464

#SPJ4

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134 grams of nitric acid is added to 512 grams of water. Calculate the molality of nitric acid.
Nana76 [90]

Answer:   4.15234 m

512 g H2O * \frac{1 kg}{1000 g} = 0.512 kg H2O

Nitric Acid:  HNO3 = 1.008 + 14.007 + 3(15.999) = 63.012 g/mol

H = 1.008 g/mol

N = 14.007 g/mol

O3 = 3*15.999

134 g HNO₃ * \frac{mol}{63.012 g} = 2.126 mol

m = \frac{2.126  mol}{0.512  kg} = 4.15234 m

6 0
3 years ago
Read 2 more answers
Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reactionP4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°r
weqwewe [10]

Answer:

Therefore  \bigtriangledown H^\circ_{rxn}= -1835 KJ

Explanation:

Enthalpy is denoted by H.

Enthalpy: Total heat change in a chemical reaction is called enthalpy.

The change of entalpy of a reaction is denoted by \bigtriangledown H^\circ_{rxn}

Hass's Law:The change in enthalpy of any process can be determined by calculating the sum of change in enthalpy of each of the steps involved in the process.

g= gas

S= solid

P₄(g)+10Cl₂(g)→ 4Cl₅(s)       \bigtriangledown H^\circ_{rxn}=?

PCl₅(s)→ PCl₃(g)+Cl₂(g) .......(1)       \bigtriangledown H^\circ_{rxn}= +157KJ

P₄(g)+6Cl₂(g)→  4PCl₃(g).............(2)     \bigtriangledown H^\circ_{rxn}= -1207 KJ

If we flip a reaction the value of enthalpy will be change positive to negative or nagative to positive but the numerical value will be remain same.

We need rearrange the equation (1) because in the required equation Cl₂ is on the left side. So we flip the first equation.

PCl₃(g)+Cl₂(g)→PCl₅(s)......(3)          \bigtriangledown H^\circ_{rxn}= -157KJ

Multiplying 4 with equation (3)

4 PCl₃(g)+4Cl₂(g)→4PCl₅(s)......(4)          \bigtriangledown H^\circ_{rxn}=4×( -157)KJ= -628 KJ

Adding equation (2) and (4) we get

P₄(g)+6Cl₂(g)+4 PCl₃(g)+4Cl₂(g)→4PCl₃(g)+4PCl₅(s)    \bigtriangledown H^\circ_{rxn}=( -1207-628)KJ

⇒P₄(g)+10Cl₂(g)→4PCl₃(g)-4PCl₃(g)+4PCl₅(s)      \bigtriangledown H^\circ_{rxn}= - 1835KJ

⇒P₄(g)+10Cl₂(g)→ 4Cl₅(s)       \bigtriangledown H^\circ_{rxn}= -1835 KJ

Therefore  \bigtriangledown H^\circ_{rxn}= -1835 KJ

5 0
2 years ago
Which substance contains a bond with the greatest ionic character?
agasfer [191]
Hello!

The correct answer is 1. KCI.

I really hope this helped you out! c:
4 0
3 years ago
Which of the following best helps to explain why the electron affinity of Br has a greater magnitude than that of I?
tankabanditka [31]

The reason why  Br has a greater magnitude of electron affinity than that of I is that there is a greater attraction between an added electron and the nucleus in Br than in I.

In the periodic table, there are trends that increase down the group and across the period. Electron affinity is a trend that increases across the period but decreases down the group.

Recall that the ability of an atom to accept an electron depends on the size of the atom. The smaller the atom, the greater the attraction between an added electron and the nucleus.

Since Br is smaller than I, there is a greater attraction between an added electron and the nucleus in Br than in I which explains why Br has a greater magnitude of electron affinity than  I.

Learn more: brainly.com/question/17696329

7 0
2 years ago
When 1. 0 l of 0. 00010 m naoh and 1. 0 l of 0. 0014 m mgso4 are mixed, would a precipitate be formed? show work
koban [17]

When 1. 0 l of 0. 00010 m NaOH and 1. 0 l of 0. 0014 m mgso4 are mixed, there will be no precipitate formed.

<h3>What is a precipitate?</h3>

The precipitate is the solid concentration of a substance that is collected over a solution.

First, we determine the concentration of magnesium and hydroxide

(Mg2+) = 7.00 × 10⁻⁴

(OH−) = 5.00 × 10⁻⁵

Now, we calculate the solubility quotient

Qc = (Mg2+) (OH−) ²

Qc = 7.00 × 10⁻⁴ x (5.00 × 10⁻⁵)²

Qc = 1.75 x 10⁻¹²

The solubility product of the magnesium hydroxide is 1.80 x 10⁻¹¹ which is more than the solubility quotient. Thus, there will be no precipitate form.

Thus, there will be no precipitate formed because the solubility quotient we calculated is less than the solubility product.

To learn more about precipitate, refer to the below link:

brainly.com/question/16950193

#SPJ4

5 0
1 year ago
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