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yaroslaw [1]
2 years ago
10

Which of the molecules represented below contains carbon with sp2 hybridization?

Chemistry
1 answer:
Darya [45]2 years ago
3 0

Answer:

Explanation:

sp² hybridization is found in those compounds having double bond .

Out of the given compounds only C₂H₂Cl₂ has double bond so this compound contains carbon with sp² hybridization .

Rest have sp³ hybridization because they are saturated compounds .

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Paige heated 3.00 g mercury (II) oxide (HgO, 216.59 g/mol) to form mercury (Hg, 200.59 g/mol) and oxygen (O2, 32.00 g/mol). She
stepan [7]

the balanced chemical equation for decomposition of HgO is as follows

2HgO --> 2Hg + O₂

stoichiometry of HgO to O₂ is 2:1

number of HgO moles heated are - 3.00 g / 216.59 g/mol = 0.0139 mol

according to stoichiometry of reaction -

number of O₂ moles formed = 0.0139 mol/ 2 = 0.00695 mol

mass of O₂ to be formed - 0.00695 mol x 32.00 g/mol = 0.2224 g

but the actual yield = 0.195 g

percent yield = actual yield / theoretical yield x 100 %

percent yield = 0.195 g / 0.2224 g x 100 % = 87.7 %

answer is 87.7 %

8 1
2 years ago
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A 50.0 mL sample of 0.600 M calcium hydroxide is mixed with 50.0 mL sample of 0.600 M hydrobromic acid in a Styrofoam cup. The t
TEA [102]

Explanation:

The reaction equation will be as follows.

     Ca(OH)_{2}(aq) + 2HBr(aq) \rightarrow CaBr_{2}(aq) + 2H_{2}O(l)

So, according to this equation, 1 mole Ca(OH)_{2} = 2 mol HBr = 1 mol CaBr_{2}

Therefore, calculate the number of moles of calcium hydroxide as follows.

     No. of moles of Ca(OH)_{2} = V \times Molarity

                                    = 50 \times 0.6

                                    = 30 mmol

Similarly, calculate the number of moles of HBr as follows.

        No. of moles of HBr = M \times V

                                          = 50 \times 0.6

                                          = 30 mmol

This means that the limiting reactant is HBr.

So, no. of moles of CaBr_{2} = 30 \times \frac{1}{2}

                                                     = 15 mmol

Hence, calculate the amount of heat released as follows.

                Heat released in the reaction(q) = m \times s \times \Delta T

as,    m = mass of solution

and,             Density = \frac{mass}{volume}

or,                  mass = Density × Volume

                               = 1.08 g/ml \times (50 + 50) ml

                               = 108 g

where,    s = specific heat of solution = 4.18 j/g.k

and,        change in temperature \Delta T = (26 - 23)^{o}C

                                                                 = 3
^{o}C

Hence, the heat released will be as follows.

                   q = m \times s \times \Delta T

                        q = 108 \times 4.18 \times 3^{o}C

                           = 1354.32 joule

or,                        = 1.354 kJ       (as 1 kJ = 1000 J)    

Also,          \Delta H_{rxn} = \frac{-q}{n}

                              = \frac{-1.354}{15 \times 10^{-3}}

                              = -90.267 kJ/mol

Thus, we can conclude that the enthalpy change for the given reaction is -90.267 kJ/mol.

6 0
2 years ago
Which formula represents the compound commonly known as phosphine <br> PH<br> PH2<br> PH3<br> PH4
AveGali [126]
The answer is C). PH3
4 0
2 years ago
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Americium-242 has a half-life of 6 hours. If you started with 24 g and you now have 3 g, how much time
kogti [31]
How many times has it halved?

24/2 = 12
12/2 = 6
6/2 = 3

It halved three times.
It halves once every 6 hours.

18 hours have passed.
7 0
2 years ago
What mass of nickel (Ni) is in a 2.4 Kg sample of propanol if the concentration is 20 ppb ? (atomic mass of Ni = 58.69)
asambeis [7]

Answer:

The mass of nickel is 48μg

Explanation:

Parts per billion is a way to describe small concentrations and is defined as the ratio between μg of solute and kg of solvent.

If a solution of nickel in propanol is 20ppb, contains 20μg of nickel in 1 kg of propanol.

Thus, a sample of 2.4kg of propanol will contain:

2.4kg × (20μg nickel / 1kg) = 48μg nickel

<h3>The mass of nickel is 48μg</h3>
8 0
2 years ago
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