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Marizza181 [45]
1 year ago
10

Hydrogen cyanide is poisonous liquid that has a faint, almond-like smell. One molecule of hydrogen cyanide (HCN) is made up of o

ne hydrogen atom, one carbon atom, and one nitrogen atom, shown below. Which of the answer choices shows the correct Lewis dot diagram for hydrogen cyanide?

Chemistry
2 answers:
sergeinik [125]1 year ago
5 0
"<span>H-C=N:" is the one answer among the choices given in the question that shows the correct dot diagram. The correct option among all the options that are given in the question is the fourth option or option "D". The other choices can be neglected. I hope that this is the answer that has come to your help.</span>
Artemon [7]1 year ago
5 0

Answer :  The correct Lewis-dot diagram is shown below.

Explanation :

Lewis-dot structure : Lewis-dot structure shows the bonding between the atoms of a molecule. Lewis-dot structure also shows the unpaired electrons present in the molecule.  In the Lewis-dot structure, the dot represent the number of valence electrons.

The given molecule is, Hydrogen cyanide (HCN)

As we know that carbon has '4' valence electrons, hydrogen has '1' valence electron and nitrogen has '5' valence electrons.

Therefore, the total number of valence electrons in hydrogen cyanide, HCN = 1 + 4 + 5 = 10

According to Lewis-dot structure, there are 10 number of bonding electrons and 0 number of non-bonding electrons.

The Lewis-dot structure of hydrogen cyanide (HCN) is shown below.

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Silver is often extracted from ores such as k[ag(cn)2] and then recovered by the reaction 2k ⎡ ⎣ag(cn)2 ⎤ ⎦(aq) + zn(s) ⟶ 2ag(s)
lions [1.4K]
Take note of the whole number beside each substance in the reaction because these will be used in the stoichiometric calculations below:

Molar mass of <span>k[ag(cn)2]:199 g/mol
Molar mass of Zn(Cn)2: 117.44 g/mol
Avogadro's number: 6.022</span>×10²³ molecules/mol

a.)

35.27 g*(1 mol/199 g)*(1 mol Zn(Cn)₂/ 2 mol K[Ag(CN)₂])*(6.022×10²³ molecules/mol) = <em>5.34×10²² molecules of Zn(Cn)₂</em>

b.)

35.27 g*(1 mol/199 g)*(1 mol Zn(Cn)₂/ 2 mol K[Ag(CN)₂])*(117.44 g/mol) = <em>10.41 g of Zn(Cn)₂</em>
5 0
2 years ago
How many mg does an 830 kg sample contain?
Aleksandr-060686 [28]

Answer:

A sample of 830 kg would contain 830000000 mg

Explanation:

3 0
1 year ago
According to the following reaction, how many moles of Fe(OH)2 can form from 175.0 mL of 0.227 M LiOH solution? Assume that ther
IgorC [24]

Answer:

0.020 moles of Fe(OH)_{2} can be formed

Explanation:

1. First determine the number of moles of LiOH.

Molarity is given by the following expression:

M=\frac{molesofsolute}{Litersofsolution}

Solving for moles of solute:

moles of solute = M * Liters of solution

Converting 175.0mL to L:

175.0mL*\frac{1L}{1000mL}=0.175L

Replacing values:

moles of solute = 0.227M*0.175L

moles of solute = 0.040

Therefore there are 0.040 moles of LiOH

2. Then write the balanced chemical reaction and use the stoichiometry of the reaction to calculate the number of moles of Fe(OH)_{2} produced:

FeCl_{2}(aq)+2LiOH(aq)=Fe(OH)_{2}(s)+2LiCl(aq)

As the problem says that there are excess of FeCl_{2}, the limiting reagent is the LiOH.

0.040molesLiOH*\frac{1molFe(OH)_{2}}{2molesLiOH}=0.020molesFe(OH)_{2} can be formed

3 0
2 years ago
Explain why the spectrum produced by a 1-gram sample of element Z would have the same spectral lines at the same wavelengths as
TEA [102]
The reason why the spectrum produced by a 1 gram sample of element z would have the same spectral lines at the same wavelengths as the spectrum produced by a 2gram is because  Mass does not determine the spectral lines

hope this helps
6 0
2 years ago
Read 2 more answers
When 1.04g of cyclopropane was burnt in excess oxygen in a bomb calorimeter, the temperature rose by 3.69K. The total heat capac
STatiana [176]

Answer:

\Delta _{comb}H=-2,093\frac{kJ}{mol}

Explanation:

Hello!

In this case, since these calorimetry problems are characterized by the fact that the calorimeter absorbs the heat released by the combustion of the substance, we can write:

Q_{rxn}+Q_{cal}=0

Thus, given the temperature change and the total heat capacity, we obtain the following total heat of reaction:

Q_{rxn}=-14.01kJ/K*3.69K\\\\Q_{rxn}=-51.70kJ

Now, by dividing by the moles in 1.04 g of cyclopropane (42.09 g/mol) we obtain the enthalpy of combustion of this fuel:

n=\frac{1.04g}{42.09g/mol}=0.0247mol\\\\\Delta _{comb}H=\frac{Q_{rxn}}{n}\\\\  \Delta _{comb}H=-2,093\frac{kJ}{mol}

Best regards!

4 0
1 year ago
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