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sergiy2304 [10]
2 years ago
15

A sample of ch4o with a mass of 32.0 g contains __________ molecules of ch4o. group of answer choices 5.32 ´ 10-23 1.00 1.88 ´ 1

022 6.02 ´ 1023 32.0
Chemistry
1 answer:
LekaFEV [45]2 years ago
3 0

The correct option is (C) 6.02 X 10²³

A sample of CH₄O with a mass of 32.0 g contains <u>6.02 X 10²³</u> molecules of CH₄O.

To calculate the number of moles;

Molar mass of CH₄O = C + 4(H) + O

          = 12.01 + 4(1.008) + 16

          = 32.04 g/mol

So, 1 mol of CH₄O = 32.04 g of CH₄O

Given, 32.0 g of CH₄O

According to Avagadro's constant 1 mole of a substance contains 6.022× 10^23 particles (molecules, atoms or ions).

 = (32.0 g/1)(1 mol CH₄O/32.04 g CH₄O)(6.02x10²³/1 mol CH₄O)

 = 6.02 X 10²³ molecules of CH₄O

Hence, a sample of  will contain  number of molecules 6.02 X 10²³ molecules.

Learn more about the Moles calculation with the help of the given link:

brainly.com/question/21085277

#SPJ4

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A double displacement reaction in which one of the product is formed as a solid is called as precipitation reaction.  

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K_2SO_4(aq)+BaCl_2(aq)\rightarrow BaSO_4(s)+2KCl(aq)

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<u>Answer:</u> The heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

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The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(1\times \Delta H_{(C_4H_{10})})]-[(1\times \Delta H_{(C_4H_6)})+(2\times \Delta H_{(H_2)})]

We are given:

\Delta H_{(C_4H_{10})}=-2877.6kJ/mol\\\Delta H_{(C_4H_6)}=-2540.2kJ/mol\\\Delta H_{(H_2)}=-285.8kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(1\times (-2877.6))]-[(1\times (-2540.2))+(2\times (-285.8))]\\\\\Delta H_{rxn}=234.2J

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Answer :  The Lewis-dot structure for the following molecules are shown below.

Explanation :

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

Now we have to determine the Lewis-dot structure for the following molecules.

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As we know that iodine and chlorine have '7' valence electrons.

Therefore, the total number of valence electrons in ICl = 7 + 7 = 14

According to Lewis-dot structure, there are 2 number of bonding electrons and 12 number of non-bonding electrons.

(b) The given molecule is, PH_3

As we know that phosphorous has '5' valence electrons and hydrogen has '1' valence electrons.

Therefore, the total number of valence electrons in PH_3 = 5 + 3(1) = 8

According to Lewis-dot structure, there are 6 number of bonding electrons and 2 number of non-bonding electrons.

(c) The given molecule is, P_4

As we know that phosphorous has '5' valence electrons.

Therefore, the total number of valence electrons in P_4 = 4(5) = 20

According to Lewis-dot structure, there are 6 number of bonding electrons and 14 number of non-bonding electrons.

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As we know that sulfur has '6' valence electrons and hydrogen has '1' valence electrons.

Therefore, the total number of valence electrons in H_2S = 6 + 2(1) = 8

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Therefore, the total number of valence electrons in N_2H_4 = 2(5) + 4(1) = 14

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(f) The given molecule is, HClO_3

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As we know that bromine has '7' valence electrons, oxygen has '6' valence electrons and carbon has '4' valence electrons.

Therefore, the total number of valence electrons in COBr_2 = 4 + 6 + 2(7) = 24

According to Lewis-dot structure, there are 8 number of bonding electrons and 16 number of non-bonding electrons.

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