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tankabanditka [31]
3 years ago
14

How many moles of methane (CH4) is 6.70 x 1022 atoms of methane?

Chemistry
1 answer:
suter [353]3 years ago
8 0

Answer:

0.1113  mol

Explanation:

Data Given:

no. of atoms of CH₄= 6.70 x 10²² atoms

no. of moles of methane (CH₄) = ?

Solution:

we will find no. of moles of methane (CH₄)

Formula used

          no. of moles = no. of atoms / Avogadro's number

Where

Avogadro's number = 6.022 x 10²³

Put values in above equation

         no. of moles = 6.70 x 10²² atoms / 6.022 x 10²³ (atoms/mol)

         no. of moles = 0.1113  mol

So,

There are 0.1113 moles of methane.

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Methane (CH4) burns in oxygen to produce carbon dioxide and water. Which reaction is occurring here?
dusya [7]
It is a combustion reaction because when methane burns with oxyzen it produces carbon dioxide,water and heat and light.
8 0
3 years ago
Radium decays to form radon. Which equation correctly describes this decay? Superscript 226 Subscript 88 Baseline Upper R a righ
KiRa [710]

Answer: 226 Subscript 88 Baseline Upper R a right arrow Superscript 222 Subscript 86 Baseline Upper R n + Superscript 4 Subscript 2 Baseline Upper H e

Explanation:

Alpha decay : When a larger nuclei decays into smaller nuclei by releasing alpha particle. In this process, the mass number and atomic number is reduced by 4 and 2 units respectively.

The general representation of alpha decay reaction is:

^{A}_{Z}\textrm{X}\rightarrow ^{A-4}_{Z-2}\textrm {Rn}+ ^{4}_{2}\textrm{He}

Representation of Radium decays to form Radon

^{226}_{88}\textrm{Ra}\rightarrow ^{222}_{86}\textrm {Rn}+ ^{4}_{2}\textrm{He}

Thus 226 Subscript 88 Baseline Upper R a right arrow Superscript 222 Subscript 86 Baseline Upper R n + Superscript 4 Subscript 2 Baseline Upper H e represents alpha decay.

         

4 0
3 years ago
Draw the Lewis structure for the polyatomic formate anion. Be sure to include all resonance structures that satisfy the octet ru
givi [52]

Answer:

Lewis structure of polyatomic formate anion.

Explanation:

To draw Lewis structure for any chemical species,

1)Count the total number of valence electrons present in it.

This can be obtained by adding valence electrons of each constituent atom.

2)Arrange those valence electrons in such a way that each atom should attain eight electrons around it to satisfy octet theory.

The structure of formate ion and its Lewis structure are shown below:

HCOO- is the formate ion.

It has total:

1+4+6+6+1 = 18 valence electrons.

Since, hydrogen has one, carbon has four and oxygen has six valence electrons and the charge of the anion is one.

Arrange this 18 electrons in such a way that each atom should get 8 electrons around it.

Resonance structures of formate ion:

8 0
2 years ago
Hemoglobin ___________. (Select all that apply).
alexandr1967 [171]

Answer:

Option B and A

Explanation:

  • First our lungs take purified oxygen from air then provide it to heart
  • Our heart gives oxygen to hemoglobin which is present in our blood
  • It carries oxygen to all cells
  • and comes back with cO_2
5 0
1 year ago
If 15 grams of Carbon dioxide is produced in a chemical reaction, how many grams of Carbon must be consumed in the reaction if w
irinina [24]

Answer:

4.13 g

Explanation:

Data Given:

Amount of CO₂ Produced = 15 g

Amount of Oxygen = 11 g

Amount of Carbon used = ?

Solution:

Suppose Carbon dioxide (CO₂) is formed by the reaction of carbon and oxygen then the reaction will be as below

                            C   +   O₂    -------------> CO₂

                          1 mol    1 mol                  1 mol

we come to know from the above reaction that

1 mole of carbon react with 1 mole of oxygen to produce 1 mol of carbon dioxide.

molar mass of C = 12 g/mol

molar mass of O₂ = 32 g/mol

molar mass of CO₂ = 12 + 2(16) = 44 g/mol

if we represent mole in grams then

           C               +                        O₂                     ------------->        CO                 1 mol (12 g/mol)                      1 mol (32 g/mol)                      1 mol (44 g/mol)

                   

              C   +   O₂    -------------> CO₂

            12 g       32 g                   44 g

So,

we come to know that 32 g of Oxygen combine with 12 g  of oxygen produce 44 g CO₂

So now how much of Carbon will be combine with 11 g of oxygen

apply unity formula

                32 g of  O₂ ≅ 12 g of  C

                  11 g of O₂  ≅  g of  C

by doing cross multiplication

           g of C = 12 g x 11 g / 32 g

           g of C = 132 g / 32 g

           g of C = 4.13 g

So,

4.13 g of carbon will consume to produce 15 g of Carbon dioxide.

to check this answer

we use the above information

                     12 g of  C ≅ 44 g of CO₂

                     4.13 g of C ≅  g of  CO₂

by doing cross multiplication

                    g of  CO₂ = 44 g x 4.13 g / 12 g

                    g of CO₂ = 15g

So it is confirmed that

4.13 g of carbon will consume to produce 15 g of Carbon dioxide.

4 0
3 years ago
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