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stiv31 [10]
3 years ago
8

Carbon atoms have four electrons in their outer shell. This means that a single carbon atom can form up to _______ bonds with ot

her atoms.
two
eight
four
six
Chemistry
2 answers:
MrMuchimi3 years ago
5 0

The answer is: four.

For example compound carbon dioxide (CO₂):

Electron configuration of carbon: ₆C 1s² 2s² 2p².  

Electron configuration of oxygen: ₈O 1s² 2s² 2p⁴.

Carbon atom is sp2 hybridized, it has three sp2 orbitals and one p orbital, they form four bonds.

Oxygen has two p orbitals, they form two bonds (one sigma and one pi bond).

p orbitals from carbon and oxygen overlap and form pi bond.

kakasveta [241]3 years ago
4 0
If my memory serves me well, the correct answer looks like this: Carbon atoms have four electrons in their outer shell. This means that a single carbon atom can form up to <span>four</span> bonds with other atoms.
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A sample of carbon-12 has a mass of 6.00 g. How many atoms of carbon-12 are in the sample?
Bingel [31]
Carbon molar mass=12
carbon mass in the above question =6
mole=number of atoms/6.02x10²³
6/12= number of atoms/6.02x10²³
number of atoms=1/2 x6.02x10²³
number of atoms= 6.02 x10²³/2
number of atoms=3.01x10²³
4 0
3 years ago
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Which of the following has the greatest mass?
Annette [7]

Answer:

I guess one mole. of mercury

Explanation:

because mercury is denser

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How many total atoms are in CaOH?
Lostsunrise [7]

Answer:

i think it's 3 because there aren't any indexes so that leaves us with one atom of Ca, one atom of O, and one atom of H

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What name is given to substances in which atoms of two or more elements are chemically combined?
anygoal [31]

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Compounds

Explanation:

Substances in which atoms of two or more elements are chemically combined are called compounds.

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3 years ago
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The chemical reaction for the formation of syngas is: CH4 + H2O -&gt; CO + 3 H2 What is the rate for the formation of hydrogen,
grin007 [14]

Answer :  The rate for the formation of hydrogen is, 1.05 M/s

Explanation :

The general rate of reaction is,

aA+bB\rightarrow cC+dD

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}

\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}

\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}

\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}

Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

CH_4+H_2O\rightarrow CO+3H_2

The expression for rate of reaction :

\text{Rate of disappearance of }CH_4=-\frac{d[CH_4]}{dt}

\text{Rate of disappearance of }H_2O=-\frac{d[H_2O]}{dt}

\text{Rate of formation of }CO=+\frac{d[CO]}{dt}

\text{Rate of formation of }H_2=+\frac{1}{3}\frac{d[H_2]}{dt}

The rate of reaction expression is:

\text{Rate of reaction}=-\frac{d[CH_4]}{dt}=-\frac{d[H_2O]}{dt}=+\frac{d[CO]}{dt}=+\frac{1}{3}\frac{d[H_2]}{dt}

As we are given that:

+\frac{d[CO]}{dt}=0.35M/s

Now we to determine the rate for the formation of hydrogen.

+\frac{1}{3}\frac{d[H_2]}{dt}=+\frac{d[CO]}{dt}

+\frac{1}{3}\frac{d[H_2]}{dt}=0.35M/s

\frac{d[H_2]}{dt}=3\times 0.35M/s

\frac{d[H_2]}{dt}=1.05M/s

Thus, the rate for the formation of hydrogen is, 1.05 M/s

6 0
3 years ago
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