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3 years ago

How much larger are carbon atoms compared to hydrogen atoms?

Chemistry

1 answer:

aleksandrvk [35]3 years ago

5 0

Answer:

Carbon is bigger

Explanation:

There are twice as many number of hydrogen-to-oxygen atoms. This applies to carbon atoms as well when compared to hydrogen atoms.

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Decide whether you would use a roman numeral in the name of the following compounds and then name them accordingly:

natka813 [3]

Answer:

its C

Explanation:

6 0

3 years ago

#7) The atom on the left (pink) in the video is shown in the diagram below. From

ruslelena [56]

Answer:

Explanation:

Remember the number of valence electrons represent the group number. As shown in figure, there is only one electron in the valence shell. Therefore, this element belongs to group 1 elements also called as Alkali metals.

Furthermore, the number of electrons in a neutral atom equals number of protons. Hence, there are 11 protons in this element.

Also, number of protons equal atomic number therefore, the atomic number of this element is 11. And, by checking periodic table, the element having atomic number 11 is Sodium (Na).

6 0

3 years ago

Bromine (Br2) is produced by reacting HBr with O2, with water as a byproduct. The O2 is part of an air (21 mol % O2, 79 mol % N2

Karolina [17]

Answer:

The mole fractions:

$x_{HBr}=\frac{100mol}{318.5}=0.314$

$x_{Br_2}=\frac{78mol}{318.5}=0.245$

$x_{H_2O}=\frac{78mol}{318.5}=0.245$

$x_{O_2}=\frac{62.5mol}{318.5}=0.196$

Explanation:

The reaction described is:

$2 HBr (g) + 1/2 O_2 (g) \longrightarrow Br_2 (g) + H_2O (g)$

The limiting reactant is the HBr (oxygen is in excess).

a) The mass (in moles) balance for this sistem:

$n_{Br_2}=\frac{ 1 mol Br_2}{1 mol HBr} *n_{HBr}*0.78$

(the 0.78 is because of the fractional conversion)

$n_{O_2}=\frac{ 0.5 mol O_2}{1 mol HBr} *n_{HBr}*1.25$

(the 1.25 is because of the oxygen excess)

$n_{H_2O}=\frac{ 1 mol H_2O}{1 mol Br_2} *n_{Br_2}$

There is only one degree of freedom in this sistem, you can either deffine the moles of HBr you have or the moles of Br2 you want to produce. The other variables are all linked by the equations above.

b) Base of calculation 100 mol of HBr:

$nn_{HBr}=100 mol HBr$