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2 years ago

Why are there groups, in the periodic table?

2 answers:

7 0

Groups help organize the elements. Elements of the same group have similar chemical qualities. So looking at an element’s group on the periodic table can help you identify some of it’s characteristics such it’s as reactivity with other elements or of it’s a good conductor.

zepelin [54]2 years ago

6 0

To classify the elements and make them easier to understand

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explain using diagrams how potassium forms the compound potassium flouride when it reacts with flourine

Katarina [22]

Answer:

The answer to your question is given below.

Explanation:

Potassium (K) has 19 electrons with electronic configuration of 2, 8, 8, 1.

Fluorine (F) has 9 electrons with electronic configuration of 2, 7.

Fluorine needs 1 electron to complete it's octet configuration.

Hence, potassium (K), will lose 1 electron to fluorine (F) to form potassium ion (K+) with electronic configuration of 2, 8, 8. The fluorine atom (F) will receive the 1 electron from potassium to form the fluoride ion (F-) with electronic configuration of 2, 8.

**** Please see attached photo for further details.

7 0

2 years ago

How many moles are present in

olasank [31]

Answer: (a) There are 0.428 moles present in 12 g of $N_{2}$ molecule.

(b) There are 2 moles present in $12.044 \times 10^{23}$ particles of oxygen.

Explanation:

(a). The mass of nitrogen molecule is given as 12 g.

As the molar mass of $N_{2}$ is 28 g/mol so its number of moles are calculated as follows.

$No. of moles = \frac{mass}{molar mass}\\= \frac{12 g}{28 g/mol}\\= 0.428 mol$

So, there are 0.428 moles present in 12 g of $N_{2}$ molecule.

(b). According to the mole concept, 1 mole of every substance contains $6.022 \times 10^{23}$ atoms.

Therefore, moles present in $12.044 \times 10^{23}$ particles are calculated as follows.

$Moles = \frac{12.044 \times 10^{23}}{6.022 \times 10^{23}}\\= 2 mol$

So, there are 2 moles present in $12.044 \times 10^{23}$ particles of oxygen.

4 0

2 years ago

Gaseous methane (CH4) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO) and gaseous water (H2O) . Suppo

Leto [7]

Answer:

0 g.

Explanation:

Hello,

In this case, since the reaction between methane and oxygen is:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:

$n_{CH_4}=0.963gCH_4*\frac{1molCH_4}{16gCH_4}=0.0602molCH_4\\ \\n_{CH_4}^{consumed}=7.5gO_2*\frac{1molO_2}{32gO_2}*\frac{1molCH_4}{2molO_2} =0.117molCH_4$

Thus, since oxygen theoretically consumes more methane than the available, we conclude the methane is the limiting reactant, for which it will be completely consumed, therefore, no remaining methane will be left over.

$left\ over=0g$

Regards.

7 0

2 years ago

can you help me with #7? #7 says "A substance enters into and is altered through the course of a chemical change. it is called a

Zigmanuir [339]

Reactant hope this helps

3 0

2 years ago

Read 2 more answers

The diagrams show gases that are stored in two separate but similar containers.

Nesterboy [21]

Answer: The diagrams show gases that are stored in two separate but similar containers. If both gases are at the same temperature, which one has the greater pressure? Gas 2 because it has more particles that are colliding.

Explanation:

8 0

2 years ago