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3 years ago

13

Does “valence electrons” have a charge?

1 answer:

Alborosie3 years ago

6 0

Answer:

Yes, negative

Explanation:

Yes, electrons have negative charge. "Valence electrons" means the electrons of the outer shell, but as electrons the charge is negative

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which atom has a change in oxidation number of -3 in the following redox reaction K2Cr2O7 + H2O +S --> KOH + Cr2O3 +SO2

yaroslaw [1]

You have to calculate the oxidation estates of the atoms in each compound.

I will start with K2Cr2O7 because I believe that Cr is the best candidate to reduce its oxidation number in 3 units.

In K2Cr2O7:

- K has oxidation state of 1+, then K2 has a charge of 2* (1+) = 2+.

- O has oxidation state of 2*, then O7 has a charge of 7* (2-) = 14-.

That makes that Cr2 has charge of 14 - 2 = +12, so each Cr has +12/2 = +6 oxidation state.

In Cr2O3:

- O has oxidation state of 2-, then O3 has charge 3 * (2-) = - 6

- Then, Cr2 has charge 6+, and each Cr has charge 6+ / 2 = 3+.

So, we have seen that Cr reduced its oxidation state in 3 units, from 6+ to 3+.

Answer: Cr has a change in oxidation number of - 3.

6 0

3 years ago

Will give brainliest

slamgirl [31]

I will help you with answering this question.

3 0

3 years ago

What elegant has 2 valence electrons

scZoUnD [109]

Explanation:

Calcium is the element that has 2 valence electrons.

8 0

3 years ago

18 An important environmental consideration is the appropriate disposal of cleaning solvents. An environmental waste treatment c

Katyanochek1 [597]

Answer:

a) Percentage by mass of carbon: 18.3%

Percentage by mass of hydrogen: 0.77%

b) Percentage by mass of chlorine: 80.37%

c) Molecular formula: $C_{2} H Cl_{3}$

Explanation:

Firstly, the mass of carbon must be determined by using a conversion factor:

$0.872g CO _{2} *\frac{12g C}{44g CO_{2} } = 0.238g CO_{2}$

The same process is used to calculate the amount of hydrogen:

$0.089g H_{2}O*\frac{2g H}{18g H_{2}O } = 0.010g H$

The percentage by mass of carbon and hydrogen are calculated as follows:

%C $\frac{0.238g}{1.3g} *100%$ = 18.3%

%H $\frac{0.010g}{1.3g} *100%$ =0.77%

From the precipation data it is possible obtain the amount of chlorine present in the compound:

$1.75 AgCl*\frac{35.45g Cl}{143.45g AgCl}$ = 0.43g AgCl

Let's calculate the percentage by mass of chlorine:

%Cl= $\frac{0.43g}{0.535g} * 100%$ = 80.37%

Assuming that we have 100g of the compound, it is possible to determine the number of moles of each element in the compound:

$18.3g C*\frac{1mol C}{12g C} = 1.52mol C$

$0.77g H*\frac{1mol H}{1g H} = 0.77mol H$

$80.37gCl*\frac{1molCl}{35.45g Cl} = 2.27mol Cl$

Dividing each of the quantities above by the smallest (0.77mol), the subscripts in a tentative formula would be

$C=\frac{1.52}{0.77} = 1.97$ ≈ 2

$H = \frac{0.77}{0.77} = 1$

$Cl =\frac{2.27}{0.77}=2.94$ ≈3

The empirical formula for the compound is:

$C_{2} H Cl_{3}$

The mass of this empirical formula is:

mass of C + mass of H + mass of Cl= 24g +1+ 106.35 =131.35g

This mass matches with the molar mass, which means that the supscript in the molecular formula are the same of the empirical one.

5 0

3 years ago

When the moon appears larger than 1/2 but less than full

Diano4ka-milaya [45]

It is called a waxxing gibbous, pls brainliest

5 0

3 years ago