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gtnhenbr [62]
3 years ago
6

Lewis structure of graphite

Chemistry
1 answer:
Andreas93 [3]3 years ago
6 0
Its a C with 4 dots, 1 on top,1 on bottom, 1 on the left and 1 on the right

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If element x forms the oxides xo and x2o3 the oxidation numbers of element x are
Deffense [45]

Answer:

b) +2 and +3.

Explanation:

Hello,

In this case, given the molecular formulas:

XO

And:

X_2O_3

We can relate the subscripts with the oxidation states by knowing that they are crossed when the compound is formed, for that reason, we notice that oxygen oxidation state should be -2 for both cases and the oxidation state of X in the first formula must be +2 since both X and O has one as their subscript as they were simplified:

X^{+2}O^{-2}

Moreover, for the second case the oxidation state of X should be +3 in order to obtain 3 as the subscript of oxygen:

X_2^{+3}O_3^{-2}

Thus, answer is b)+2 and +3

Best regards.

3 0
2 years ago
The table shows the students data from an investigation about a plant growth and five plants of the same type what is the best w
NARA [144]

Answer:

This question is incomplete

Explanation:

The question is incomplete because of the absence of the table but since the question says there are data from an investigation about a plant growth and five other plants (making six) of the same type, the best way to display this type of data for analyst is to use the grouped bar chart. <u>The grouped bar chart will display the data obtained (from an investigation on plant growth) from different students on each of the six plants (of the the same type)</u>.

Colours are usually used to identify the bars (of a group) or could be used to separate the group from other groups but in this case, colours are better used to identify the bars of a group.

8 0
3 years ago
The decomposition of a compound at 400⁰C is first order with half-life of 1570 seconds. what fraction of an initial amount of th
kirill115 [55]

Answer: After 4710 seconds, 1/8 of the compound will be left

Explanation:

Using the formulae

Nt/No = (1/2)^t/t1/2

Where

N= amount of the compound  present at time t

No= amount of compound present at time t=0

t= time taken for N molecules of the compound to remain = 4710 seconds

t1/2 = half-life of compound  = 1570 seconds

Plugging in the values, we have  

Nt/No = (1/2)^(4710s/1570s)

Nt/No = (1/2)^3

Nt/No= 1/8

Therefore after 4710 seconds, 1/8 molecules of the compound will be left

5 0
2 years ago
Predict the empirical formula of these compounds. a. C2N2. b. P4O10. c. N2O5. d. NaCl. e. C9H20. f. B2H6. g. K2Cr2O7. h. Al2Br6.
Stella [2.4K]
The empirical formula is the simplest form of the formula expressed in the lowest ratio. In this case, we just have to divide each subscript by the greatest common factor. Hence.
a. CN
b. P2O5
c.N2O5
d.NaCl
e. C9H20
f. BH3
g.K2Cr2O7
h.AlB3
i.CH
j.SiCl4
5 0
2 years ago
Draw the structure of 1,4-hexanediamine.
brilliants [131]

Answer:

1,4-hexanediamine contains two -NH_{2} functional groups.

Explanation:

1,4-hexanediamine is an organic molecule which contains two -NH_{2} functional groups at C-1 and C-4 position.

The longest carbon chain in 1,4-hexanediamine contains six carbon atoms.

Molecular formula of 1,4-hexanediamine is C_{6}H_{16}N_{2}.

1,4-hexanediamine used as a bidentate ligand in organometallic chemistry.

The structure of 1,4-hexanediamine is shown below.

5 0
3 years ago
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