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r-ruslan [8.4K]
3 years ago
7

If an atom has 16 protons, 14 neutrons, and 18 electrons. What is the charge of the nucleus?

Chemistry
1 answer:
aalyn [17]3 years ago
8 0

. The atom has 2 more electrons than protons giving it a negative charge (2-)

You might be interested in
Suppose Orville measures the length to be 1.55 cm. If the correct value is 1.32 cm, find Orville's % error.
Step2247 [10]

Answer:

Error in measure= 1.55-1.32= 0.23

Orville's % error = 0.23/1.32*100% = 17.42%

Explanation:

8 0
3 years ago
Write structures for the following compounds.(a) 3-ethyl-4-methylhexane (b) 3-ethyl-5-isobutyl-3-methylnonane(c) 4-tert-butyl-2-
frutty [35]

Answer:

Find attached answer to this question.

Explanation:

a. 4-Isopropyloctane has a chain of eight carbons, with an isopropyl group on the fourth carbon.

5-tert-Butyldecane has a chain of ten carbons, with a tert-butyl group on the fifth.

CH3

4-isopropyloctane

CHCH3 CH3

CH3 CH3CH2CH2CH2CH2CH2 CH

5-tert-butyldecane

CH3 CH3CH2CH2CH2CH2CH2CH2CH2 CH

CCH3 CH3

3-3 Nomenclature of Alkanes 93

di- means 2 tetra- means 4 hexa- means 6

tri- means 3 penta- means 5 hepta- means 7

b. Using this rule, we can construct names for some complicated structures. Let\u2019s fin-

ish naming the heptane on p. 90, shown here in the margin. This compound has an ethyl

group on C3 and three methyl groups on C2, C4, and C5. List the ethyl group alphabeti-

cally before the methyl groups, and give each of the four substituents a location number.

Problem-solving Hint

When substituents are alphabetized,

iso- is used as part of the alkyl group

name, but the hyphenated prefixes

are not. Thus, isobutyl is alphabetized

with i, but n-butyl, tert-butyl, and sec-

butyl are alphabetized with b. The

number prefixes di-, tri-, tetra-, etc.

are ignored in alphabetizing.

CH3 CH

2-bromobutane

Br

CH2CH3 CH3 CH

3-chloro-2-methylpentane

CH3 Cl

CH CH2CH3 CH3 CH

1,2-difluoropropane

F

CH2F

c. CH2CH(CH3)2CH2CH3

CH CHCH2 CH2 CH2CH3 CH3

S O LV E D P R O B L E M 3-2

Give a systematic (IUPAC) name for the following compound.

SOLUTION

The longest carbon chain contains eight carbon atoms, so this compound is named as an octane.

Numbering from left to right gives the first branch on C2; numbering from right to left gives

the first branch on C3, so we number from left to right.

CH CH2

CH3

CH CH CH2CH3

CH CH3

CH3

CH3

CH3

3-ethyl-2,4,5-trimethylheptane

7

65

4 3

2

1

CH3

CH3

CH3 CH3

CH3

CH3

CH3

CH

CH CH CH

CH2CH3

CH2

C

d. CH3 CH3

CH3

CH2 CH2 CH2 CH2 CH2

CH2

CH

CH3 CH3CH

CH

CH3 CH

Br CH2CH3

CH CH3(b)

CH3CH2

CH2CHCH3 CH3

8 0
3 years ago
Read 2 more answers
A container with a volume of 893L contains how may moles of air at STP?
Vadim26 [7]

Answer:

i think 54

Explanation:

7 0
3 years ago
If a molecule diffusing through extracellular fluid travels 1 mm in 1 sec, how long will it take that molecule to diffuse 1 cm
hichkok12 [17]
Given that the rate of diffusion is 1 mm per 1 second. then the time it travels in 1 cm can be solve using the formula
t = d / r
where d is the distance
r is the rate

first, 1 cm is equal to 10 mm

t = 10 mm / ( 1 mm / s )
t = 10 s
4 0
3 years ago
Consider the hypothetical reaction 4A + 3B → C + 2D Over an interval of 3.00 s the average rate of change of the concentration o
pogonyaev

Answer:

Final concentration of C at the end of the interval of 3s if its initial concentration was 3.0 M, is 3.06 M and if the initial concentration was 3.960 M, the concentration at the end of the interval is 4.02 M

Explanation:

4A + 3B ------> C + 2D

In the 3s interval, the rate of change of the reactant A is given as -0.08 M/s

The amount of A that has reacted at the end of 3 seconds will be

0.08 × 3 = 0.24 M

Assuming the volume of reacting vessel is constant, we can use number of moles and concentration in mol/L interchangeably in the stoichiometric balance.

From the chemical reaction,

4 moles of A gives 1 mole of C

0.24 M of reacted A will form (0.24 × 1)/4 M of C

Amount of C formed at the end of the 3s interval = 0.06 M

If the initial concentration of C was 3 M, the new concentration of C would be (3 + 0.06) = 3.06 M.

If the initial concentration of C was 3.96 M, the new concentration of C would be (3.96 + 0.06) = 4.02 M

3 0
3 years ago
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