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morpeh [17]
3 years ago
13

Convert 112°C to Kelvin.

Chemistry
2 answers:
podryga [215]3 years ago
5 0
112°C is 385.15 Kelvin
geniusboy [140]3 years ago
4 0

Answer:

112 °C = 385 K

Explanation:

The relation between Kelvin and Celsius degrees is

0°C = 273.15 K

To convert the temperature from Celsius to Kelvin we must add 273.15:

112 °C + 273.15 = 385.15 K

With the correct significant figures the answer would be 385 K

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What is the smallest part of an element?
GenaCL600 [577]

An atom is the smallest unit of matter that has the properties of an element. It is composed of a dense core called the nucleus and a series of outer shells occupied by orbiting electrons. The nucleus, composed of protons and neutrons, is at the center of an atom.

Explanation:

8 0
3 years ago
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The principal component of mothballs is naphthalene, a compound with a molecular mass of about 130 amu, containing only carbon a
DIA [1.3K]

Answer:

Empirical formula = C5H4

Molecular formula = C10H8

Explanation:

When the 3000 mg of naphthalene are burned they produce 10.3 mg of CO2. Knowing the unbalanced equation of the combustion of naphthalene, we have:

CxHy + O2 = CO2 + H2O

We calculate the molar composition of the sample. We look for the molecular weights in the periodic table:

CO2 = 12,011 + 2 (15,999) = 44,009 g

Mol C = 10.3 mg * (1 mol CO2 / 44.009 g CO2) * (1 mol C / 1 mol CO2) = 0.234 mmol C

Mass C = 0.234 mmol C * (12.011 g C / 1 mol C) = 2.8105 mg C

Mass H = 3 mg - 2.8105 mg = 0.1895 mg H

Mol H = 0.1895 mg H * (1 mol H / 1,008 g H) = 0.188 mmol H

To calculate the empirical formula, we must divide the number of moles of each element by the smallest number of moles, in this case, of hydrogen:

C = 0.2340 mmol C / 0.1895 mol H = 1.25

H = 0.1895 mmol H / 0.1895 mmol H = 1

We multiply the coefficients by 4, and we have the empirical formula:

C1.25 * 4H1 * 4 = C5H4

The molecular formula is equal to (C5H4)m, where m is calculated by the molecular and empirical mass ratio, as follows:

Empirical mass = (5 * 12.011) + (4 * 1.008) = 64.09 g

m = 130 g / 64.09 g = 2.02 = 2

Therefore we have the molecular formula:

(C5H4)2 = C10H8

4 0
3 years ago
If 100.0 grams of na3n decompose to form sodium and nitrogen, how many moles of sodium are formed? write and balance equation be
Yakvenalex [24]
The balanced equation is:
2Na_3N-\ \textgreater \ 6Na + N_2

Then proceed with the following equations.


100g Na_3N*(\frac{1molNa_3N}{82.98gNa_3N})*(\frac {6mol Na}{2molNa_3N})*(\frac {22.99gNa}{1molNa})=83.12gNa

The answer is 83.12gNa.
7 0
3 years ago
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Please help! Help me solve problems about naming structures with IUPAC rules
lianna [129]
A. The longest carbon chain is eight, and it has two methyl groups attached to carbon three, and a special group attached to carbon five. Its two names could be:

3-dimethyl-5-(1-methylethyl)octane
3-dimethyl-5-isopropyloctane

Both of these are correct. This is an alkane, because it has all single bonds.

B. This has a triple bond contained between carbons 2 and 3, and has a methyl group off carbon 4. The longest chain is 5. It’s name is:

4-methyl-2-pentyne

This is an alkene, because of the double bond.

C. This has a double bond contained between carbons 2 and 3, and has a methyl off of four and an methyl off of six. The longest chain is eight (follow the longest chain of carbons).

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This is an alkene, because of the double bond.

D. This has an ethyl group at 1 and a methyl group at 2 (rotate the compound to make it as clean as possible, in this case, the ring is flipped and rotated to make it alphabetical with the smallest numbers possible). The two names are:

1-ethyl-2-methylbenzene
ortho-ethylmethylbenzene

Both are correct, the ortho prefix telling the location of the ethyl and methyl groups. This is an aromatic structure because of its double bonded ring.

E. The longest chain is nine, and has methyls at three, five, and seven, along with a propyl at five. The name is:

3,5,7-trimethyl-5-propylnonane

This is an alkane, due to the single bonds.

Hope this helps!
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3 years ago
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The surrounding ecosystem in and around the water
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