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Nataliya [291]
4 years ago
11

Question 3 (1 point)

Chemistry
1 answer:
pentagon [3]4 years ago
5 0

Full question:

The IUPAC name for  CH3CH2C≡CCH3 is:

Answer:

2-pentyne

Explanation:

To name hydrocarbons, you first you have to identify the longest carbon chain.  There are 5 carbons in this chain, so we know the name is "pent".

You then have to identify the presence of any double or triple bonds. If double bonds, it is an alkene, if triple bonds, it is an alkyne. In this case there is a triple bond, so we know the hydrocarbon is pentyne.

You then number the chain to give the lowest number to the triple bond. It could either be 4 (countnig carbons from left to right) or 2 (from right to left). Therefore, the answer is 2-pentyne.

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An object has a density of 16.3 g/mL and a volume of 0.46 L. Calculate the
deff fn [24]

Answer:

16.53 pounds

Explanation:

this type of problem needs convertion method we need to convert from grams to pound.

3 0
3 years ago
Convert 6mol NO2 into grams Convert 800 grams of LiO into moles! Convert 4500 grams of SO2 into molecules! Convert 30 mol H2O in
Nezavi [6.7K]

Answer:

1. 276 g of NO₂

2. 34.8 moles of LiO

3. 4.23×10²⁵ molecules of SO₂

4. 540 g of H₂O

5. 224 g CO

Explanation:

Let's define the molar mass of the compound to define the moles or the grans of each.

Molar mass . moles = Mass

Mass (g) / Molar mass = Moles

1. 6 mol . 46 g / 1 mol = 276 g of NO₂

2. 800 g . 1mol / 22.94 g = 34.8 moles of LiO

3. To determine the number of molecules, we convert the mass to moles and then, we use the NA (1 mol contains 6.02×10²³ molecules)

4500 g . 1mol / 64.06 g = 70.2 moles of SO₂

70.2 mol . 6.02×10²³ molecules / 1 mol = 4.23×10²⁵ molecules of SO₂

4. 30 mol . 18g / 1 mol = 540 g of H₂O

5. 8 mol . 28g /  1mol = 224 g CO

3 0
3 years ago
Which quantity is equivalent to 39 grams of LiF?
Nutka1998 [239]
I think that it is 1.5 mole it might not be
3 0
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Read 2 more answers
Which mixture is homogeneous? a.soil b.salad dressing c.granola cereal d.milk
Hoochie [10]
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A chemist must dilute 97.1 ml of aqueous magnesium fluoride solution until the concentration falls to 389 microMolarity . He'll
mojhsa [17]

Answer:

0.302L

Explanation:

<em>...97.1mL of 1.21m M aqueous magnesium fluoride solution</em>

<em />

In this problem the chemist is disolving a solution from 1.21mM = 1.21x10⁻³M, to 389μM = 389x10⁻⁶M. That means the solution must be diluted:

1.21x10⁻³M / 389x10⁻⁶M = 3.11 times

As the initial volume of the original concentration is 97.1mL, the final volume must be:

97.1mL * 3.11 = 302.0mL =

0.302L

6 0
3 years ago
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