Give the formula for the alkene containing 15 carbons.

chubhunter [2.5K]

Answer:

A, B, and C are the correct answers.

Explanation:

The reason why solid, liquid, and gas are the most common of the elements because usually in school they teach you more about those 3 elements more than plasma ( from where I am ). Also, all of these states behave differently. Liquid is a substance that flows freely but is of constant volume, having a consistency like that of water or oil. Solid is firm and stable in shape. And gas is an air like substance. I'm not sure how to explain this more, it sounded better in my head. Hope this helped though!

5 0

3 years ago

How do you do problem #15?

SVETLANKA909090 [29]

You would convert the cg to g. 1.66 cg is 0.0166 g. then you add. 0.398 + 0.0166 = 0.4146. the answer is 0.4146 grams.

7 0

3 years ago

How do you find the formula for potassium citrate? I need the steps

deff fn [24]

Potassium citrate (also known as tripotassium citrate) is a potassium salt of citric acid with the molecular formula K3C6H5O7.

8 0

3 years ago

The mass unit associated with density is usually grams. if the volume (in ml or cm3) is multiplied by the density (g/ml or g/cm3

jok3333 [9.3K]

The weight in grams = 7.93 g

Given volume = 2.00 $in^{3}$

Given density = 0.242 g/ $cm^{3}$

We need to find the Mass(weight) in grams.

To find the weight in grams we need to keep in mind that the volume and density must use the same volume unit for cancellation. So that the volume units will cancel out, leaving only the mass units.

The unit of given volume is $in^{3}$ and unit of volume in density is $cm^{3}$ , so first we need to change the unit of volume from $in^{3}$ to $cm^{3}$ so that the volume units will cancel out, leaving only the mass units.

1 $in^{3}$ = 16.39 $cm^{3}$ (given conversion)

$2 in^{3}\times \frac{(16.39 cm^{3})}{(1 in^{3})}$

$in^{3}$ units get cancel out leaving the $cm^{3}$ unit.

$2 in^{3} = 32.78 cm^{3}$

Mass = Density X Volume.

Density = 0.242 g/ $cm^{3}$ and Volume = 32.78 $cm^{3}$

$Mass =0.242\frac{g}{cm^{3}}\times 32.78 cm^{3}$

Mass = 7.93 grams (g)

3 0

3 years ago

What is the molar concentration of a sucrose solution prepared by dissolving 350.25 g of sucrose is enough deionized water to yi

vovangra [49]

The molar concentration of a sucrose solution prepared by dissolving 350.25 g of sucrose is enough deionized water to yield a final solution volume of 500.00 mL is 2.048 M.

We must first obtain the number of moles of sucrose in the solution as follows;

Number of moles = mass/ molar

Mass of sucrose = 350.25 g

Molar mass of sucrose = 342 g/mol

Number of moles of sucrose= 350.25 g / 342 g/mol

= 1.024 moles

Recall that;

Number of moles = concentration × volume

concentration = Number of moles/volume

volume of solution = 500.00 mL or 0.5 L

concentration = 1.024 moles/0.5 L

concentration = 2.048 M

Learn more: brainly.com/question/13385951

3 0

3 years ago