How much stronger is methane than carbon dioxide as a greenhouse gas?

if you are curious at to why so many of ur family members have certain hair or eye color, you might want to look at... A. Astrol

densk [106]

Answer:

c) hereditary

Explanation:

This means that if your parents had one eye colour, you'll most likely have the same one because you share the same genes, and it's passed down to you.

7 0

2 years ago

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How many molecules of propane were in the erlenmeyer flask? Avogadro's number is 6. 022 × 10^23 molecules/mol

Ede4ka [16]

3.74× $10^{21}$

3.74 × $10^{21}$ molecules of propane were in the erlenmeyer flask.

number of moles of propane can be calculated as moles of propane.

mass of propane = 0.274 g

molar mass of propane = 44.1

So this gives us the value of 6.21× $10^{-3}$ moles of propane

No one mole of propane As a 6.0-2 × $10^{23}$

so, 6.21 × $10^{-3}$ × 6. 022 × 10^23

= 3.74 × $10^{21}$

Therefore, molecules of propane were in the erlenmeyer flask is found to be 3.74 × $10^{21}$

<h3>What is erlenmeyer flask?</h3>

A laboratory flask with a flat bottom, a conical body, and a cylindrical neck is known as an Erlenmeyer flask, sometimes known as a conical flask or a titration flask.
It bears the name Emil Erlenmeyer after the German chemist.

<h3>What purpose does an Erlenmeyer flask serve?</h3>

Liquids are contained in Erlenmeyer flasks, which are also used for mixing, heating, chilling, incubating, filtering, storing, and other liquid-handling procedures.

For titrations and boiling liquids, their sloped sides and small necks make it possible to whirl the contents without worrying about spills.

To learn more about calculating total molecules visit:

brainly.com/question/8933381

#SPJ4

4 0

11 months ago

Vanillin (used to flavor vanilla ice cream and other foods) is the substance whose aroma the human nose detects in the smallest

balu736 [363]

Answer:

Cost to supply enough vanillin is $3.2\$$

Explanation:

Threshold limit of vanillin in air is $2.0\times 10^{-11}g$ per litre means there should be $2.0\times 10^{-11}g$ of vanillin in 1L of air to detect aroma of vanillin.

$1ft^{3}=28.32L$

So, $5.0\times 10^{7}ft^{3}=(5.0\times 10^{7}\times 28.32)L$

So amount of vanillin should be present to detect = $(2.0\times 10^{-11}\times 5.0\times 10^{7}\times 28.32)g$

As cost of 50 g vanillin is $112\$$ therefore cost of $(2.0\times 10^{-11}\times 5.0\times 10^{7}\times 28.32)g$ vanillin = $(2.0\times 10^{-11}\times 5.0\times 10^{7}\times 28.32\times 112)\$ = 3.2\$$

5 0

2 years ago

Ethyl acetate reacts with H2, in the presence of a catalyst to yield ethyl alcohol: C4H8O2 + 2H2--> 2C2H6O. A. How many grams

Anni [7]

The balanced chemical reaction is:

<span>C4H8O2 + 2H2--> 2C2H6O

</span>A. How many grams of ethyl alcohol are produced by reaction of 2.7 mol of ethyl acetate with H2? 2.7 mol C4H8O2 ( 2 mol C2H6O / 1 mol C4H8O2) (46.07 g / 1 mol) = 248.78 g ethyl alcohol

B. How many grams of ethyl alcohol are produced by reaction of 13.0g of ethyl acetate with H2?
13.0 g C4H8O2 ( 1 mol / 88.11 g) ( 2 mol C2H6O / 1 mol C4H8O2) (<span>46.07 g / 1 mol) = 13.59 g ethyl alcohol</span>

C. How many grams of H2 are needed to react with 13.0g of ethyl acetate?
13.0 g C4H8O2 ( 1 mol / 88.11 g) ( 2 mol H2 / 1 mol C4H8O2) (2.02 g / 1 mol) = 0.5961 g H2

6 0

2 years ago

What volume of lead (of density 11.3 g/cm3) has the same mass as 330 cm3 of a piece of redwood (of density 0.38 g/cm3)?

zimovet [89]

Answer:

The sample of lead has a volume of 11.1 cm³

Explanation:

<u>Step 1:</u> Data given

x cm³ lead has a density of 11.3 g/cm³

it has the same mass as 330cm³ of a piece of redwood with density 0.38g/cm³

<u>Step 2</u>: Calculate mass of the piece of redwood

Density = mass/volume

mass = density * volume

Mass of the piece of redwood = 0.38 g/cm³ * 330cm³ = 125.4 grams

Since the sample of lead has the same mass, it also has a mass of 125.4 grams

<u>Step 3</u>: Calculate volume of the lead

Density = mass/ volume

Volume = mass/ density

Volume of lead = 125.4g / 11.3g/cm³ = 11.097 cm³≈11.1 cm³

The sample of lead has a volume of 11.1 cm³

3 0

2 years ago