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professor190 [17]

3 years ago

7

Which one of these is not an example of intercellular communication?

2 answers:

g100num [7]3 years ago

8 0

The answer is "B"......

Marianna [84]3 years ago

4 0

B)A molecule of glucose is produced in a chloroplast.

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Ahat [919]

Acids<span> and </span>bases<span> are two types of corrosive substances. Any substance with a pH value </span>between<span> 0 up to 7 is considered acidic, whereas a pH value of 7 to 14 is a </span>base<span>. </span>Acids<span> are ionic compounds that break apart in water to form a hydrogen ion (H+).

Hope this helped!</span>

5 0

4 years ago

Some metals cannot be welded in air because hot metal reacts with oxygen and nitrogen in the air. How do you think such metals a

marishachu [46]

Explanation:

please can you send me the pic of Faith because I am not able to understand any of your location is too bad I can't answer sorry for stamp in which class you are in

7 0

3 years ago

For which blocks of elements are outer electrons the same as valence electrons? For which are d electrons often included among v

Amanda [17]

Answer:

1. Group 1 — 3

2. Transition metals

Explanation:

3 0

4 years ago

Wewaii [24]

Answer:

A.Light +Energy

Explanation:

it is because photo means light and synthesis is more likely referred as energy

8 0

3 years ago

Read 2 more answers

The total mass of the atmosphere is about 5.00 x 1018 kg. How many moles each of air, O2, and CO2 are present in the atmosphere?

n200080 [17]

<u>Answer:</u> The moles of oxygen and carbon dioxide in air is $3.63\times 10^{19}mol$ and $7.18\times 10^{16}mol$ respectively

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of atmosphere = $5.00\times 10^{18}kg=5.00\times 10^{21}g$

Average molar mass of atmosphere = 28.96 g/mol

Putting values in above equation, we get:

$\text{Moles of atmosphere}=\frac{5.00\times 10^{21}g}{28.96g/mol}=1.73\times 10^{20}mol$

We know that:

Percent of oxygen in air = 21 %

Percent of carbon dioxide in air = 0.0415 %

Moles of oxygen in air = $\frac{21}{100}\times 1.73\times 10^{20}=3.63\times 10^{19}mol$

Moles of carbon dioxide in air = $\frac{0.0415}{100}\times 1.73\times 10^{20}=7.18\times 10^{16}mol$

Hence, the moles of oxygen and carbon dioxide in air is $3.63\times 10^{19}mol$ and $7.18\times 10^{16}mol$ respectively

6 0

3 years ago