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2 years ago

12

Refer to the table . Which plants would most likely be found in the rain forest

1 answer:

OlgaM077 [116]2 years ago

6 0

Answer: I would think 4 because there are plants every were there like literally there is lots of flowers green bushes trees vines there's lots of wonderful beautiful stuff there so I think the number 4

Explanation:

A rainforest orchid. Orchids are very common plants in the tropical rainforest. The Amazon Rainforest itself is home to more than 40,000 plant species! The most common tree in the Amazon Rainforest is the açai here are some of the plants there are there

Bromeliads Plant (Bromeliaceae)

Emergent Plant

Heliconia Flower (Lobster-Claw)

Orchid Plant

Passion flowers (Passiflora spp.)

Lianas

Vines

Rubber Tree (Hevea brasiliensis)

Cacao (Theobroma cacao)

Giant Water Lilies (Victoria amazonica)

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What is the percent by mass of Fluorine in Nitrogen triflouride NF3?

Kipish [7]

Answer:

The answer to your question is 80.3%

Explanation:

Data

Percent by mass of F

molecules NF₃

Process

1.- Calculate the molar mass of nitrogen trifluoride

molar mass = (1 x 14) + (19 x 3)

= 14 + 57

= 71 g

2.- Use proportions and cross multiplications to find the percent by mass of F. The molar mass of NF₃ is equal to 100%.

71 g of NF₃ ------------------ 100%

57 g of F ------------------- x

x = (57 x 100)/71

x = 5700 / 71

x = 80.3%

3.- Conclusion

Fluorine is 80.3% by mass of the molecule NF₃

3 0

3 years ago

Calculate a reasonable amount (mass in g) of your unknown acid to use for a titration. You will want about 30 mL of titrant to g

Vinil7 [7]

Answer:

"0.60 g" is the appropriate solution.

Explanation:

The given values are:

Volume of base,

= 30 ml

Molarity of base,

= 0.05 m

Molar mass of acid,

= 400 g/mol

As we know,

⇒ $Molarity=\frac{Number \ of \ moles \ of \ base}{Number \ of \ solution}$

On substituting the values, we get

⇒ $0.05=\frac{Number \ of \ moles \ of \ base}{30\times 10^{-3}}$

⇒ $Number \ of \ moles \ of \ base=0.05\times 30\times 10^{-3}$

⇒ $=1.5\times 10^{-3}$

hence,

⇒ $Moles \ of \ acid=\frac{Mass \ of \ acid}{Molar \ mass \ of \ acid}$

On substituting the values, we get

⇒ $1.5\times 10^{-3}=\frac{Mass \ of \ acid}{400}$

⇒ $Mass \ of \ acid=1.5\times 10^{-3}\times 400$

⇒ $=0.60 \ g$

8 0

3 years ago

2.<br> An alkane has at least on C=C<br> bond.<br> ut of<br> Select one:<br> O True<br> O False

garik1379 [7]

I think it’s false?????

8 0

3 years ago

If 5.55 moles of gas have a volume of 12.2 L, what is the new volume when the moles of

KatRina [158]

Answer:

$12.61 + 5.55) \div 12.2 =$

Explanation:

it's too easy

8 0

1 year ago

Where do producers get matter and energy to live and grow ?

bixtya [17]

Answer:c

Explanation:

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2 years ago