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2 years ago

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

Chemistry

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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How many moles of Co2 are produced when 0.2 moles of sodium carbonate reacts with excess HCl

Pie

To determine the number of moles of carbon dioxide that is produced, we need to know the reaction of the process. For the reaction of HCl and sodium carbonate, the balanced chemical equation would be expressed as:

2HCl + Na2CO3 = 2NaCl + H2O + CO2

From the initial amount given of sodium carbonate and the relation of the substances from the balanced reaction, we calculate the moles of carbon dioxide as follows:

0.2 moles Na2Co3 ( 1 mol CO2 / 1 mol Na2Co3 ) = 0.2 moles CO2

Therefore, the amount in moles of carbon dioxide that is produced from 0.2 moles sodium carbonate would be 0.2 moles as well.

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3 years ago

The molecule NH3 contains all single bonds.<br><br> true <br> false

dsp73

<h3><u>Answer;</u></h3>

True

<h3><u>Explanation</u>;</h3>

The molecule NH3 contains all single bonds.
NH3 has a three single covalent bond among its nitrogen and hydrogen atoms,because one valence electron of each of three atom of hydrogen is shared with three electron.
There are three covalent bonds are in NH3 . Each hydrogen make a single bond with nitrogen and there is also a pair of electron which is unpaired from nitrogen.

8 0

3 years ago

Read 2 more answers

Balance the following equations by inserting the proper coefficients.

Montano1993 [528]

1212. 1212. 1323........

6 0

3 years ago

Read 2 more answers

Find percent yield:

saveliy_v [14]

<u>Answer:</u> The percent yield of the reaction is 91.8 %

<u>Explanation:</u>

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For $B_5H_9$ :</u>

Given mass of $B_5H_9$ = 4.0 g

Molar mass of $B_5H_9$ = 63.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of }B_5H_9=\frac{4g}{63.12g/mol}=0.0634mol$

<u>For oxygen gas:</u>

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{10g}{32g/mol}=0.3125mol$

The chemical equation for the reaction of $B_5H_9$ and oxygen gas follows:

$2B_5H_9+12O_2\rightarrow 5B_2O_3+9H_2O$

By Stoichiometry of the reaction:

12 moles of oxygen gas reacts with 2 moles of $B_2H_5$

So, 0.3125 moles of oxygen gas will react with = $\frac{2}{12}\times 0.3125=0.052mol$ of $B_2H_5$

As, given amount of $B_2H_5$ is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

12 moles of oxygen gas produces 5 moles of $B_2O_3$

So, 0.3125 moles of oxygen gas will produce = $\frac{5}{12}\times 0.3125=0.130moles$ of water

Now, calculating the mass of $B_2O_3$ from equation 1, we get:

Molar mass of $B_2O_3$ = 69.93 g/mol

Moles of $B_2O_3$ = 0.130 moles

Putting values in equation 1, we get:

$0.130mol=\frac{\text{Mass of }B_2O_3}{69.63g/mol}\\\\\text{Mass of }B_2O_3=(0.130mol\times 69.63g/mol)=9.052g$

To calculate the percentage yield of $B_2O_3$ , we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of $B_2O_3$ = 8.32 g

Theoretical yield of $B_2O_3$ = 9.052 g

Putting values in above equation, we get:

$\%\text{ yield of }B_2O_3=\frac{8.32g}{9.052g}\times 100\\\\\% \text{yield of }B_2O_3=91.8\%$

Hence, the percent yield of the reaction is 91.8 %

6 0

4 years ago

A mixture is made of 40 ml of salt water to 200 ml of solution. What percent of the solution is salt water?

Serggg [28]

Answer:

16.7%

Explanation:

40 ml of salt water + 200 ml of solution = 240 ml

40/240 = 4/24 = 1/6=16.7%

5 0

3 years ago