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solniwko [45]
3 years ago
8

Choose all the answers that apply.

Chemistry
2 answers:
elena-14-01-66 [18.8K]3 years ago
8 0

Answer:

have waxy leaves and stems

&

produce seeds quickly

Hope this helps good luck :)

Vitek1552 [10]3 years ago
7 0

Answer: They Have Waxy Leaves And Stems

Explanation:

:)

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ANSWER ASAP IF YOU KNOW THIS ......What is the mass number of this atom?
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The correct answer is 7 I just took the test :)
6 0
2 years ago
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What are the steps of natural selection?
blagie [28]

Answer:

The five steps involved in the process of natural selection are

Variation • Inheritance • Selection • Time • Adaptation

☆anvipatel77☆

•Expert•

Brainly Community Contributor

5 0
2 years ago
What is the solution to the problem expressed to the correct number of significant figures? 2.13 + 1 = ?
Paha777 [63]

So,

With addition, we the last digit we keep will be the one which is known for both individual values.

We know 2.13 to the hundredths, but we only know 1 to the ones.  Therefore, we will round off in the ones place.

2.13 + 1 = 3.13 (unrounded)

= 3 (rounded)

Hope this helps!

7 0
3 years ago
Gaseous methane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water. If 2.59 g of water is produc
max2010maxim [7]

<u>Answer:</u> The percent yield of the water is 31.98 %

<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 methane:</u>

Given mass of methane = 6.58 g

Molar mass of methane = 16 g/mol

Putting values in equation 1, we get:

\text{Moles of methane}=\frac{6.58g}{16g/mol}=0.411mol

  • <u>For oxygen gas:</u>

Given mass of oxygen gas = 14.4 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

\text{Moles of oxygen gas}=\frac{14.4g}{32g/mol}=0.45mol

The chemical equation for the combustion of methane is:

CH_4+2O_2\rightarrow CO_2+2H_2O

By Stoichiometry of the reaction:

2 moles of oxygen gas reacts with 1 mole of methane

So, 0.45 moles of oxygen gas will react with = \frac{1}{2}\times 0.45=0.225mol of methane

As, given amount of methane 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

2 moles of oxygen gas produces 2 moles of water

So, 0.45 moles of oxygen gas will produce = \frac{2}{2}\times 0.45=0.45 moles of water

  • Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 0.45 moles

Putting values in equation 1, we get:

0.45mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(0.45mol\times 18g/mol)=8.1g

  • To calculate the percentage yield of water, we use the equation:

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

Experimental yield of water = 2.59 g

Theoretical yield of water = 8.1 g

Putting values in above equation, we get:

\%\text{ yield of water}=\frac{2.59g}{8.1g}\times 100\\\\\% \text{yield of water}=31.98\%

Hence, the percent yield of the water is 31.98 %

4 0
3 years ago
Why is the composition of the vapor different from the composition of the solution?
BaLLatris [955]
If there is solution with nonvolatile solute (<span>substance that does not readily </span>evaporate<span> into a </span>gas) <span>only the pure vapor of the solvent is present above the solution and solute stays in solution and do not enters vapor above solution. This is because nonvolatile solute has slow rate of evaporation and low vapore pressure.
If solution has two volatile components, composition of the vapor depends on vapor pressures of the components according </span><span>Raoult's Law.</span>
8 0
3 years ago
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