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lukranit [14]
3 years ago
6

Where is pollen produced

Chemistry
2 answers:
nexus9112 [7]3 years ago
6 0
I looked this up for you "Stamen: The pollen producing part of a flower"
AlexFokin [52]3 years ago
5 0
Pollen is produced in the stamen.
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PLEASE HELP ME ASAP!!! ILL GIVE BRAINLIEST ITS PLATO!!!
BARSIC [14]

A. Fission creates new elements from which electricity can be generated.

Hope this helps!

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3 years ago
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Which best illustrates the way in which radiation? Giving brainy.
Ilya [14]

Answer:

I dont really understand your question. but maybe this might help.

https://quizlet.com/283377931/radiation-flash-cards/

Explanation:

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2 years ago
Fritz Haber, a German chemist, discovered a way to synthesize ammonia gas (NH3) by combining hydrogen and nitrogen gases accordi
Westkost [7]
1) Write the balanced equation to state the molar ratios:

<span>3H2(g) + N2(g) → 2NH3(g)

=> molar ratios = 3 mol H2 : 1 mol N2 : 2 mol NH3

What volume of nitrogen is needed to produce 250.0 L of ammonia gas at STP?

First, convert the 250.0 L of NH3 to number of moles at STP .

Use the fact that 1 mole of gas at STP occupies 22.4 L

=> 250.0 L * 1mol/22.4 L = 11.16 L

Second, use the molar ratio to find the number of moles of N2 that produces 11.16 L of NH3

=> 11.16 L NH3 * [1 mol N2 / 2 mol NH3] = 5.58 mol N2

Third, convert 5.58 mol N2 into liters at STP

=> 5.58 mol N2 * [22.4 L/mol] = 124.99 liters

Answer: 124,99 liters

What volume of hydrogen is needed to produce 2.50 mol NH3 at STP?
 

First, find the number of moles of H2 that produce 2.50 mol by using the molar ratios:

2.50 mol NH3 * [3mol H2 / 2 mol NH3] = 3.75 mol H2

Second, convert the number of moles to liters of gas at STP:

3.75 mol * 22.4 L/mol =  84 liters of H2

Answer: 84 liters

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6 0
3 years ago
Read 2 more answers
= 25 X 5 = (use the correct number of sig figs)
Anton [14]

Answer:

1.25 *10^2

Explanation:

25*5 = 125

= 1.25 *10^2

5 0
3 years ago
A student reacts 5.0 g of sodium with 10.0 g of chlorine and collect 5.24 g of sodium chloride. What is the percent yield of thi
Ede4ka [16]

Answer: The percent yield of this combination reaction is 41.3 %

Explanation : Given,

Mass of Na = 5.0 g

Mass of Cl_2 = 10.0 g

Molar mass of Na = 23 g/mol

Molar mass of Cl_2 = 71 g/mol

First we have to calculate the moles of Na and Cl_2.

\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}

\text{Moles of }Na=\frac{5.0g}{23g/mol}=0.217mol

and,

\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}

\text{Moles of }Cl_2=\frac{10.0g}{71g/mol}=0.141mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

2Na+Cl_2\rightarrow 2NaCl

From the balanced reaction we conclude that

As, 2 mole of Na react with 1 mole of Cl_2

So, 0.217 moles of Na react with \frac{0.217}{2}=0.108 moles of Cl_2

From this we conclude that, Cl_2 is an excess reagent because the given moles are greater than the required moles and Na is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of NaCl

From the reaction, we conclude that

As, 2 mole of Na react to give 2 mole of NaCl

So, 0.217 mole of HCl react to give 0.217 mole of NaCl

Now we have to calculate the mass of NaCl

\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl

Molar mass of NaCl = 58.5 g/mole

\text{ Mass of }NaCl=(0.217moles)\times (58.5g/mole)=12.7g

Now we have to calculate the percent yield of this reaction.

Percent yield = \frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100

Actual yield = 5.24 g

Theoretical yield = 12.7 g

Percent yield = \frac{5.24g}{12.7g}\times 100

Percent yield = 41.3 %

Therefore, the percent yield of this combination reaction is 41.3 %

4 0
3 years ago
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