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2 years ago

17. Which of the following statement best explains what pollination is?

Chemistry

2 answers:

babunello [35]2 years ago

8 0

Answer:

Explanation:

Have a nice day☺gsjsjamsvwksk

Anna35 [415]2 years ago

4 0

Answer:

ıt is the projesi where the polen of one flower reaches the sığmaz of another flower.

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Automotive air bags inflate when sodium azide, NaN3, decomposes explosively to its constituent elements. How many moles of nitro

attashe74 [19]

Answer:

2.29 g of N2

Explanation:

We have to start with the chemical reaction:

$NaN_3~->~Na~+~N_2$

The next step is to balance the reaction:

$2NaN_3~->~2Na~+~3N_2$

We can continue with the mol calculation using the molar mass of

$NaN_3$ (65 g/mol), so:

$3.55~g~NaN_3\frac{1~mol~NaN_3}{65~g~NaN_3}=0.054~mol~NaN_3$

Now, with the molar ratio between $NaN_3$ and $N_2$ we can calculate the moles of $N_2$ (2:3), so:

$0.054~mol~NaN_3\frac{3~mol~N_2}{2~mol~NaN_3}=0.0819~mol~N_2$

With the molar mass of $N_2$ we can calculate the grams:

$0.0819~mol~N_2=\frac{1~mol~N_2}{28~g~N_2}=2.29~g~N_2$

I hope it helps!

5 0

2 years ago

Which is an example of a compound ? sand , gold , water , iodine

OleMash [197]

Which is an example of a compound? sand, gold, water, or iodine? well, the answer is water, water is a compound.

4 0

3 years ago

Read 2 more answers

A gas of unknown identity diffuses at a rate of 155 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate o

Ostrovityanka [42]

Answer:

19.07 g mol^-1

Explanation:

The computation of the molecular mass of the unknown gas is shown below:

As we know that

$\frac{Diffusion\ rate\ of unknown\ gas }{CO_{2}\ diffusion\ rate} = \frac{\sqrt{CO_{2\ molar\ mass}} }{\sqrt{Unknown\ gas\ molercular\ mass } }$

where,

Diffusion rate of unknown gas = 155 mL/s

CO_2 diffusion rate = 102 mL/s

CO_2 molar mass = 44 g mol^-1

Unknown gas molercualr mass = M_unknown

Now placing these values to the above formula

$\frac{155mL/s}{102mL/s} = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ 1.519 = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ {\sqrt{M_{unknown}} } = \frac{\sqrt{44 g mol^{-1}}}{1.519} \\\\ {\sqrt{M_{unknown}} } = \frac{44 g mol^{-1}}{(1.519)^{2}}$

After solving this, the molecular mass of the unknown gas is

= 19.07 g mol^-1

4 0

3 years ago

Which of the following indicates the cathode rays has a large charge to mass ratio?

Anna007 [38]

Answer:

Gggg

Explanation:

8 0

2 years ago

Which of the following word best describe the diagram

adoni [48]

does it have multiple choice

3 0

3 years ago