Circle which of the following depend on the amount of matter present:

If one mole of a substance has a mass of 56.0 g, what is the mass of 11 nanomoles of the substance? Express your answer in nanog

9966 [12]

Answer:

616,0 ng is the right answer.

Explanation:

You should know that 1 mole = 1 .10^9 nanomoles

Get the rule of three.

1 .10^9 nanomoles ...................... 56.0 gr

11 nanomoles .....................

(11 x 56) / 1 .10^9 nanomoles = 6.16 x 10^-7 gr

Let's convert

6.16 x 10^-7 gr x 1 .10^9 = 616 ngr

8 0

3 years ago

How many molecules are there in 4dm³ of nitrogen gas

daser333 [38]

There are 1.078 x 10²³ molecules

<h3>Further explanation</h3>

Given

4 dm³ = 4 L Nitrogen gas

Required

Number of molecules

Solution

Assumptions on STP (1 atm, 273 K), 1 mol gas = 22.4 L, so for 4 L :

mol = 4 : 22.4

mol = 0.179

1 mol = 6.02 x 10²³ particles(molecules, atoms)

For 0.179 :

= 0.179 x 6.02 x 10²³

= 1.078 x 10²³

5 0

2 years ago

Please help!

Dahasolnce [82]

Answer:

reproduction

Explanation:

reproduction, process by which organisms replicate themselves

6 0

2 years ago

HELP ASAP!!

Harman [31]

I would think it is a heterogeneous mixture since it can't be an element since there are more than one type of atom, it can't be a compound since the leaves are not bonded together, and it can not be a homogeneous mixture since the leaves don't all blended together (the pile is not uniform) and you can distinguish all the different parts of the mixture. It can be considered a heterogeneous mixture since the leaves are mixed together (along with other things like dirt) in a non-uniform way so that you can point out the parts of the mixture and it does not look like one thing.

I hope this helps. Let me know in the comments if anything is unclear.

4 0

3 years ago

How many grams of material is lost in the aqueous phase if two extractions are carried out on 100 mL of a 5% (m/v) aqueous solut

san4es73 [151]

Answer:

4.94g of material

Explanation:

Partition coefficient (Kp) of a substance is defined as the ratio between concentration of organic solution and aqueous solution, that is:

Kp = 8 = Concentration in Ethyl acetate / Concentration in water

100mL of a 5% solution contains 5g of material in 100mL of water. Thus:

8 = X / 100mL / (5g-X) / 100mL

Where X is the amount of material in grams that comes to the organic phase.

8 = X / 100mL / (5g-X) / 100mL

8 = 100X / (500-100X)

4000 - 800X = 100X

4000 = 900X

4.44g = X

Thus, in the first extraction you will lost 4.44g of material from the aqueous phase.

And will remain 5g-4.44g = 0.56g.

In the second extraction:

8 = X / 100mL / (0.56g-X) / 100mL

8 = 100X / (56-100X)

448 - 800X = 100X

448 = 900X

0.50g = X

In the second extraction, you will extract 0.50g of material

Thus, after the two extraction you will lost:

4.44g + 0.50g = 4.94g of material

6 0

3 years ago