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4 years ago

14

Which organisms are capable of converting gaseous nitrogen in the air into a form that other living organisms can use?

1 answer:

Alexeev081 [22]4 years ago

7 0

The answer is A() nitrogen-fixing bacteria

it converts nitrogen it kind of says it in the name

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PLEASE HELP URGET!!!!

strojnjashka [21]

Answer:

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Explanation:

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5 0

3 years ago

How many grams of H3PO4 are in 175mL of a 2.50M solution of H3PO4?

solmaris [256]

<span>(0.1875 moles)(98.004 g/mole) = 18.37575 g </span>
<span>In correct number of significant figures: 18.4 </span>

5 0

3 years ago

Read 2 more answers

Purification of copper can be achieved by electrorefining copper from an impure copper anode onto a pure copper cathode in an el

Dovator [93]

Answer: 406 hours

Explanation:

$Q=I\times t$

where Q= quantity of electricity in coloumbs

I = current in amperes = 39.5 A

t= time in seconds = ?

The deposition of copper at cathode is represented by:

$Cu^{2+}+2e^-\rightarrow Cu$

$96500\times 2=193000$ Coloumb of electricity deposits 1 mole of copper

i.e. 63.5 g of copper is deposited by = 193000 Coloumb

Thus 19.0 kg or 19000 g of copper is deposited by = $\frac{193000}{63.5}\times 19000=57748032$ Coloumb

$57748032=39.5\times t$

$t=1461975sec=406hours$ (1hour=3600s)

Thus it will take 406 hours to plate 19.0 kg of copper onto the cathode if the current passed through the cell is held constant at 39.5 A

3 0

3 years ago

The edge length of the unit cell of KCl (NaCl-like structure, FCC) is 6.28 Å. Assuming anion-cation contact along the cell edge,

Mila [183]

Answer:

1.33 Å

Explanation:

Given that the edge length , a of the KCl which forms the FCC lattice = 6.28 Å

Also,

For the FCC lattice in which the anion-cation contact along the cell edge , the ratio of the radius of the cation to that of anion is 0.731.

Thus,

$\frac {r^+}{r^-}=0.731$ .................1

Also, the sum of the radius of the cation and the anion in FCC is equal to half of the edge length.

Thus,

$r^++r^-=\frac {a}{2}$ ...................2

Given that:

$Cl^-\ (r^-) = 1.82\ \dot{A}$

To find,

$K^+\ (r^+) = ? \dot{A}$

Using 1 and 2 , we get:

$1.731\ r^+=0.731\times \frac {6.28}{2}$

<u>Size of the potassium ion = 1.33 Å</u>

4 0

3 years ago

Use the IUPAC nomenclature rules to give the name for this compound - Al2S3.

jonny [76]

Aluminium Sulfide
According to rules the positive specie is named first and the negative specie is named last.

6 0

3 years ago

Read 2 more answers