_____ NaPO4+____

What length of time does the geologic time scale cover?

Nezavi [6.7K]

I would say you are correct so b 4.6

8 0

3 years ago

Read 2 more answers

Explain why the catalyst is used as a very fine powder and larger pieces of iron are not used.

Allushta [10]

Answer:

<h2><u>Reason:</u></h2>

Catalyst is used as a very fine powder and larger pieces of iron are not used. This is because the surface area of catalyst needs to be large so that more of the surface is exposed to the substrate and more of the substrate is catalyzed.

<h2><u>Important Info:</u></h2>

=> Larger Pieces of Iron has a smaller surface area than the fine particles.

=> Larger the surface area of catalysts/enzymes , more will be the reaction rate and vice versa.

Hope this helped!

<h2>~AnonymousHelper1807</h2>

7 0

3 years ago

Nickel has an FCC structure and an atomic radius of 0.124 nm. (12 points) a. What is the coordination number? b. Calculate the A

LekaFEV [45]

Answer:

a)CN=12

b)APF=74 %

c)a=0.35 nm

d)ρ=9090.9 $Kg/m^3$

Explanation:

Given that

Nickel have FCC structure

We know that in FCC structure ,in FCC 8 atoms at corner with 1/8 th part in one unit cell and 6 atoms at faces with 1/2 part in one unit cell .

Z=8 x 1/8 + 1/2 x 6 =4

Z=4

Coordination number (CN)

The number of atoms which touch the second atoms is known as coordination number.In other word the number of nearest atoms.

CN=12

Coordination number of FCC structure is 12.These 12 atoms are 4 atoms at the at corner ,4 atoms at 4 faces and 4 atoms of next unit cell.

APF

$APF=\dfrac{Z\times \dfrac{4}{3}\pi R^3}{a^3}$

We know that for FCC

$4R=\sqrt{2}\ a$

Now by putting the values

$APF=\dfrac{4\times \dfrac{4}{3}\pi R^3}{\left(\dfrac{4R}{\sqrt2}\right)^3}$

APF=0.74

APF=74 %

$4R=\sqrt{2}\ a$

$4\times 0.124=\sqrt{2}\ a$

a=0.35 nm

Density

$\rho=\dfrac{Z\times M}{N_A\times a^3}$

We know that M for Ni

M=58.69 g/mol

a=0.35 nm

$N_A=6.023\times 10^{23}\ atom/mol$

$\rho=\dfrac{Z\times M}{N_A\times a^3}$

$\rho=\dfrac{4\times 58.69}{6.023\times 10^{23}\times( 0.35\times 10^{-9})^3}\ g/m^3$

ρ=9090.9 $Kg/m^3$

3 0

3 years ago

How many moles of co2 are produced when 5.70 mol of ethane are burned in an excess of oxygen

sweet-ann [11.9K]

I believe the answer is 6

4 0

4 years ago

A chemist adds of a sodium thiosulfate solution to a reaction flask. Calculate the mass in grams of sodium thiosulfate the chemi

schepotkina [342]

This is a incomplete question.The complete question is:

A chemist adds 180.0 ml of a 1.77 mol/L of sodium thiosulfate solution to a reaction flask. Calculate the mass in grams of sodium thiosulfate the chemist has added to the flask. Be sure your answer has the correct number of significant digits.

Answer: 50.4 g

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Moles of solute}={\text{Molarity of the solution}}\times{\text{Volume of solution (in L)}}$ .....(1)

Molarity of sodium thiosulfate solution = 1.77 M

Volume of sodium thiosulfate solution = 180.0 mL = 0.1800 L

Putting values in equation 1, we get:

$\text{Moles of sodium thiosulfate}={1.77}\times{0.1800}=0.319moles$

Mass of sodium thiosulfate = $moles\times {\text {Molar mass}}=0.319moles\times 158.11g/mol=50.4g$

Thus 50.4 g of sodium thiosulfate the chemist has added to the flask.

6 0

3 years ago

_____ NaPO4+_____KOH

_ NaPO4+_KOH