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3 years ago

What is the ionic compound for FePO4?

1 answer:

8 0

The ionic compound of FePO4 is the iron III phosphate. Or you can call it ferric phosphate or ferric orthophosphate.
Just for general informations, It's normally used in steel and metal manufacturing processes, in organic farming, and many other uses.

Hope this Helps :)

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Which of these solutions are basic at 25 Â°C? Solution A: [OHâˆ’]=3.13Ã—10âˆ’7 M Solution C: [H3O+]=0.000747 M Solution B: [H3O+

solong [7]

Answer:

Are basic:

[OH⁻] = 3.13x10⁻⁷M and [H₃O⁺] = 9.55x10⁻⁹M

Explanation:

A solution is basic when pH = - log [H₃O⁺] is higher than 7.

It is possible to convert [OH⁻] to [H₃O⁺] using:

[H₃O⁺] = 1x10⁻¹⁴ / [OH⁻]

a. [OH⁻] = 3.13x10⁻⁷M

[H₃O⁺] = 1x10⁻¹⁴ / [3.13x10⁻⁷M]

[H₃O⁺] = 3.19x10⁻⁸M

pH = - log [H₃O⁺] = 7.50

[OH⁻] = 3.13x10⁻⁷M is basic

b. pH = -log [H₃O⁺] = - log 0.000747M = 3.13.

This solution is not basic

c. [H₃O⁺] = 9.55x10⁻⁹M

pH = 8.02

This solution is also basic.

8 0

2 years ago

Can someone help me out please ? I am very stuck here

Alex17521 [72]

Hey there!

Atomic Masses :

H = 1.00794 a.m.u
N = 14.0067 a.m.u
O = 15.9994 a.m.u

Therefore:

HNO3 = 1.00794 + 14.0067 + ( 15.9994 * 3 ) => 63.0128 g/mol

7 0

3 years ago

Which planet likely has flowing liquid metallic hydrogen as the source of its magnetic field?

uysha [10]

Your answer is Saturn.

8 0

3 years ago

Valence electrons are: O a. Electrons that have been lost in an atom O b. Electrons in the nucleus of an atom O c. electrons in

mestny [16]

C ) outer most shell of atom

7 0

2 years ago

Question 26 Suppose a flask is filled with of and of . The following reaction becomes possible: The equilibrium constant for thi

Blizzard [7]

Answer:

0.36 M

Explanation:

There is some info missing. I think this is the complete question.

Suppose a 250 mL flask is filled with 0.30 mol of N₂ and 0.70 mol of NO. The following reaction becomes possible:

N₂(g) +O₂(g) ⇄ 2 NO(g)

The equilibrium constant K for this reaction is 7.70 at the temperature of the flask. Calculate the equilibrium molarity of O₂. Round your answer to two decimal places.

Initially, there is no O₂, so the reaction can only proceed to the left to attain equilibrium. The initial concentrations of the other substances are:

[N₂] = 0.30 mol / 0.250 L = 1.2 M

[NO] = 0.70 mol / 0.250 L = 2.8 M

We can find the concentrations at equilibrium using an ICE Chart. We recognize 3 stages (Initial, Change, and Equilibrium) and complete each row with the concentration or change in the concentration.

N₂(g) +O₂(g) ⇄ 2 NO(g)

I 1.2 0 2.8

C +x +x -2x

E 1.2+x x 2.8 - 2x

The equilibrium constant (K) is:

$K=7.70=\frac{[NO]^{2}}{[N_{2}][O_{2}]} =\frac{(2.8-2x)^{2} }{(1.2+x).x}$

Solving for x, the positive one is x = 0.3601 M

[O₂] = 0.3601 M ≈ 0.36 M

7 0

3 years ago