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vichka [17]
3 years ago
5

Another metal phosphate is cobalt phosphate. It will behave similar to calcium phosphate in an acid solution. What is the net io

nic equation including phases for CoPO4(s) dissolving in H3O^+(aq)?
Chemistry
2 answers:
ryzh [129]3 years ago
8 0
We are given with
Cobalt phosphate - CoPO4

We are asked for the net ionic equation for the phosphate dissolving in H3O+

The net ionic equation is
CoPO4 (s) + H3O+ (aq) ----->  HPO42- (aq) + Co3+ (aq) + H2O *(l)
Sidana [21]3 years ago
7 0

Answer: The equation is given below.

Explanation:

Calcium phosphate does not dissolve in water but in acid it does.Similarly cobalt phosphate is also soluble in acidic solution as it get dispersed in the acid solution.

CoPO_4(s)+H_3O^(aq)+\rightarrow HPO_{4}^{2-}(aq)+Co^{3+}(aq)+H_2O(l)

When cobalt phosphate dissolves in acidic solution as it get strongly attracted by the acid molecules and dissociates into cobalt(III) ions, hydrogen phosphate and water molecule.

Hence, the net ionic equation for cobalt phosphate in hydronium ion is given above.

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Balance the following skeleton reaction and identify the oxidizing and reducing agents: Include the states of all reactants and
N76 [4]

Answer:

Oxidizing agent - CrO4^2-

Reducing agent- N2O

Explanation:

Let us look at the equation closely;

CrO4^2- (aq) + 3N2O(g) ------------> Cr^3+ (aq) + 3NO(g) [acidic]

The reduction half equation is;

CrO4^2- (aq) + 3e -------->Cr^3+ (aq)

Oxidation half equation is;

3N2O(g) ------>3 NO(g) +3 e

Note that the oxidizing agent participates in the reduction half equation while the reducing agent participates in the oxidation half equation as seen above.

8 0
3 years ago
A weather balloon is filled with helium that occupies a volume of 5.00 × 10^4 L at 0.995 atm and 32.0°C. After it is released, i
DaniilM [7]

Answer:

The new volume is 5.913*10^4 L

Explanation:

Step 1: Write out the formula to be used:

Using general gas equation;

P1V1 / T1 =P2V2 /T2

V2 = P1V1T2 / P2T1

Step 2: write out the values given and convert to standard unit's where necessary

P1 = 0.995atm

P2 0.720atm

V1 = 5*10^4 L

T1 = 32°C = 32+ 273 = 305K

T2 = -12°C = -12 + 273 = 261K

Step 3: Equate your values and do the calculation:

V2 = 0.995 * 5*10^4 * 261 / 0.720 * 305

V2 = 1298.475 * 10^4 / 219.6

V2 = 5.913 * 10^4 L

So the new volume of the balloon is 5.913*10^4 L

3 0
3 years ago
Use the image to answer the question.
Lena [83]

Answer:

1. the process of plants taking in sunlight and creating oxygen

2. outputs are oxygen

3. inputs are sunlight

Explanation:

7 0
3 years ago
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I need an answer now please asap and you will me marked brainiest please it's missing I need it now.
BaLLatris [955]
I really hope this helps I’ve been there too and I’m just doing this so I get the 20 words and I have the picture

8 0
2 years ago
What is the molarity (M) of the following solutions?
Dennis_Churaev [7]

Answer:

The molarity (M) of the following solutions are :

A. M = 0.88 M

B. M = 0.76 M

Explanation:

A. Molarity (M) of 19.2 g of Al(OH)3 dissolved in water to make 280 mL of solution.

Molar mass of Al(OH)3 = Mass of Al + 3(mass of O + mass of H)

                                      = 27 + 3(16 + 1)

                                      = 27 + 3(17) = 27 + 51

                                      = 78 g/mole

Al(OH)_3 = 78 g/mole

Given mass= 19.2 g/mole

Mole = \frac{Given\ mass}{Molar\ mass}

Mole = \frac{19.2}{78}

Moles = 0.246

Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution(L)}

Volume = 280 mL = 0.280 L

Molarity = \frac{0.246}{0.280)}

Molarity  = 0.879 M

Molarity  = 0.88 M

B .The molarity (M) of a 2.6 L solution made with 235.9 g of KBr​

Molar mass of KBr = 119 g/mole

Given mass = 235.9 g

Mole = \frac{235.9}{119}

Moles = 1.98

Volume = 2.6 L

Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution(L)}

Molarity = \frac{1.98}{2.6)}

Molarity = 0.762 M

Molarity = 0.76 M

4 0
3 years ago
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