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ycow [4]
3 years ago
13

Explain why aluminum does not react with potassium nitrate (KNO3) although it reacts with copper nitrate.

Chemistry
1 answer:
Kipish [7]3 years ago
4 0

Answer:

Potassium nitrate react with aluminum to produce potassium aluminate, aluminum oxide and nitrogen. This reaction takes place at a temperature near 400°C.

Explanation:

6KNO3 + 10Al → 6KAlO2 + 2Al2O3 + 3N2

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Which substances will make a salt when combined?
BartSMP [9]
Vinegar and tea for sure
3 0
2 years ago
A company wants to develop audio speakers using an inexpensive type of plastic that has a very high quality of sound output. Whi
olga nikolaevna [1]

Answer:

a

Explanation:

Just took the test

6 0
3 years ago
g Determine the empirical formula for a compound that contains C, H and O. It contains 40.92% C, 4.58% H, and 54.50% O by mass.
Lady bird [3.3K]

Answer:

The empirical formula for the compound is C3H4O3

Explanation:

The following data were obtained from the question:

Carbon (C) = 40.92%

Hydrogen (H) = 4.58%

Oxygen (O) = 54.50%

The empirical formula for the compound can be obtained as follow:

C = 40.92%

H = 4.58%

O = 54.50%

Divide by their molar mass

C = 40.92/12 = 3.41

H = 4.58/1 = 4.58

O = 54.50/16 = 3.41

Divide by the smallest i.e 3.41

C = 3.41/3.41 = 1

H = 4.58/3.41 = 1.3

O = 3.41/3.41 = 1

Multiply through by 3 to express in whole number

C = 1 x 3 = 3

H = 1.3 x 3 = 4

O = 1 x 3 = 3

The empirical formula for the compound is C3H4O3

6 0
3 years ago
What is the molar solubility of marble (i.e., [ca2 ] in a saturated solution in normal rainwater, for which ph=5.60? express you
Brut [27]
Missing in your question :

Ksp of(CaCO3)= 4.5 x 10 -9

Ka1 for (H2CO3) =  4.7 x 10^-7

Ka2 for (H2CO3) = 5.6 x 10 ^-11

1) equation 1 for Ksp = 4.5 x 10^-9 

CaCO3(s)→ Ca +2(aq)    +  CO3-2(aq)  

2) equation 2 for Ka1 = 4.7 x 10^-7

 H2CO3 + H2O → HCO3- + H3O+

3) equation 3 for Ka2 = 5.6 x 10^-11

 HCO3-(aq) + H2O(l) → CO3-2 (aq)  + H3O+(aq)

so, form equation 1& 2&3 we can get the overall equation:
CaCO3(s)  +  H+(aq)  → Ca2+(aq)   + HCO3-(aq)

note: you could get the overall equation by adding equation 1 to the inverse of equation 3 as the following:
when the inverse of equation 3 is :

CO3-2 (aq) + H3O+ (aq) ↔ HCO3- (aq) + H2O(l)  Ka2^-1 = 1.79 x 10^10
when we add it to equation 1
CaCO3(s) ↔ Ca2+(aq)  +  CO3-2(aq)   Ksp = 4.5 x 10^-9

∴ the overall equation will be as we have mentioned before:
when H3O+ = H+

CaCO3(s) + H+(aq)  ↔ Ca2+ (aq) + HCO3-(aq)   K= 80.55

from the overall equation:

∴K = [Ca2+][HCO3-] / [H+]

when we have [Ca2+] = [HCO3-] so we can assume both = X

∴K = X^2 / [H+]

when we have the PH = 5.6 so we can get [H+]

PH = - ㏒[H+]
5.6 = -㏒[H]
∴[H] = 2.5 x 10^-6

so, by substitution on K expression:

∴ 80.55 = X^2 / (2.5 x10^-6)

∴X = 0.0142

∴[Ca2+] = X = 0.0142 
6 0
3 years ago
What happend if NH3+ HJ
Liula [17]
Are you sure you copied that down correctly? I know what NH3 is but HJ I am comming up blank on. The only thing HJ I know is hebdojoule which is unit system.
Are you doing chemical balancing?
6 0
3 years ago
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