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Naya [18.7K]
3 years ago
9

What happen when the nitrate of an alkali metal is heatad?​

Chemistry
2 answers:
Art [367]3 years ago
8 0
Most nitrates tend to decompose on heating to give the metal oxide, brown fumes of nitrogen dioxide, and oxygen.
sergey [27]3 years ago
7 0

answer

Gives brown colour gas

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How many kilograms of a fertilizer are made of pure P2O5 would be required to supply 1.69 kilogram of phosphorus to the soil?
Alekssandra [29.7K]

Answer:

3.89 kg P2O5 must be used to supply 1.69 kg Phosphorus to the soil.

Explanation:

The molecular mass of P2O5 is

P2 = 2* 31 =           62

O5 = 5 *<u> 16 =         80</u>

Molecular Mass = 142

Set up a Proportion

142 grams P2O5 supplies 62 grams of phosphorus

x    kg P2O5        supplies 1.69 kg of phosphorus

Though this might be a bit anti intuitive, you don't have to convert the units for this question. The ratio is all that is important.

142/x = 62/1.69            Cross multiply

142 * 1.69 = 62x           combine the left

239.98 = 62x               Divide by 62

239.98/62 = x

3.89 kg of P2O5 must be used.

3 0
3 years ago
Pcl5 exist but ncl5 does not.why?
jeyben [28]
N is as electronegative as Cl is. So Cl doesnt react with O, N
6 0
3 years ago
How can scientists determine what kind of atoms are in stars and nebulas?
statuscvo [17]
Each element absorbs light at specific wavelengths unique to that atom. When astronomers look at an object's spectrum, they can determine its composition based on these wavelengths.
Not sure if this is 100% but I hope it helps.
4 0
3 years ago
What is the experimental group? on edpuzzle
RideAnS [48]
An experimental group, also known as a treatment group, receives the treatment whose effect researchers wish to study, whereas a control group does not. They should be identical in all other ways
3 0
3 years ago
Read 2 more answers
In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide Al2O3 dissolved in molten c
gtnhenbr [62]

Answer:

1.13 × 10⁶ g

Explanation:

Let's consider the reduction of aluminum (III) from Al₂O₃ to pure aluminum.

Al³⁺ + 3 e⁻ → Al

We can establish the following relations:

  • 1 Ampere = 1 Coulomb / second
  • The charge of 1 mole of electrons is 96,468 c (Faraday's constant)
  • 1 mole of Al is produced when 3 moles of electrons circulate
  • The molar mass of Al is 26.98 g/mol.

The mass of aluminum produced under these conditions is:

90.0 s \times \frac{1s}{620c} \times \frac{96,468c}{1mole^{-} } \times \frac{3mole^{-}}{1molAl} \times \frac{26.98gAl}{1molAl} =1.13 \times 10^{6} g Al

6 0
3 years ago
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