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frozen [14]
3 years ago
13

__NH4NO3 ---> _N2 + _H2O + _O2

Chemistry
1 answer:
Sveta_85 [38]3 years ago
6 0

Answer:

It is decomposition synthesis Combustion.

Explanation:

It is used to produce dinitrogen monoxide and water. The reaction usually takes place in a place in a temperature of 200-260 degrees Celsius.

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Which statement correctly uses a description of time?
Eva8 [605]

Answer:

A

Explanation:

it is because it began a new millennium

3 0
3 years ago
A student dissolves 3.9g of aniline (C6H5NH2) in 200.mL of a solvent with a density of 1.05 g/mL . The student notices that the
Tresset [83]

Answer:

a. Molarity= M =2.1x10^{-1}M

b. Molality= m=2.0x10^{-1}m

Explanation:

Hello,

In this case, given the information about the aniline, whose molar mass is 93g/mol, one could assume the volume of the solution is just 200 mL (0.200 L) as no volume change is observed when mixing, therefore, the molarity results:

M=\frac{n_{solute}}{V_{solution}} =\frac{3.9g*\frac{1mol}{93g} }{0.2L} =2.1x10^{-1}M

Moreover, the molality:

m=\frac{n_{solute}}{m_{solvent}} =\frac{3.9g*\frac{1mol}{93g} }{0.2L*\frac{1.05kg}{1L} } =2.0x10^{-1}m

Best regards.

7 0
3 years ago
What happens when a metal is reduced. Give an example of a metal being reduced
AlexFokin [52]

Answer:

The oxidation number of the metal decreases

2 Al  + Fe₂O₃ → Al₂O₃ + 2 FeO

The metal element iron, is reduced from Fe⁺³ in Fe₂O₃ to Fe⁺² in FeO

Explanation:

When an element gains electron, the element becomes reduced, hence when a metal is reduced, the metal gains electrons, which reduces the oxidation number of the metal

An example of a metal being reduced is;

2 Al  + Fe₂O₃ → Al₂O₃ + 2 FeO

In the above reaction, the iron (III) oxide is reduced to iron (II) oxide by aluminium metal.

6 0
3 years ago
What is the name of B2(SeO4)3
Gnoma [55]

This compound is Boron selenate. Molar mass of B2(SeO4)3 is 450.4948 g/mol.

4 0
3 years ago
Read 2 more answers
PLEASE HELP!! GIVING BRAINLIEST
RSB [31]

Answer:

The answer is "0.00172172603".

Explanation:

Given:

Mass (M) = 1.60 \times  10^{-3} \ g\\\\Density (D) = 9.293 \times 10^{-1} \ \frac{g}{cm^3}\\\\Volume (V) =  ?

Formula:

\to \bold{V = \frac{M}{D}}

       = \frac{1.60 \times  10^{-3}}{9.293 \times  10^{-1}} \\\\  = \frac{1.60 \times  10^{1}}{9.293 \times  10^{3}} \\\\ = \frac{1.60 \times  10}{9.293 \times  1000} \\\\ = \frac{1.60 }{9.293 \times  100} \\\\ = \frac{1.60 }{929.3 } \\\\= 0.00172172603

4 0
3 years ago
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