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2 years ago

Salt is a pure compound. Yes No

1 answer:

mihalych1998 [28]2 years ago

7 0

Yes, salt is considered a pure substance. The definition of a pure substance is something that cannot be broken down further without losing the...

Salt is another compound. Its chemical formula is NaCl, which stands for sodium chloride. Tooth fillings may look like pure silver, but they are actually a solution of mercury in silver.

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Atoms of two adjacent elements in the third period are in the ground state. An atom of element A has only s electrons in its val

Dennis_Churaev [7]

Answer:

Explanation:

A is Magnesium, B is Aluminium both are adjacent element and lie in third period.

Magnesium with atomic number of 12 consist two s electrons in it's valence shell in ground state whereas, Aluminium which has atomic number of 13 consist three electrons in it's valence shell in the ground state out of which two are s electrons and only one p electron.

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3 years ago

Write a balanced equation for the oxidation-reduction reaction that occurs when hydrogen peroxide reacts with ferrous ion

Alex17521 [72]

H₂O₂ + 2FeSO₄ + H₂SO₄ → Fe₂(SO₄)₃ + 2H₂O

H₂O₂ + 2H⁺ + 2e⁻ → 2H₂O k=1
Fe²⁺ → Fe³⁺ + e⁻ k=2

H₂O₂ + 2H⁺ + 2Fe²⁺ → 2H₂O + 2Fe³⁺

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3 years ago

what does a set of four quantum numbers tell you about an electron. compare and contrast the locations and properties of two ele

daser333 [38]

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3 years ago

What is empirical and moleculer formula?.

den301095 [7]

The molecular formula shows the exact number of molecules. Therefor, the empirical formula is the simplest formula of the molecular formula

4 0

3 years ago

I NEED HELP PLEASE!!!! CHEMISTRY QUESTION: If 38 g of Li3P and 15 grams of Al2O3 are reacted, what total mass of products will r

maksim [4K]

Answer:

21.5 g.

Explanation:

Hello!

In this case, since the reaction between the given compounds is:

$2Li_3P+Al_2O_3\rightarrow 3Li_2O+2AlP$

We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:

$n_{Li_3P}^{reacted}=38gLi_3P*\frac{1molLi_3P}{51.8gLi_3P}=0.73molLi_3P$

Now, the moles of Li3P consumed by 15 g of Al2O3:

$n_{Li_3P}^{consumed \ by \ Al_2O_3}=15gAl_2O_3*\frac{1molAl_2O_3}{101.96gAl_2O_3} *\frac{2molLi_3P}{1molAl_2O_3} =0.29molLi_3P$

Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:

$m_{Li_2O}=0.29molLi_3P*\frac{3molLi_2O}{2molLi_3P} *\frac{29.88gLi_2O}{1molLi_2O} =13gLi_2O\\\\m_{AlP}=0.29molLi_3P*\frac{2molAlP}{2molLi_3P} *\frac{57.95gAlP}{1molAlP} =8.5gAlP$

Therefore, the total mass of products is:

$m_{products}=13g+8.5g\\\\m_{products}=21.5g$

Which is not the same to the reactants (53 g) because there is an excess of Li₃P.

Best Regards!

7 0

2 years ago