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Bess [88]
3 years ago
6

Set up the ion formation equations, with ionization energy values for each electron in the valence layer, of the atoms of the ch

emical elements below: a) Na z=11 b) Ca z= 20 c) Sr z 38 d) Li z= 3 e) Cs z= 55 f) Be z= 43

Chemistry
1 answer:
devlian [24]3 years ago
6 0

Explanation:

here is your answer.........

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Explain why elements and compounds are pure substances
Anna11 [10]
<span>N2 - Element; Pure substance O2 - Element; Pure substance N2O - Compound; Pure Substance Air - Homogeneous; Solution</span>
8 0
3 years ago
Read 2 more answers
Describe an experiment for the preparation and collection of Oxygen sodium peroxide​
vitfil [10]

Answer:

Making oxygen

Oxygen can be made from hydrogen peroxide, which decomposes slowly to form water and oxygen:

hydrogen peroxide → water + oxygen

2H2O2(aq) → 2H2O(l) + O2(g)

The rate of reaction can be increased using a catalyst, manganese(IV) oxide. When manganese(IV) oxide is added to hydrogen peroxide, bubbles of oxygen are given off.

Apparatus arranged to measure the volume of gas in a reaction. Reaction mixture is in a flask and gas travels out through a pipe in the top and down into a trough of water. It then bubbles up through a beehive shelf into an upturned glass jar filled with water. The gas collects at the top of the jar, forcing water out into the trough below.

To make oxygen in the laboratory, hydrogen peroxide is poured into a conical flask containing some manganese(IV) oxide. The gas produced is collected in an upside-down gas jar filled with water. As the oxygen collects in the top of the gas jar, it pushes the water out.

Instead of the gas jar and water bath, a gas syringe could be used to collect the oxygen.

5 0
3 years ago
Metric conversions.<br> Please help ASAP.
lesya692 [45]

Answer:

14. 13.2cg = 1.32dg

15. 3.8m = 0.0038km

16. 24.8L = 24800mL

17. 0.87kL = 870L

18. 26.01cm = 0.0002601km

19. 0.001hm = 10cm

Explanation:

14. 13.2/10 = 1.32

15. 38/1000 = 0.0038

16. 24.8(1000) = 24,800

17. 0.87(1000) = 870

18. 26.01/100000 = 0.0002601

19. 0.001hm(10000) = 10

An easy way to do these by yourself is to familiarize yourself with what each prefix means. Once you do this, you can multiply the value of the prefix when converting from a smaller unit of measurement to a larger one and divide the value of the prefix when converting from a large unit of measurement to a smaller one.

7 0
3 years ago
Fe(NO3)2 not sure how to get the oxidation numbers of all elements
Shtirlitz [24]
You must remember that oxidation number of hydrogen in acids is always +1, oxidation number of oxygen in oxides & acids is always -2... metals has always oxidation number on plus!

group NO3 comes from HNO3...and oxidation number of whole acid group is always on minus and equal to the amount of hydrogen atoms in this acid... so oxidation number of NO3 = -1

we have 2 NO3 groups so 2*(-1) = -2 and that is the reason why oxidation number of Fe in this formula must be +2... because sum of all elements always gives 0!

Now we could count of oxidation number for nitrogen... we write HNO3 and start counting from right to left:
3*(-2) from oxygens + 1 from hydrogen = -5
so nitrogen must have +5 oxidation number... because sum all in formula must be 0.


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
3 years ago
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