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Ne4ueva [31]
3 years ago
12

Five alloys at home and it’s usage, pls help :(

Chemistry
1 answer:
kogti [31]3 years ago
8 0
Omg I’ll help you shawty:)
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What do I write? Please hurry
avanturin [10]

Answer:

I'd say Water

Explanation:

The plant absorbs more water through the roots not sure if i'm correct.

<h3 />
4 0
3 years ago
Covalent bonding is the ____ of electrons​
taurus [48]

Answer: sharing

Reason: They do this to gain stability. The reason they don’t actually transfer is because the difference in electronegativity values are above a certain value.

8 0
3 years ago
What is the valency of dioxide?
nasty-shy [4]
Thus, one atom of carbon will combine with two atoms of oxygen to form the molecule CO 2 ( carbon dioxide ). (b) From the chemical formula of the compound aluminium oxide Al 2 O 3 , the valency of aluminium can be determined. ... Hence the valency of aluminium is 3
7 0
4 years ago
Read 2 more answers
A 9.03 g sample of magnesium reacts completely with 3.48 g of nitrogen to form a compound of
hjlf

Answer:

12.5 g

Explanation:

<em>12.5 g of the compound would be formed.</em>

First, let us look at the balanced equation of reaction.

3 Mg(s) + N_2(g) --> Mg_3N_2(s)

3 moles of Mg is required to react with 1 mole of N2 to produce 1 mole of product.

<em>Recall that: mole = mass/molar mass</em>

9.03 g of Mg = 9.03/24.3 = 0.3716 mole

3.48 g of N2 = 3.48/28 = 0.1243 mole

Mole ratio of Mg/N2 = 3:1

<em>Hence, there is no limiting reactant.</em>

3 moles of Mg is required for 1 mole of product.

0.3716 mole of Mg will therefore require:

       0.3716 x 1/3 = 0.1239 moles of product.

Molar mass of product Mg_3N_2 = 100.9 g/mol

Mass of 0.1239 mole Mg_3N_2 = mole x molar mass

                      = 0.1239 x 100.9 = 12.5 g

8 0
3 years ago
A silver nitrate, AgNO3, solution of unknown concentration was discovered in the lab. To determine the concentration of the solu
harina [27]

Answer:

0.174 M

Explanation:

The same can be solve by using Nernst's equation

The Nernst's equation is:

Ecell=E^{0}_{cell}-\frac{0.0592}{n}log\frac{[anodic]}{[cathodic]}

For silver cell

n= 1

As both the compartments have silver nitrate solution,t the standard emf of cell will be zero.

Given

Ecell = 0.045

[Anodic compartment]= 1 M

Putting values

0.045 = 0 -\frac{0.0592}{1}log(\frac{1}{x} )

0.760=log(\frac{1}{x})

Taking antilog and solving

[AgNO3]=0.174 M

3 0
4 years ago
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