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Bezzdna [24]
3 years ago
6

In which atmosphere layer do humans live most of their life

Chemistry
1 answer:
marysya [2.9K]3 years ago
3 0
Troposphere, this is the layer of the atmosphere closest to the earths crust.
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How many atoms are in 1.0 formula units of magnesium acetate?
Ray Of Light [21]

Answer:

15

Explanation:

Magnesium Acetate Mg(C2H3O2)2

Number of atoms:

Carbon = 4

Hydrogen = 6

Magnesium = 1

Oxygen = 4

Total = 15

7 0
3 years ago
Brainliest is you answer quick!!!
Diano4ka-milaya [45]

Answer:

The inside layer of skin is the Insulator layer.

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8 0
1 year ago
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How to change an emerson thermostat from celsius to fahrenheit.
Nana76 [90]

How to change an emerson thermostat from celsius to fahrenheit.

Answer at the picture above.

5 0
2 years ago
A careless student forgets to label one of their reagent containers. They know it contains one of the following solutions: 0.1 M
antiseptic1488 [7]

Answer:

0.1 M NaCl

Explanation:

This question reminds of of the solubility rules. Let us recall that all chlorides are soluble except those of lead, mercury II and silver which are insoluble in water.

The following reaction will occur leading to the formation of a precipitate;

Pb(NO3)2(aq) + 2NaCl(aq) -------> 2NaNO3(aq) + PbCl2(s)

The white precipitate formed is  PbCl2.

4 0
2 years ago
For the following reaction, 53.7 grams of iron(III) oxide are allowed to react with 22.8 grams of aluminum. iron(III) oxide (s)
ZanzabumX [31]

Answer:

34.23 grams is the maximum amount of aluminum oxide that can be formed.

Iron (III) oxide is a limiting reagent i.e Fe_2O_3.

4.6764 grams is the amount of the aluminum which remains after the reaction is complete

Explanation:

iron(III) oxide (s) + aluminum (s) → aluminum oxide (s) + iron (s)

Fe2O_3(s)+2Al(s)\rightarrow Al_2O_3(s)+2Fe(s)

Moles of  iron(III) oxide : \frac{53.7 g}{160 g/mol}=0.3356 mol

Moles of aluminium : \frac{22.8 g}{27 g/mol}=0.8444 mol

According to recation, 1 mole of iron(III) oxide reacts with 2 moles of aluminum.

Then 0.3356 moles of iron(III) oxide will react with:

\frac{2}{1}\times 0.3356 mol=0.6712 mol of aluminum.

As we can see that moles of iron(III) are in limiting amount.Hence iron(III) oxide is a limiting reagent i.e Fe_2O_3 and aluminum in the an excessive reagent.

Amount of aluminum oxide will depend upon moles of limiting reagent that is iron(III) oxide.

According to reaction , 1 mole iron(III) oxide gives 1 moles of aluminum oxide.

Then 0.3356 moles will give:

\frac{1}{1}\times 0.3356 mol=0.3356 mol of aluminum oxide

Mass of 0.3356 moles of aluminum oxide:

0.3356 mol × 102 g/mol = 34.23 g

34.23 grams is the maximum amount of aluminum oxide that can be formed.

Moles of excessive reagent left = 0.8444 mol - 0.6712 mol = 0.1732 mol

Mass of 0.1732 moles of aluminum :

0.1732 mol × 27 g/mol = 4.6764 g

4.6764 grams is the amount of the aluminum which remains after the reaction is complete

7 0
3 years ago
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