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Blizzard [7]
3 years ago
8

How does latitude affect the temperature in an area

Chemistry
1 answer:
Elina [12.6K]3 years ago
6 0

Answer:

Latitudes are used to indicate a precise location of any feature on the earth's surface.

Temperature changes with latitudes .At higher latitudes, the Sun's rays are less direct. The farther an area is from the Equator, the lower its temperature. At the poles, the Sun's rays are least direct.Near the equator sun's rays are perpendicular resulting into very high temperatures. Generally, around the world, it gets warmer towards the equator and cooler towards the poles.

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How many atoms of oxygen are in 5Li2SO4
olga55 [171]

Answer:

20 oxygen atoms

Explanation:

Li2SO4 has 4 oxygen atoms, so if there are 5 of that compound then there will be 20 oxygen atoms.

8 0
3 years ago
10) an object has mass of 50g 500mg and 0.1g. express the total mass of the object in grams
Pavel [41]
500 mg in g :

1 g ----------- 1000 mg
? -------------- 500 mg

500 x 1 / 1000 => 0.5 g

total mass:

50 g + 0.5 g + 0.1 g => 50.6 g

hope this helps!
8 0
3 years ago
Do the following statement describe physical or chemical property!
alexandr402 [8]
This is a chemical property due to radioactivity being a chemical characteristic of uranium. This fact describes its chemical structure.
8 0
2 years ago
Complete the table by writing correct formulas for the compounds formed by combining positive and negative ions. then name each
Gwar [14]
NH₄NO₃ - ammonium-nitrate.
(NH₄)₂CO₃ - ammonium carbonate.
NH₄CN - ammonium cyanide.
(NH₄)₃PO₄- ammonium phosphate.
Sn(NO₃)₄ - tin(IV) nitrate.
Sn(CO₃)₂ - tin(IV) carbonate.
Sn(CN)₄ - tin(IV) cyanide.
Sn₃(PO₄)₄ - tin(IV) phosphate.
Fe(NO₃)₃ - iron(III) nitrate.
Fe₂(CO₃)₃ - iron(III) carbonate.
Fe(CN)₃ - iron(III) cyanide.
FePO₄ - iron(III) phosphate.
Mg(NO₃)₂ - magnesium nitrate.
MgCO₃ - magnesium carbonate.
Mg(CN)₂ - magnesium cyanide.
Mg₃(PO₄)₂ - magnesium phosphate.
4 0
3 years ago
Use the standard half-cell potentials listed below to calculate the standard cell potential for the following reaction occurring
iren2701 [21]

Answer:

1.20 V

Explanation:

The standard cell potential is calculated from the expression

ε⁰ cell = ε⁰ oxidation + ε⁰  reduction

The species that will be reduced is the one with the higher standard reduction potential and the species that will be oxidized will be the one with the more negative reduction potential.

Thus for our question we will have

oxidation:

Pb(s)  →   Pb2+(aq) + 2 e-       ε⁰ oxidation       =  -   ε⁰  reduction

                                                                          =   - ( - 0.13 V ) = + 0.13 V

reduction    

Br2(l) + 2 e- → 2 Br-(aq)           ε⁰  reduction     = +1.07 V

ε⁰ cell = ε⁰ oxidation + ε⁰  reduction = + 0.13 V + 1.07 V  = 1.20 V

4 0
3 years ago
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