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Kruka [31]
3 years ago
8

Which of the following occurs because of unequal heating of the Earth's surface?

Chemistry
2 answers:
lilavasa [31]3 years ago
6 0
It is winds because tides gravity and moon phases are all due either forces of nature or light wind currents are effected by different heated fronts of the earth going against each other. 
gladu [14]3 years ago
5 0
The answer is Winds ..................
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Iron and carbon monoxide are made by heating 5.53 kg of iron ore, Fe₂O₃, and carbon. What is the theoretical yield of iron in ki
vfiekz [6]

Answer:

Mass of Fe produced = 3.86785211069  ≈ 3.87 kg

Explanation:

From the question the chemical reaction can be written as follow :

Fe2O3 + C → Fe + CO  

Balance the equation

Fe2O3 + 3C → 2Fe + 3CO  

compute the molecular mass of Fe2O3 and atomic mass of Iron(Fe)

Molecular mass of Fe2O3 = 55.845(2) + 15.999(3) = 111.69 + 47.997 = 159.687 g

Atomic mass of iron = 55.845 g

From the balanced equation

159.687 g of Fe2O3  produces 2 ×  55.845 = 111.69 g of Fe(iron)

Convert the 5.53 kg to gram

1 kg = 1000 g

5.53 kg  = 5.53 × 1000 = 5530 g

since,

159.687 g of Fe2O3  produces 2 ×  55.845 = 111.69 g of Iron(Fe)

5530 g of  Fe2O3  will produce  

cross multiply

Mass of Fe produced =5530 × 111.69/159.687

Mass of Fe produced = 617645.7

/159.687

Mass of Fe produced = 3867.85211069  g

convert to kg

1000 g = 1 kg

3867.85211069   = 3867.85211069/1000

Mass of Fe produced = 3.86785211069  ≈ 3.87 kg

3 0
3 years ago
Which ones are wrong please tell the right answer it Hlep me a lot
DiKsa [7]
13.) would be cytoskeleton and number 16.) would be lysosomes
6 0
3 years ago
Calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting tota
Ghella [55]

Answer:

(a) ΔS_{sys}  = 2.881 J/K; ΔS_{sur}  = -2.881 J/K; total change in entropy = 0

(b)ΔS_{sys}  = 2.881 J/K; ΔS_{sur}  = 0 ; total change in entropy = 2.881 J/K

(c) ΔS_{sys}  = 0 ; ΔS_{sur}  = 0 ; total change in entropy = 0

Explanation:

In the given problem, we need to calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting total change in entropy, when a sample of nitrogen gas of mass 14 g at 298 K and 1.00 bar doubles its volume. We have the following variable:

mass (m) = 14 g

Temperature = 298 K

Pressure = 1.00 bar

Initial volume = V_{1}

Final volume = V_{2} = 2V_{1}

(a) Change in entropy of the system ΔS_{sys} = nRIn\frac{V_{2} }{V_{1} }

where R = 8.314 J/(mol*K)

n = number of moles = mass/molar mass = 14/ 28 = 0.5 moles

ΔS_{sys} = 0.5*8.314*ln2 = 2.881 J/K

Change in entropy of the surrounding ΔS_{sur} = -2.881 J/K

Therefore, for a reversible process, the total change in entropy = ΔS_{sys}+ΔS_{sur} = 2.881 - 2.881 = 0

(b) Because entropy is a state function, we use the same procedure as in part (a). Thus, ΔS_{sys}  = 2.881 J/K

Since surrounding does not change in this process ΔS_{sur} = 0.

total change in entropy = ΔS_{sys}+ΔS_{sur} = 2.881 - 0 = 2.88 J/K

(c) For an adiabatic reversible expansion, q(rev) = 0, thus:

ΔS_{sys}  = 0

Since heat energy is not transferred from the system to the surrounding

ΔS_{sur}  = 0

total change in entropy = ΔS_{sys}+ΔS_{sur} = 0

6 0
3 years ago
How can you find the charge on po4​
Anvisha [2.4K]

Answer:

Answer:

Explanation:

I hope it's helpful!

7 0
3 years ago
Read 2 more answers
Select the molecule where free rotation around one or more bonds is not possible. Group of answer choices N2H4 C2Cl4 NH3 C2H6 CC
Alexandra [31]

Answer:

C₂Cl₄

Explanation:

To know if free rotation around a bond in a compound is possible, we need to see the structure of the compound (picture in attachment).

In single bonds, which are formed by σ bonds, the atoms are not fixed in a single position, and free rotation is permitted.

Double and triple bonds are formed by a σ bond and one or two π bonds, respectively. These bonds do not allow rotation, since it is not possible to twist the ends without breaking the π bond.

The chloroethylene (C₂Cl₄) has two carbons with an sp2-sp2 hybridization, they are bonded together by a double bond. <u>Free rotation on this bond is not possible, because six atoms, including the carbon atoms, doubly bonded and the four chlorine atoms bonded to them, must be on the same plane. </u>

4 0
3 years ago
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