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2 years ago

Help pls I really need the answer I will brainlest

Chemistry

2 answers:

zlopas [31]2 years ago

6 0

Explanation:

latitude at horizontal circular line and show anguler distance from the equator and Longitudes are semi-circular lines and show agular distance from the prime Meridian .

jeka57 [31]2 years ago

3 0

Answer:

Latitude: the angular distance of a place north or south of the earth's equator, or of a celestial object north or south of the celestial equator, usually expressed in degrees and minutes. (Left to right)

Longitude: the angular distance of a place east or west of the meridian at Greenwich, England, or west of the standard meridian of a celestial object, usually expressed in degrees and minutes. (Up and down)

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During fusion _____. hydrogen atoms are split apart U-235 atoms are created helium atoms are created uranium atoms are fused tog

musickatia [10]

The correct answer is the second option. During fusion, uranium atoms are fused together. Fusion reaction happens when two or more nuclei combine or collide to form an element with a higher atomic number. In this process, some of the matter of the fusing nuclei is converted to energy.

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3 years ago

How many moles are represented by 118g of cobalt? Cobalt had an atomic mass of 58.93amu

skad [1K]

N=m/M
n=118/58.93=2
Answer: 2 moles

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3 years ago

Pls answer these and show work

Assoli18 [71]

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3 years ago

9. A gas canister can tolerate internal pressures up to 210 atmospheres. If a 2.0 L

Reika [66]

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2 years ago

The most useful ore of aluminum is bauxite, in which Al is present as hydrated oxides, Al2O3 * xH2O.

lesantik [10]

Answer:

44.7 kWh

Explanation:

Let's consider the reduction of Al₂O₃ to Al in the Bayer process.

6 e⁻ + 3 H₂O + Al₂O₃ → 2 Al + 6 OH⁻

We can establish the following relations:

The molar mass of Al is 26.98 g/mol.
2 moles of Al are produced when 6 moles of e⁻ circulate.
1 mol of e⁻ has a charge of 96468 c (Faraday's constant).
1 V = 1 J/c
1 kWh = 3.6 × 10⁶ J

When the applied electromotive force is 5.00 V, the energy required to produce 3.00 kg (3.00 × 10³ g) of aluminum is:

$3.00 \times 10^{3} gAl.\frac{1molAl}{26.98gAl} .\frac{6mole^{-}}{2molAl}.\frac{96468c}{1mole^{-}}.\frac{5.00J}{c}.\frac{1kWh}{3.6 \times 10^{6}J} =44.7kWh$

6 0

3 years ago

Help pls I really need the answer I will brainlest​

Help pls I really need the answer I will brainlest