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Basile [38]
3 years ago
7

What object has 13 protons

Chemistry
1 answer:
bekas [8.4K]3 years ago
5 0

Answer:

The isotope Al-27 has 13 protons

Explanation:

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The volume of 10 grams of frozen water is more than the volume of 10 grams of liquid water
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3 years ago
If 3.3167 moles of Al are produced, how many moles of<br> AlCl3 were reacted?<br> I
tresset_1 [31]

Answer:

3.3167 moles Of AlCl3

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

3Ca + 2AlCl3 —> 3CaCl2 + 2Al

From the balanced equation above,

2 moles of AlCl3 reacted to produce 2 moles of Al.

Finally, we shall obtained the number of moles of AlCl3 that reacted to produce 3.3167 moles of Al as follow:

From the balanced equation above,

2 moles of AlCl3 reacted to produce 2 moles of Al.

Therefore, 3.3167 moles Of AlCl3 will also react to produce 3.3167 moles of Al.

Thus, 3.3167 moles Of AlCl3 is needed for the reaction.

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Another student whose name is geraldine brought 2 litro packs of orange juice and 6 liters of bottled water. the ideal mixture t
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Answer:

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7 0
3 years ago
A cylinder with a moveable piston is completely filled with a small amount (100 millimoles) of liquid water at a pressure of 1.0
svet-max [94.6K]

Answer:

Region B, because the pressure inside the cylinder is equal to the vapor pressure of water at 80∘C when both liquid and gas phases are present.

Explanation:

As expansion occurs, liquid water evaporates reversibly, holding the pressure constant at the equilibrium vapor pressure of water at 80∘C(0.47atm) 80∘C (0.47 atm). When all of the liquid has evaporated, the pressure drops and follows the ideal gas law.

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Determine the heat energy needed to raise the temperature of 120 grams of ice at -5 to steam at 115°
CaHeK987 [17]

Answer:

Q = 30355.2 J

Explanation:

Given data:

Mass of ice = 120 g

Initial temperature = -5°C

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Solution:

Specific heat capacity of ice is = 2.108 j/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Q = m.c. ΔT

ΔT = T2 -T1

ΔT = 115 - (-5°C)

ΔT = 120 °C

Q = 120 g × 2.108 j/g.°C × 120 °C

Q = 30355.2 J

5 0
3 years ago
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