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3 years ago

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If you want to make 8.00 moles of AlF₃ how many moles of F₂ will you need, using the following balanced chemical equation? 2 Al

+ 3 F₂ → 2 AlF₃

1 answer:

Zinaida [17]3 years ago

7 0

Answer:

You will need 12 moles of F2 if you want to make 8 moles of AlF3.

Explanation:

It takes 3 moles F2 to make 2 moles of AlF3 (this will be our mole ratio)

2 moles AlF3/3 moles F2 =8 moles AlF3/x moles AlF3.

x=12 moles AlF3

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Suppose an object is moving through space in a straight line. What could cause the object to start moving in a circle?

MAVERICK [17]

It could pass by a large enough object that has enough gravity to pull the object into its orbit and the object would stay in orbit because it has centripetal force.

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3 years ago

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Which formula equation represents the burning of sulfur to produce sulfur dioxide?

Sunny_sXe [5.5K]

Answer:

The balanced reaction that describes the reaction between sulfur and oxygen to produce sulfur dioxide is expressed S(s) + O2 (g) = SO2 (g). In many manufacturing plants, sulfur dioxide is furhter reacted to oxygen to form sulfur trioxide then added with water to produce sulfuric acid.

Explanation:

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3 years ago

At 100°C, Kp = 60.6 for the reaction2NOBr(g) ⇄ 2NO(g) + Br₂(g) In a given experiment, 0.10 atm of each component is placed in a

Mariana [72]

The system is not at equilibrium.

The reaction will proceed to the right to attain the equilibrium.

Let's consider the following reaction.

2 NOBr(g) ⇌ 2 NO(g) + Br₂(g)

The pressure equilibrium constant (Kp) is 60.6. To determine if the system is at equilibrium when the pressure of each component is 1.75 atm, we have to calculate the reaction quotient (Q) and compare it with Kp.

Q = [NO]².[Br₂] / [NOBr]²

Q = (1.75)².(1.75) / (1.75)²

Q = 1.75

Since Q ≠ Kp, the system is not at equilibrium.

Since Q < Kp, the reaction will proceed to the right to attain the equilibrium.

To know more about equilibrium.

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1 year ago

The milliliter is a metric unit for

Anon25 [30]

Answer:

<u>d. Volume</u>

Explanation:

<em>The milliliter is a metric unit used for measuring the capacity or volume of an object.</em>

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2 years ago

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Sodium phosphate is added to a solution that contains 0.0030 M aluminum nitrate and 0.016 M calcium chloride. The concentration

gavmur [86]

Explanation:

It is given that aluminium nitrate and calcium chloride are mixed together with sodium phosphate.

And, $K_{sp} of AlPO_{4} = 9.84 \times 10^{-21}$

$K_{sp} of Ca_{3}(PO_{4})_{2} = 2.0 \times 10^{-29}$

Let us assume that the solubility be "s". And, the reaction equation is as follows.

$AlPO_{4} \rightleftharpoons Al^{3+} + PO^{3-}_{4}$

$9.84 \times 10^{-21} = s \times s$

s = $9.92 \times 10^{-11}$

Also, $Ca_{3}(PO_{4})_{2} \rightleftharpoons 3Ca^{2+} + 2PO^{3-}_{4}$

$2 \times 10^{-29} = (3s)^{3} \times (2s)^{2}$

s = $7.14 \times 10^{-7}$

This means that first, aluminium phosphate will precipitate.

Now, we will calculate the concentration of phosphate when calcium phosphate starts to precipitate out using the $K_{sp}$ expression as follows.

$K_{sp} = [Ca^{2+}]^{3}[PO^{3-}_{4}]^{2}$

$2.0 \times 10^{-29} = (0.016)^{3}[PO^{3-}_{4}]^{2}$

$2.0 \times 10^{-29} = 4.096 \times 10^{-6} \times [PO^{3-}_{4}]^{2}$

$[PO^{3-}_{4}]^{2}$ = $4.88 \times 10^{-24}$

= $2.21 \times 10^{-12}$ M

Similarly, calculate the concentration of aluminium at this concentration of phosphate as follows.

$AlPO_{4} \rightleftharpoons Al^{3+} + PO^{3-}_{4}$

$K_{sp} = [Al^{3+}][PO^{3-}_{4}]$

$9.84 \times 10^{-21} = [Al^{3+}] \times 2.21 \times 10^{-12}$

$[Al^{3+}] = 4.45 \times 10^{-9}$ M

Thus, we can conclude that concentration of aluminium will be $4.45 \times 10^{-9}$ M when calcium begins to precipitate.

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3 years ago