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Naddika [18.5K]
3 years ago
11

How many moles of chlorine are used up in a reaction that produces 0.35kg of BCl3

Chemistry
1 answer:
maw [93]3 years ago
4 0

4.48 mol Cl2. A reaction that produces 0.35 kg of BCl3 will use 4.48 mol of Cl2.

(a) The <em>balanced chemical equation </em>is

2B + 3Cl2 → 2BCl3

(b) Convert kilograms of BCl3 to moles of BCl3

MM: B = 10.81; Cl = 35.45; BCl3 = 117.16

Moles of BCl3 = 350 g BCl3 x (1 mol BCl3/117.16 g BCl3) = 2.987 mol BCl3

(c) Use the <em>molar ratio</em> of Cl2:BCl3 to calculate the moles of Cl2.

Moles of Cl2 = 2.987 mol BCl3 x (3 mol Cl2/2 mol BCl3) = 4.48 mol Cl2

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How many sulfur atoms are there in 3.90 mol of sulfur?
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3 years ago
What is the kinetic energy in kJ of 1 mole of water molecules (mass=18) if the average velocity is 590 m/s (1300 mph)
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= 3132.9 Joules

Explanation:

  • Kinetic energy is the energy possessed by a body when in motion.
  • Kinetic energy is calculated by the formula; K.E = 1/2 mV², where m is the mass of the body or object, and V is the velocity.
  • Therefore kinetic energy depends on the mass and the velocity of the body or the object in motion.

In this case;

Kinetic energy = 0.5 × 0.018 kg × 590²

                        <u>= 3132.9 Joules</u>

7 0
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In the water cycle diagram above, which letter represents
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3 years ago
Before any reaction occurs, the concentration of A in the reaction below is 0.0510 M. What is the equilibrium constant if the co
Lilit [14]

Answer:

0.0119

Explanation:

There was a part missing. I think this is the whole question:

<em>Before any reaction occurs, the concentration of A in the reaction below is 0.0510 M. What is the equilibrium constant if the concentration of A at equilibrium is 0.0153 M?</em>

A (aq) ⇌ 2B (aq) + C(aq)

<em>Remember to use correct significant figures in your answer. Do not include units in your response.</em>

First, we have to make an ICE Chart, which stands for initial, change and equilibrium. We will call "x" unknown concentrations.

        A (aq) ⇌ 2B (aq) + C (aq)

I       0.0510       0            0

C         -x          +2x          +x

E    0.0510-x     2x            x

Since the concentration at equilibrium of A is 0.0153 M, we get

0.0510 - x = 0.0153 \\x = 0.0357

We can use the value of x to calculate the concentrations at equilibrium.

[A]e = 0.0153 M \\[B]e = 2x = 2(0.0357) = 0.0714 M \\[C]e = x = 0.0357 M \\

The equilibrium constant, Kc, is the ratio of the equilibrium concentrations of products over the equilibrium concentrations of reactants each raised to the power of their stoichiometric coefficients.

Kc = \frac{[B]^{2}  \times [C]}{[A]} = \frac{0.0714^{2}  \times 0.0357}{0.0153} = 0.0119

The equilibrium constant for this reaction at equilibrium is 0.0119.

You can learn more about equilibrium here: brainly.com/question/4289021

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2 years ago
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