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cestrela7 [59]
3 years ago
15

Potassium (K) combines with magnesium bromide (MgBr2) to form potassium bromide (KBr) and magnesium (Mg) during a chemical react

ion.
Which of the following shows the balanced chemical equation for this reaction?

a) K + MgBr2 = KBr + Mg
b) 2 K + MgBr2 → 2 KBr + Mg
c) 2 K + MgBr2 = KBr + 2 Mg
d) 2 K +2 MgBr2 = 2 KBr + 2 Mg
Chemistry
1 answer:
amm18123 years ago
5 0

Answer: option B - 2 K + MgBr2 = 2 KBr + Mg

Explanation:

2K + MgBr2 --> 2 KBr + Mg

On Left side: 2moles of Potassium (K) reacts with 1 mole of MgBr2

On Right side: 2 moles of KBr and 1 mole of Mg.

You will observe that the coefficient of each element on left side and Right side is balanced in option B, unlike other options.

So, option B is the answer.

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Planck's constant relates the energy in one photon of electromagnetic radiation to the frequency of the radiation, varying:
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How many molecules of lactate can one molecule of lactate dehydrogenase act on in 21 minutes?
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4 years ago
The equilibrium constant, K., for the following reaction is 83.3 at 500 K. PC13(g) + Cl2(g) = PC13(E) Calculate the equilibrium
lord [1]

Answer:

The equilibrium concentration of PCl_5=0.228 M.

The equilibrium concentration of PCl_3=0.280 M -0.228 M=0.052M.

The equilibrium concentration of Cl_2=0.280 M -0.228 M=0.052M.

Explanation:

Answer:

The equilibrium concentration of HCl is 0.01707 M.

Explanation:

Equilibrium constant of the reaction = K_c=83.3

Moles of PCl_3 = 0.280 mol

Concentration of  [PCl_3]=\frac{0.280 mol}{1.00 L}=0.280 M

Moles of Cl_ = 0.280 mol

Concentration of [Cl_2]=\frac{0.280 mol}{1.00 L}=0.280M

           PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)

Initial:            0.280      0.280                            0

At eq'm:         (0.280-x)   (0.280-x)                     x

We are given:

[PCl_3]_{eq}=(0.280-x)

[Cl_2]_{eq}=(0.280-x)

[PCl_5]_{eq}=x

Calculating for 'x'. we get:

The expression of K_{c} for above reaction follows:

K_c=\frac{[PCl_5]}{[PCl_3][Cl_2]}

Putting values in above equation, we get:

83.3=\frac{x}{(0.280-x)\times (0.280-x)}

On solving this quadratic equation we get:

x = 0.228, 0.344

0.228 M < 0.280 M< 0.344 M

x = 0.228 M

The equilibrium concentration of PCl_5=0.228 M.

The equilibrium concentration of PCl_3=0.280 M -0.228 M=0.052M.

The equilibrium concentration of Cl_2=0.280 M -0.228 M=0.052M.

4 0
3 years ago
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