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3 years ago

Question 2 (1 point)

Chemistry

1 answer:

SVEN [57.7K]3 years ago

7 0

Answer:

B. Unfilled d orbitals

Explanation:

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Plz help this was due yesterday butttttt i didn’t want to do it...

sergejj [24]

So what am I suppose to answer here?

8 0

3 years ago

Read 2 more answers

Na2CrO4 + PbCL2<br> Formula in double replacement

Marta_Voda [28]

Na₂CrO₄ + PbCl₂ → PbCrO₄ + 2 NaCl

<u>Explanation:</u>

In a double displacement reaction, the reactants which are involved in the reaction exchanging their ions thereby produces 2 new compounds. Here sodium chromate and lead chloride are undergoing double displacement reaction, the ions exchanges their position there by forming sodium chloride and lead chromate. So the double displacement reaction is given as,

Na₂CrO₄ + PbCl₂ → PbCrO₄ + 2 NaCl

6 0

3 years ago

If the temperature on 244 mL of a gas is changed to 488 mL and 6 atm, at constant

Fittoniya [83]

Answer:

Explanation:

To find the initial pressure we use the formula for Boyle's law which is

$P_1V_1 = P_2V_2$

Since we are finding the initial pressure

$P_1 = \frac{P_2V_2}{V_1} \\$

From the question we have

$P_1 = \frac{488 \times 6}{244} = \frac{2928}{244} \\$

We have the final answer as

Hope this helps you

6 0

3 years ago

Consider the reaction: 2 SO2(g) + O2(g) <----> 2 SO3. If, at equilibrium at a certain temperature, [SO2] = 1.50 M, [O2] =

vlada-n [284]

Answer:

5.79

Explanation:

The following data were obtained from the question:

Concentration of SO2, [SO2] = 1.50 M

Concentration of O2, [O2] = 0.120 M

Concentration of SO3, [SO3] = 1.25 M

Equilibrium constant, (K) =..?

The balanced equation for the reaction is given below:

2SO2(g) + O2(g) <==> 2SO3(g)

From the balanced equation, the equilibrium constant K is written as follow:

K = [SO3]²/ [SO2]²• [O2]

With the above, the value of the equilibrium constant, K can be obtained as follow

K = [SO3]²/ [SO2]²• [O2]

K = (1.25)² / (1.50)² × 0.120

K = 1.5625 / ( 2.25 x 0.120)

K = 5.79

Therefore, the equilibrium constant is 5.79.

7 0

3 years ago

1) Given the balance equation below. Calculate how much Na3PO4 in grams you

juin [17]

Answer:

<u>136.67 g of Na3PO4 i</u>s required to create 100 gram of NaOH.

Explanation:

The balanced equation:

$Na_{3}PO_{4} + 3 KOH \rightarrow 3 NaOH + K_{3}PO_{4}$

1 mole Na3PO4 = 164 g/mole (Molar mass)

1 mole NaOH = 40 g/mole (Molar mass)

Now,

1 mole of Na3PO4 produce = 3 mole of NaOH

164 g/mol of Na3PO4 produce = 3(40) g/mol of NaOH

120 g/mol of NaOH is produced from = 164 g/mol of Na3PO4

1 g/mol of NaOH is produced from =

$\frac{164}{120}$

100 grams of NaOH is produced from =

$\frac{164}{120}\times100$ gram of Na3PO4

calculate,

= 136.67 g

8 0

3 years ago