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3 years ago

When solutions of magnesium chloride and potassium phosphate are combined write the formula for the precipitate formed

1 answer:

6 0

The products will be magnesium phosphate and potassium chloride. You then have to watch a solubility chart to see which one of these is not soluable. In this case it is magnesium phosphate.

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How many liters of 4.0 M NaOH solution will react with 1.2 mol H2SO4? (Remember to balance the equation.)

nydimaria [60]

Answer:

The answer is D. 0.60 L

Explanation:

The balanced reaction equation including states of matter is;

H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)

More simple:

H2SO4 + 2NaOH → Na2SO4 + 2H2O

Now, we can see from this reaction equation that the mole ratio of NaOH to H2SO4 is 2:1

Number of moles of H2SO4 reacted = 1.2 moles

Hence;

2 moles of NaOH reacts with 1 mole of H2SO4

x moles of NaOH reacts with 1.2 moles of H2SO4

x = 2 * 1.2/1 = 2.4 moles of NaOH

Recall that;

Number of moles = Concentration * Volume

Volume = number of moles/concentration

Volume of NaOH is obtained from;

Volume = 2.4 moles/ 4.0 M

Volume = 0.60 L

6 0

3 years ago

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Vikki [24]

Yes what is this tho like

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2 years ago

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Any one from Michigan on here????

Andrew [12]

Answer:

dont go to any taco bells in Michigan!! There have been numerous cases of Ligma that have been traced back to Taco Bells across Michigan

Explanation:

7 0

2 years ago

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2AgNO3 + CaCl2 → 2AgCl + Ca(NO3)2

Harrizon [31]

It can be found that 337.5 g of AgCl formed from 100 g of silver nitrate and 258.4 g of AgCl from 100 g of CaCl₂.

<u>Explanation:</u>

2AgNO₃ + CaCl₂ → 2 AgCl + Ca(NO₃)₂

We have to find the amount of AgCl formed from 100 g of Silver nitrate by writing the expression.

$100 g \text { of } A g N O_{3} \times \frac{2 \text { mol } A g N O_{3}}{169.87 g A g N O_{3}} \times \frac{2 \text { mol } A g C l}{1 \text { mol } A g N O_{3}} \times \frac{143.32 g A g C l}{1 \text { mol } A g C l}$

= 337.5 g AgCl

In the same way, we can find the amount of silver chloride produced from 100 g of Calcium chloride.

It can be found as 258.4 g of AgCl produced from 100 g of Calcium chloride.

4 0

3 years ago

In the Haber process for ammonia synthesis, K " 0.036 for N 2 (g) ! 3 H 2 (g) ∆ 2 NH 3 (g) at 500. K. If a 2.0-L reactor is char

lisabon 2012 [21]

Answer : The partial pressure of $N_2,H_2\text{ and }NH_3$ at equilibrium are, 1.133, 2.009, 0.574 bar respectively. The total pressure at equilibrium is, 3.716 bar

Solution : Given,

Initial pressure of $N_2$ = 1.42 bar

Initial pressure of $H_2$ = 2.87 bar

$K_p$ = 0.036

The given equilibrium reaction is,

$N_2(g)+H_2(g)\rightleftharpoons 2NH_3(g)$

Initially 1.42 2.87 0

At equilibrium (1.42-x) (2.87-3x) 2x

The expression of $K_p$ will be,

$K_p=\frac{(p_{NH_3})^2}{(p_{N_2})(p_{H_2})^3}$

Now put all the values of partial pressure, we get

$0.036=\frac{(2x)^2}{(1.42-x)\times (2.87-3x)^3}$

By solving the term x, we get

$x=0.287\text{ and }3.889$

From the values of 'x' we conclude that, x = 3.889 can not more than initial partial pressures. So, the value of 'x' which is equal to 3.889 is not consider.

Thus, the partial pressure of $NH_3$ at equilibrium = 2x = 2 × 0.287 = 0.574 bar

The partial pressure of $N_2$ at equilibrium = (1.42-x) = (1.42-0.287) = 1.133 bar

The partial pressure of $H_2$ at equilibrium = (2.87-3x) = [2.87-3(0.287)] = 2.009 bar

The total pressure at equilibrium = Partial pressure of $N_2$ + Partial pressure of $H_2$ + Partial pressure of $NH_3$

The total pressure at equilibrium = 1.133 + 2.009 + 0.574 = 3.716 bar

6 0

3 years ago