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2 years ago

Balance the equation KHCO3+ H3PO4 arrow K2HPO4+H20 C02.

Chemistry

1 answer:

kherson [118]2 years ago

4 0

Answer:

2KHCO3+ H3PO4 arrow K2HPO4 + 2 H20 + 2C02

Explanation:

2KHCO3+ H3PO4 arrow K2HPO4 + 2 H20 + 2C02

Start with the Ks. Don't start on the right. Just balance the Ks on the left. The reason is that the CO3 is going to break down as well.

2KHCO3+ H3PO4 arrow K2HPO4 + H20 + C02

Leave the PO4 alone. Now look at the Hs that are on the right. There are 5 on the left. Leave the K2HPO4 alone. If you put a 2 in front of tthe H2O then you have a total of 5 Hs on the right.

2KHCO3+ H3PO4 arrow K2HPO4 + 2H20 + C02

There are 2 carbons on the left. You need 2 on the right.

2KHCO3+ H3PO4 arrow K2HPO4 + 2H20 + 2C02

That should do it.

Let's make a table to show balance.

Species K H C PO4- O

Left 2 5 2 1 6 from CO3^(2-)

Right 2 5 2 1 6 from H2O/CO2

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Describe the sources of water pollution in hydrological cycle.

Bingel [31]

Explanation:

The hydrological cycle is the continuous cycling of water between land, open water surfaces and the sea. This cycle begins with evaporation, sunlight evaporates water from the surface of earth, next condensation happens, the water absorbed is now used to form clouds, after these clouds are filled to the maximum, precipitation happens, this can be in the form of rainfall and snow, this cycle finalizes when the precipitation of water runs off the land and back into water sources.

Sources of water pollution:

<em>During precipitation: </em>Smog can be gathered in the atmosphere, during precipitation this pollution can turn into acid rain.
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I hope you find this information useful and interesting! Good luck!

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2 years ago

What would happen to an object if the force of gravity on Erath decreased by half

abruzzese [7]

Answer:

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Explanation:

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3 years ago

The equilibrium constant Kc for the reaction PCl3(g) + Cl2(g) ⇌ PCl5(g) is 49 at 230°C. If 0.70 mol of PCl3 is added to 0.70 mol

Ymorist [56]

Answer : The correct option is, (B) 0.11 M

Solution :

First we have to calculate the concentration $PCl_3$ and $Cl_2$ .

$\text{Concentration of }PCl_3=\frac{\text{Moles of }PCl_3}{\text{Volume of solution}}$

$\text{Concentration of }PCl_3=\frac{0.70moles}{1.0L}=0.70M$

$\text{Concentration of }Cl_2=\frac{\text{Moles of }Cl_2}{\text{Volume of solution}}$

$\text{Concentration of }Cl_2=\frac{0.70moles}{1.0L}=0.70M$

The given equilibrium reaction is,

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

Initially 0.70 0.70 0

At equilibrium (0.70-x) (0.70-x) x

The expression of $K_c$ will be,

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

$K_c=\frac{(x)}{(0.70-x)\times (0.70-x)}$

Now put all the given values in the above expression, we get:

$49=\frac{(x)}{(0.70-x)\times (0.70-x)}$

By solving the term x, we get

$x=0.59\text{ and }0.83$

From the values of 'x' we conclude that, x = 0.83 can not more than initial concentration. So, the value of 'x' which is equal to 0.83 is not consider.

Thus, the concentration of $PCl_3$ at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of $Cl_2$ at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M