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PSYCHO15rus [73]
2 years ago
5

Water is a substance made up of the elements hydrogen

Chemistry
2 answers:
hichkok12 [17]2 years ago
6 0
I think it might be 3
xeze [42]2 years ago
3 0

Answer:

A. 2

Explanation:

A particle of water is made up of two hydrogen atoms and one oxygen atom. However, the question asked for how many types of atoms there are, so the answer is 2.

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2Al + 3H2SO4 -> Al2(SO4)3 + 3H2How many grams of aluminum sulfate would be formed if 250g H2SO4 completely reacted with alumi
bazaltina [42]

Answer:

290.82g

Explanation:

The equation for the reaction is given below:

2Al + 3H2SO4 -> Al2(SO4)3 + 3H2 now, let us obtain the masses of H2SO4 and Al2(SO4)3 from the balanced equation. This is illustrated below:

Molar Mass of H2SO4 = (2x1) + 32 + (16x4) = 2 + 32 +64 = 98g/mol

Mass of H2SO4 from the balanced equation = 3 x 98 = 294g

Molar Mass of Al2(SO4)3 = (2x27) + 3[32 + (16x4)]

= 54 + 3[32 + 64]

= 54 + 3[96] = 54 + 288 = 342g

Now, we can obtain the mass of aluminium sulphate formed by doing the following:

From the equation above:

294g of H2SO4 produced 342g of Al2(SO4)3.

Therefore, 250g of H2SO4 will produce = (250 x 342)/294 = 290.82g of Al(SO4)3

Therefore, 290.82g of aluminium sulphate (Al(SO4)3) is formed.

7 0
3 years ago
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.
  • <em>During runoff:</em> After acid rain hits the ground this polluted water can run into water sources (lakes, rivers, reservoirs).To some extent rivers are a self-renewing resource, if a small quantity of pollution discharges in it the river can return to a clean, unpolluted condition, unfortunately, if the pollution is too big the renewing won't be possible, another problem is even though rivers get cleaned the pollution moves to the seas. Lakes are even more vulnerable to pollution, the flushing effect in these water bodies is less evident than in rivers.

I hope you find this information useful and interesting! Good luck!

8 0
2 years ago
Sodium oxide reacts with water to make sodium hydroxide.
Vlada [557]

Answer:

So the total mass is 50 plus 150 grams the heat capacity 4.18 joules per gram per degree C. And the temperature change is 36 minus 25 and so we can calculate Delta H for the reaction that takes place.

Explanation:

3 0
2 years ago
What is the mass of 1.72 moles of sodium nitrate? Use the periodic table and the polyatomic ion resource. A. 85.0 g B. 91.2 g C.
Mrac [35]
The answer is C. 146g because you add all of the masses of the individual elements and then mulyiply by 1.72 to get your answer.
7 0
3 years ago
Read 2 more answers
Suppose a 0.025M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4−2. You'll find
FromTheMoon [43]

<u>Answer:</u> The concentration of SO_4^{2-} at equilibrium is 0.00608 M

<u>Explanation:</u>

As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.

In the second dissociation, the ions will remain in equilibrium.

We are given:

Concentration of sulfuric acid = 0.025 M

Equation for the first dissociation of sulfuric acid:

       H_2SO_4(aq.)\rightarrow H^+(aq.)+HSO_4^-(aq.)

            0.025          0.025       0.025

Equation for the second dissociation of sulfuric acid:

                    HSO_4^-(aq.)\rightarrow H^+(aq.)+SO_4^{2-}(aq.)

<u>Initial:</u>            0.025            0.025      

<u>At eqllm:</u>      0.025-x          0.025+x        x

The expression of second equilibrium constant equation follows:

Ka_2=\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}

We know that:

Ka_2\text{ for }H_2SO_4=0.01

Putting values in above equation, we get:

0.01=\frac{(0.025+x)\times x}{(0.025-x)}\\\\x=-0.0411,0.00608

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of sulfate ion = x = 0.00608 M

Hence, the concentration of SO_4^{2-} at equilibrium is 0.00608 M

4 0
2 years ago
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