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goblinko [34]
3 years ago
8

Explain how knowledge of acids/bases and neutralization can apply to the treatment of soil acidification

Chemistry
1 answer:
sammy [17]3 years ago
6 0

Explanation:

If soil consists of more acidic nature then, it does not allow plant growth.

In that situation, treat the soil with (anti-acid which is nothing but) base.

When an acid reacts with a base, a neutralization reaction takes place and salt and water are formed in this reaction.

The equation is shown below:

acid+base-> salt +water

For example, if the soil has more sulphuric acid traces then, it should be treated with lime.

H_2SO_4(aq)+Ca(OH)_2(aq)->CaSO_4(aq)+2H_2O(l)

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To a 0.0001 m solution of mg(no3)2, naoh was added to a final concentration of 0.001m did a precipitate form?
Natalija [7]

I looked on a solubility chart to answer this question, and hydroxides are generally insoluble (with some exceptions of course). However, it says to consider Mg(OH)_{2} as an insoluble substance, though it may be moderately soluble.


The answer that you are most likely looking for is: Yes, a precipitate does form - this is due to the double placement reaction:


Mg(NO_{3})_{2}_{(aq)} + 2NaOH_{(aq)} → Mg(OH)_{2} {(s)} + 2NaNO_{3}_{(aq)}

8 0
3 years ago
1) A mixture of anhydrous sodium carbonate and sodium hydrogencarbonate of mass 10.000 g was heated until it reached a constant
vodomira [7]

The masses of the components are obtained as;

  • Sodium hydrogen carbonate = 3.51 g
  • Sodium carbonate =  8.708 g
<h3>What is decomposition?</h3>

The term decomposition has to do with the breakdown of the given substance into its components. The components of sodium hydrogen carbonate could be identified as water vapor, carbon dioxide gas and sodium carbonate. Among these products that have been listed here, we can see that it is only the sodium carbonate that remains as a solid. The others are gases that move away from the system that is under study.

Now putting down the equation of the reaction, we have;

2NaHCO_{3} (s) ----- > Na_{2} CO_{3} (s) + CO_{2} (g) + H_{2} O(g)

Now, the loss in  mass must be due to the carbon dioxide and the water. Hence we obtain the loss in mass to be 10.000 g -  8.708 g = 1.292 g

Mass of sodium hydrogen carbonate = 2 * 88 g/mol * 1.292 g/62 g/mol

= 3.51 g

Learn more about anhydrous sodium carbonate :brainly.com/question/20479996

#SPJ1

6 0
1 year ago
At a certain temperature the vapor pressure of pure chloroform (CHCl3) is measured to be 91. torr. Suppose a solution is prepare
OleMash [197]

Answer:

P_{CCl_4}=52.43torr

Explanation:

Hello there!

In this case, sine the solution of this problem require the application of the Raoult's law, assuming heptane is a nonvolatile solute, so we can write:

P_{CCl_4}=x_{CCl_4}P_{CCl_4}^{vap}

Thus, we first calculate the mole fraction of chloroform, by using the given masses and molar masses as shown below:

x_{CCl_4}=\frac{140/153.81}{140/153.81+67.1/100.21}=0.576

Therefore, the partial pressure of chloroform turns out to be:

P_{CCl_4}=0.576*91torr\\\\P_{CCl_4}=52.43torr

Regards!

3 0
3 years ago
Can u explain to me how to do this? This is my first time doing this type of question so explain or give me some of the answers
WITCHER [35]
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5 0
3 years ago
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Compound is a clear liquid with a strong pleasantly fruity smell. If cooled it freezes at about. In the solid state it does not
son4ous [18]

Answer:

Compound 3 is a clear liquid with a strong pleasantly fruity smell. If cooled it freezes at about −10°C. In the solid state it does not conduct electricity. ... It dissolves slightly in water, and a solution of 2g in 100mL of water doesn't change the electrical conductivity of the water.

5 0
2 years ago
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