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Mashutka [201]
4 years ago
12

When an acid reacts with a base, salt and water is formed! Why??​

Chemistry
2 answers:
Mariulka [41]4 years ago
7 0

Answer:

☆<《HOPE IT WILL HELP YOU 》>☆

Explanation:

when a acid react with Base,salt and water is formad because when acid react with Base. It loses acdic properties &Base its basic property to form natural substances

vladimir1956 [14]4 years ago
5 0

Answer:

When n acid reacts with a base , salt and water is formed because when acid reacts with base, it loses its acidic property and base loses its basic property to form a neutral substance like salt and water..

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Imagine that one tree outside your home looks unhealthy, although all the other trees seem healthy and strong. Describe how you
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Answer:

Hypothesis---experiments----results----conclusion.

Explanation:

First we make a hypothesis means a statement about why the tree looks unhealthy. In this segment of scientific method we have to test the hypothesis through experimentation. After that we have to take the readings of various parts of the tree and analyze the data to find out the problem. In the next step, we have to made the results on the basis of the data that is obtained. In the last we have to write the conclusion of the analysis and see the hypothesis.

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A certain liquid has a normal boiling point of and a boiling point elevation constant . A solution is prepared by dissolving som
jek_recluse [69]

The question is incomplete, the complete question is:

A certain substance X has a normal freezing point of -6.4^oC and a molal freezing point depression constant K_f=3.96^oC.kg/mol. A solution is prepared by dissolving some glycine in 950. g of X. This solution freezes at -13.6^oC . Calculate the mass of urea that was dissolved. Round your answer to 2 significant digits.

<u>Answer:</u> The mass of glycine that can be dissolved is 1.3\times 10^2g

<u>Explanation:</u>

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times m

OR

\text{Freezing point of pure solvent}=\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}           ......(1)

where,

Freezing point of pure solvent = -6.4^oC

Freezing point of solution = -13.6^oC

i = Vant Hoff factor = 1 (for non-electrolytes)

K_f = freezing point depression constant = 3.96^oC/m

m_{solute} = Given mass of solute (glycine) = ?

M_{solute} = Molar mass of solute (glycine) = 75.07 g/mol

w_{solvent} = Mass of solvent = 950. g

Putting values in equation 1, we get:

-6.4-(-13.6)=1\times 3.96\times \frac{m_{solute}\times 1000}{75.07\times 950}\\\\m_{solute}=\frac{7.2\times 75.07\times 950}{1\times 3.96\times 1000}\\\\m_{solute}=129.66g=1.3\times 10^2g

Hence, the mass of glycine that can be dissolved is 1.3\times 10^2g

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3 years ago
A gas occupies a volume of 50.0 mL at 27°C and 630 mmHg. At what
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T2+500k - 273 = 223 *C

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