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3 years ago

12

When we react a weak acid with a strong base of equal amounts and concentration, the component of the reaction that will have th

e greatest effect on the pH of the solution is:

1 answer:

Svetllana [295]3 years ago

5 0

Answer and explanation:

Initially, the pH of the solution will be determined by the dissociation of the weak acid.

Usually, this kind of solution consists of a mixture of weak acid that has not yet reacted and the salt that will be formed by the reaction of the weak acid with a strong base that has been added.

After the dissociation process, the pH will be more affected due to the excess of base in the solution.

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What are the law of 10 in energy Transfer

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When the energy is passed on from one trophic level to another, only 10 percent of the energy is passed on to the next trophic level.

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3 years ago

Read 2 more answers

How many grams of nan3 are required to produce 19.0 ft3 of nitrogen gas, about the size of an automotive air bag, if the gas has

Papessa [141]

The balanced chemical reaction is given as:

$2NaN_{3}(s)\rightarrow 2Na(s)+3N_{2}(g)$

Now, convert $19.0 ft^{3}$ into litres.

$1 ft^{3} = 28.3168$

So, $19.0 ft^{3} = 19\times 28.3168 = 538.0192 L$

Density is equal to the ratio of mass to the volume.

$D=\frac{M}{V}$

where, M = mass and V= volume $(538.0192 L)$

Substitute the value of density and volume in formula to get the value of mass.

$1.25 g/L=\frac{M}{538.0192 L}$

$1.25 g/L\times 538.0192 L= M$

$Mass = 672.524 g$

Now, number of moles of $N_{2}$ gas= $\frac{672.524 g}{28.02 g/mol}$

= $24.00 moles$

According to the reaction, 2 moles of sodium azide gives 3 moles of nitrogen gas.

Now, in 24.00 moles of nitrogen gas produced from= $\frac{2 moles of sodium azide}{3 moles of nitrogen gas}\times 24.00 moles$ of nitrogen gas, moles of sodium azide.

number of moles of sodium azide = $16 moles$

Mass of sodium azide in g = $number of moles\times molar mass of sodium azide$ .

= $16 moles\times 65.00 g/mol$

= $1040 g$

Thus, mass of sodium azide which is required to produce $19.0 ft^{3}$ of nitrogen gas = $1040 g$

3 0

2 years ago

Why does such a small decrease in pH mean such a large increase in acidity?

den301095 [7]

Answer:

pH is an index of how many protons, or hydrogen ions (H+) are dissolved and free in a solution. The pH scale goes from 0 to 14. A fluid with a pH of 7 is neutral. Below 7, it is acidic; above 7, it is alkaline.

The more below or above 7 a solution is, the more acidic or alkaline it is. The scale is not linear—a drop from pH 8.2 to 8.1 indicates a 30 percent increase in acidity, or concentration of hydrogen ions; a drop from 8.1 to 7.9 indicates a 150 percent increase in acidity. Bottom line: Small-sounding changes in ocean pH are actually quite large and definitely in the direction of becoming less alkaline, which is the same as becoming more acidic.

If you think about it, we use descriptive words like this all the time. A person who stands 5’5” tall and weighs 300 pounds isn’t thin. If he loses 100 pounds, he still won’t be thin, but he will be thinner than he was before he went on the diet. (And we are more likely to comment that he’s looking trimmer than to say he’s not as fat as he used to be.)

7 0

2 years ago

Solid NaI is slowly added to a solution that is 0.0079 M Cu+ and 0.0087 M Ag+.Which compound will begin to precipitate first?NaI

NeTakaya

Answer :

AgI should precipitate first.

The concentration of $Ag^+$ when CuI just begins to precipitate is, $6.64\times 10^{-7}M$

Percent of $Ag^+$ remains is, 0.0076 %

Explanation :

$K_{sp}$ for CuI is $1\times 10^{-12}$

$K_{sp}$ for AgI is $8.3\times 10^{-17}$

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgI has a smaller than CuI then AgI should precipitate first.

Now we have to calculate the concentration of iodide ion.

The solubility equilibrium reaction will be:

$CuI\rightleftharpoons Cu^++I^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Cu^+][I^-]$

$1\times 10^{-12}=0.0079\times [I^-]$

$[I^-]=1.25\times 10^{-10}M$

Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

$AgI\rightleftharpoons Ag^++I^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ag^+][I^-]$

$8.3\times 10^{-17}=[Ag^+]\times 1.25\times 10^{-10}M$

$[Ag^+]=6.64\times 10^{-7}M$

Now we have to calculate the percent of $Ag^+$ remains in solution at this point.

Percent of $Ag^+$ remains = $\frac{6.64\times 10^{-7}}{0.0087}\times 100$

Percent of $Ag^+$ remains = 0.0076 %

8 0

3 years ago