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kvv77 [185]
3 years ago
5

Is it possible to change the strength of an acid? Explain how you would do this.

Chemistry
1 answer:
Verdich [7]3 years ago
4 0

Answer:

Reducing or increasing the amount of H+ ions / hydronium (H3O+) ions

Explanation:

To reduce the pH (reducing the strength of the acid) can be done by adding a base (including a conjugate base such as bicarbonate ion) which will absorb the H+ ions either through adsorption or reaction.

Adding more H+ decreases the pH of the acid making it stronger. This can be done by adding HCL that will dissociate and increase the H+ ions.

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A solution is made by dissolving 4.87 g of potassium nitrate in water to a final volume of 86.4 mL solution. What is the weight/
lara31 [8.8K]

Answer:

A solution is made by dissolving 4.87 g of potassium nitrate in water to a final volume of 86.4 mL solution. The weight/weight % or percent by mass of the solute is :

<u>2.67%</u>

Explanation:

Note : Look at the density of potassium nitrate in water if given in the question.

<u><em>You are calculating </em></u><u><em>weight /Volume</em></u><u><em> not weight/weight % or percent by mass of the solute</em></u>

Here the <u>weight/weight % or percent by mass</u> of the solute is asked : So first convert the<u> VOLUME OF SOLUTION into MASS</u>

Density of potassium nitrate in water KNO3 = 2.11 g/mL

density=\frac{mass}{volume}

Density = 2.11 g/mL

Volume of solution = 86.4 mL

2.11=\frac{mass}{86.4}

mass = 2.11\times 86.4

mass=182.3grams

Mass of Solute = 4.87 g

Mass of Solution = 183.2 g

w/w% of the solute =

= \frac{mass\ of\ solute}{mass\ of\ solution}\times 100

=\frac{4.87}{183.2}\times 100

w/w%=2.67%

8 0
3 years ago
What process takes place when hydrogen-3 and hydrogen-2 combine to form
stira [4]

Answer:

A. Fusion

Explanation:

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3 years ago
The radioisotope phosphorus-32 is used in tracers for measuring phosphorus uptake by plants. The half-life of phosphorus-32 is 1
Harman [31]

Answer:

54 days

Explanation:

We have to use the formula;

0.693/t1/2 =2.303/t log Ao/A

Where;

t1/2= half-life of phosphorus-32= 14.3 days

t= time taken for the activity to fall to 7.34% of its original value

Ao=initial activity of phosphorus-32

A= activity of phosphorus-32 after a time t

Note that;

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Substituting values;

0.693/14.3 = 2.303/t log Ao/0.0734Ao

0.693/14.3 = 2.303/t log 1/0.0734

0.693/14.3 = 2.6/t

0.048=2.6/t

t= 2.6/0.048

t= 54 days

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