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OverLord2011 [107]
3 years ago
7

Eutrophication is when a lake eventually become _______.

Chemistry
1 answer:
Nitella [24]3 years ago
4 0
Over flowing with BOD, which drags down the O2 levels.
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What is 11.75 millimeters in meters
Lana71 [14]

Answer:

in meters its 0.01175

Explanation:

8 0
1 year ago
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Isotopes of the same element vary from each other in their number of:
Lina20 [59]

Answer: Neutrons  vary from each other in their element

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5 0
3 years ago
2.40 kJ of heat is added to a slug of platinum and a separate 2.40 kJ of heat is added to a slug of zinc. The heat capacity of t
leonid [27]

Final temperature of platinum :-   36.5°C

Final temperature of Zinc :-  29.17°C


3 0
3 years ago
A chemist has a 2.0 M solution of copper (II) sulfate. Is he takes 400 mL of this solution and dilutes it with 1200 mL of water,
Ksenya-84 [330]

0.66 M is the accurate molarity of the new solution of volume of 1200 ml.

Explanation:

Data given:

molarity of copper(II) sulphate, Mconc.= 2M

volume of 2M solution taken Vconc. = 400 ml

volume taken for dilution, Vdilute = 1200 ml

molarity of the diluted solution, Mdilute =?

We will use the formula for dilution as

Mconc Vconc = Mdilute x V dilute  (conc is concentrated)

putting the values in the equation:

2 x 400 = Mdilute x 1200

Mdilute = \frac{800}{1200}

   Mdilute  = 0.66 M

When the solution is diluted to the volume of 1200 ml its molarity changes to 0.66 M.

4 0
3 years ago
How much 10.0 M HNO must be added to 1.00 L of a buffer that is 0.0100 M acetic acid and 0.100 M sodium acetate to reduce the pH
Naya [18.7K]

Explanation:

It is given that molarity of acetic acid = 0.0100 M

Therefore, moles of acetic acid = molarity of acetic acid × volume of buffer

            Moles of acetic acid = 0.0100 M × 1.00 L

                                              = 0.0100 mol

Similarly, moles of acetate = molarity of sodium acetat × volume of buffer

                                           = 0.100 mol

When HNO_{3} is added, it will convert acetate to acetic acid.

Hence, new moles acetic acid = (initial moles acetic acid) + (moles HNO_{3})

                                                = 0.0100 mol + x

New moles of sodium acetate = (initial moles acetate) - (moles HNO_{3})

                                        = 0.100 mol - x

According to Henderson - Hasselbalch equation,

           pH = pK_{a} + log\frac{[conjugate base]}{[weak acid]}

             pH = pKa + log\frac{(new moles of sodium acetate)}{(new moles of acetic acid)}

           4.95 = 4.75 + log\frac{(0.100 mol - x)}{(0.0100 mol + x)}

       log\frac{(0.100 mol - x)}{(0.0100 mol + x)}  = 4.95 - 4.75

                                            = 0.20

\frac{(0.100 mol - x)}{(0.0100 mol + x)}  = antilog (0.20)

                                           = 1.6

Hence,    x = 0.032555 mol

Therefore, moles of HNO_{3} = 0.032555 mol

volume of HNO_{3} = \frac{moles HNO_{3}}{molarity of HNO_{3}}

                                = \frac{0.032555 mol}{10.0 M}

                                 = 0.0032555 L

or,                             = 3.25           (as 1 L = 1000 mL)

Thus, we can conclude that volume of HNO_{3} added is 3.26 mL.

5 0
2 years ago
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