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Paul [167]
4 years ago
7

Initial temperature of water (oC)

Chemistry
1 answer:
Ilia_Sergeevich [38]4 years ago
5 0
The burning point of water in Celsius or the cooling point?
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Calculate the molarity of the solution if 80.0 g of Kool-Aid is dissolved in 1.0<br> L of solution.
Serga [27]

Answer:

.234 M

Explanation:

Molarity, or concentration, of a substance is simply found by dividing the amount of that substance in moles by its volume. Here, we are given the mass (in grams) of Kool-Aid, so we need to convert that to moles. To do this, we simply need the chemical formula of Kool-Aid, which I have found to be C12H22O11. The molar mass of C12H22O11, using the periodic table, is 342.296 g. Next, take this molar mass and divide the given mass of Kool-Aid by it. You then get about .234 mol Kool-Aid.

Lastly, divide this by the volume of 1.0 L. You get .234 M as your final molarity.

7 0
3 years ago
The metals in Groups 1A, 2A, and 3A ____.
choli [55]
All of these metals from cations, so they lost electrons when forming ions. The answer is D.
3 0
3 years ago
Read 2 more answers
The kind of energy released when atoms are rearranged to form new bond
Murljashka [212]
The kind of energy released when atoms are rearranged to form new bonds is called Chemical energy.
6 0
3 years ago
What do you think would happen if a positive ion and a negative ion were close together? (What do opposite charges do?)
viktelen [127]

Answer: opposite charges attract

Explanation:

7 0
3 years ago
Select True or False: A mixture made from 10 mL of 1 M HCl and 20 mL of 1 M CH3COONa would be classified as a buffer solution.
DIA [1.3K]

Answer:

True.

Explanation:

There are three ways to make a buffer.

a.  Generating a solution of a weak acid and its conjugate base.

b. Adding to a solution of a weak acid a certain quantity of

strong base, so that the acid remains in excess.

c. adding a limited amount to a conjugate base solution

of strong acid so that the base remains in excess.

We are in c, in this situation.

How do you calculate, pH? We apply Henderson Hasselbach.

pH = pka + log (mmoles base - mmoles acid)/ mmoles acid

pH = pKa + log ((20 ml . 1 M - 1 ml . 10M) / 10 mmoles

pH = 4.76 + log 1 → 4.76 ⇒ pH = pKa

8 0
3 years ago
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