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melomori [17]
3 years ago
13

Que tipo de fermentacion es al mezclar el agua tibia con levadura

Chemistry
1 answer:
Lena [83]3 years ago
4 0

Answer:

kapsüsnwsibd bqvw j dkpövahfbjbjwv8kfb

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kaheart [24]
What does this even mean
7 0
3 years ago
A 50.0 mL solution of 0.133 M KOH is titrated with 0.266 M HCl . Calculate the pH of the solution after the addition of each of
shtirl [24]

Answer:Calculate The PH Of The Solution After The Addition Of The Following Amounts Of HCl. PLEASE HELP! SHOW ALL STEPS!! This problem has been solved!

Explanation:

4 0
3 years ago
Calculate the volume in liters of a silver perchlorate solution that contains of silver perchlorate . Be sure your answer has th
lbvjy [14]

Answer:

See explanation below

Explanation:

You are missing two values there. First, we need to know the concentration or initial quantity of the silver perchlorate solution, and then, the quantity in grams or moles of the solution.

Tipically these kind of questions, usually gives the original concentration of the solution, and then, the grams or moles.

In this case, you are not providing either value, but I will assume some values and do it in the longest way, and then, you just have to replace your values in the procedure to get the accurate result.

For this problem, I will assume the concentration of 3.5 mol/L and contains 180 g of perchlorate.

We need to know the volume in liters of this solution. We use the following expression:

M = n/V

From here, we solve for V:

V = n/M (1)

But we do not have the moles, but these can be calculated with:

n = m/MM (2)

So all we have to do, is to use the molar mass of perchlorate and calculate the moles. The reported molar mass for silver perchlorate is 207.32 g/mol so replacing in (2)

n = 180 / 207.32 = 0.8682 moles

Now that we have the moles, replace in (1) to get the volume:

V = 0.8682 / 3.5

<u>V = 0.2481 L or simple 248.1 mL</u>

<u>This is the volume in this solution. Now replace your own data in this procedure and you should get the accurate result.</u>

5 0
3 years ago
What is the Ph of a 3.9* 10^-8 M OH- solution
Maksim231197 [3]
Molarity = 3.9 * 10 ^ -8

pH = - log (3.9 * 10 ^ -8)

pH = - (-7.40893539)

pH = 7.409 = 7.4 ... Ans
3 0
3 years ago
I need help asap!!! At least with the first part
kondaur [170]

Answer:

The correct answer -

a. Cd and Pb(NO3)2

b. Redox reactions

c. Pb and Cd(NO3)2

Explanation:

This is the reaction known as the redox or reduction-oxidation reaction of metals. In this particular reaction, there are two reactants Cadmium (III) in solid-state and lead (II) nitrate in the aqueous state. At the end of this reaction, the products that we get are lead (II) in solid-state and Cadmium (III) nitrate in the aqueous state.

cadmium (s)+ lead nitrate (aq) = lead (s) + cadmium nitrate (aq)

Cd (s) + Pb2+(aq) → Pb(s) + Cd2+(aq)

Here, Oxidizing agent is Pb2+ and the reducing agent is Cd.

7 0
3 years ago
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