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3 years ago

Which statement is true of the two molecules below?

Chemistry

2 answers:

g100num [7]3 years ago

8 0

Answer:

Explanation:

they can bond together to form a disaccharide.

bagirrra123 [75]3 years ago

7 0

Answer: d

Explanation: They are made of four different kinds of elements.

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What do all prokaryotes and eukaryotes have in common? A. They are unicellular organisms. B. They are multicellular organisms. C

Sauron [17]

They are multicellular

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Anika [276]

Answer:

That reminds me of the avengers logo

8 0

3 years ago

Read 2 more answers

Formulate a hypothesis about the stoichiometry of the reaction between NaCl and AgNO3.

MAXImum [283]

Answer:

AgCl + NaNO3 would be the products of the reaction between sodium chloride and silver nitrate.

The stoichiometry of this reaction is written below, and it is because for this reaction to be fulfilled the products have to be in equilibrium with the reactants, since the mass in the reaction is conserved and must be balanced in the amount of molecules that they react to each other.

Explanation:

NaCl + AgNO3 -------------- AgCl + NaNO3

3 0

4 years ago

A 0.20 M solution of a weak acid has a pH of 5.40. What is the Ka for the acid?

sweet [91]

Answer:

The Ka for this weak acid is 7,92 * 10^-11

Explanation:

First of all, let's think the equation

HA + H2O <------> H3O+ + A-

When we add water to a weak acid, it dissociates in an equilibrium to generate the corresponding anion and the hydronium cation (the acid form of water)

How do you calculate Ka?? Ka is the acid equilibrium constant.

( [H3O+] . [A-] ) / [HA] where all the concentrations are in equilibrium.

We don't have the concentration in equilibrium but we have the initial concentration. So...

HA + H2O <------> H3O+ + A-

initial- 0.2 M I don't have H3O+, either A-

reaction - an specific amount reacted (X)

in equilibrium (0,2 - X) <-----> X + X

And now, how's the formula for Ka

( [H3O+] . [A-] ) / [HA] = Ka

(X . X) / (0.2-X)

X^2 / (0.2-X) = Ka

Look, that we don't have X as the [H3O+] but we know the pH, so we can know the [H3O+] indeed.

10^-pH = [H3O+]

10^-5,40 = 3,98 * 10^-6

Let's go back to Ka

( [H3O+] . [A-] ) / [HA] = Ka

(3,98 * 10^-6)^2 / (0.2 - 3,98 * 10^-6) = Ka

(3,98 * 10^-6 is an small number, soooo small that we can approximate to 0)

If we have in order 10^-6, 10^-5 we can consider that.

So now, we have

(3,98 * 10^-6)^2 / (0.2) = Ka

1,58 * 10^-11 / (0.2) = Ka = 7,92* 10^-11

5 0

3 years ago

A water treatment plant has 4 settling tanks that operate in parallel (the flow gets split into 4 equal flow streams), and each

ale4655 [162]

Answer:

a) When the 4 tanks operate in parallel the retention time is 1.26 hours.

b) If the tanks are in series, the retention time would be 0.31 hours

Explanation:

The plant has 4 tanks, each tank has a volume V = 600 $m_3$ . The total flow to the plant is Ft = 12 MGD (Millions of gallons per day)

When we use the tanks in parallel, it means that the total flow will be divided in the total number of tanks. F1 will be the flow of each tank.

Firstly, we should convert the MGD to $m_3 /day$ . In that sense, we can calculate the retention time using the tank volume in $m_3$ .

$Ft =12 \frac{MG}{day} (\frac{1*10^6 gall}{1MG} ) (\frac{3.78 L}{1 gall} ) (\frac{1 dm^3}{1L} ) (\frac{1 m^3}{10^3 dm^3} ) = 45360 \frac{m^3}{day}$

After that, we should divide the total flow by four, because we have four tanks.

$F1 = Ft/4 =(45360 \frac{m^3}{d} )/4 = 11 340 \frac{m^3}{d}$

To calculate the retention time we divide the total volume V by the flow of each tank F1.

$t1 =\frac{V1}{F1} = \frac{600 m^3}{11340 \frac{m^3}{d} } = 0.0529 day\\t1 = 0.0529 d (\frac{24 h}{1d} ) = 1.26 h$

After converting t1 to hours we found that the retention time when the four reactors are in parallel is 1.26 hours.

If the four reactors were working in series, the entire flow goes first through one tank, then the second and so on. It means the total flow will be the flow of each tank.

In that order of ideas, the flow for reactors in series will be F2, and will have the same value of F0.

F2 = F0

To calculate the retention time t2 we divide the total volume V by the flow of each tank F2.

$t2 =\frac{V1}{F2} = \frac{600 m^3}{45360 \frac{m^3}{d} } = 0.01 day\\t2 = 0.01 d (\frac{24 h}{1d} ) = 0.31 h$

After converting t2 to hours we found that the retention time when the four reactors are in series is 0.31 hours.

5 0

3 years ago