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Paraphin [41]
2 years ago
14

What is the overall (X-fold) dilution of 4 serial 2-fold dilutions?

Chemistry
1 answer:
svet-max [94.6K]2 years ago
6 0

Answer:

16-fold dilution.

Explanation:

A serial dilution is any dilution where the concentration decrease by the same quantity in each successive step. So, for a 2-fold dilution, the concentration decrease 1/2, it means that if we have a sample with 10 M of concentration, after a 2-fold dilution it will be 5 M. For the next step it will be 1/2 of 5= 2.5 M, and successively.

Then, we just multiply the factor for each dilution. After 4 serial dilutions:

1/2 x 1/2 x 1/2 x 1/2 = 1/16

So, it would be a 16-fold dilution in the end.

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A __________ has distinct properties and composition that never vary.
erik [133]
<h2>Answer </h2>

Matter

<u>Explanation </u>

A matter has distinct properties and composition that never vary. The matter is anything which possesses mass and occupies space. The matter has different characteristics and a distribution that does not change from unit to unit. Materials which cannot be disintegrated into simpler materials. Each is made of only one kind of atom in subatomic level. For example, a chair, table, and similar everything that has mass and occupies space is matter.

3 0
2 years ago
Read 2 more answers
The water in a pressure cooker boils at a temperature greater than 100°C because it is under pressure. At this higher temperatur
vazorg [7]

Answer:

the activation energy Ea = 179.176 kJ/mol

it will take  7.0245 mins for the same food to cook in an open pot of boiling water at an altitude of 10000 feet.

Explanation:

From the given information

T_1 = 100^0 C = 100+273 = 373 \ K \\ \\  T_2 = 113^0 C = 113 + 273 = 386 \ K

R_1 = \dfrac{1}{7}

R_2 = \dfrac{1}{49}

Thus; \dfrac{R_2}{R_1} = 7

Because at 113.0°C; the rate is 7 time higher than at 100°C

Hence:

In (7) = \dfrac{Ea}{8.314}( \dfrac{1}{373}- \dfrac{1}{386})

1.9459 = \dfrac{Ea}{8.314}* 9.0292  *10^{-5}

1.9459*8.314 = Ea * 9.0292*10^{-5}

16.1782126= Ea * 9.0292*10^{-5}

Ea = \dfrac{16.1782126}{ 9.0292*10^{-5}}

Ea = 179.176 kJ/mol

Thus; the activation energy Ea = 179.176 kJ/mol

b)

here;

T_2 = 386 \  K  \\ \\T_1 = (89.8 + 273)K = 362.8 \ K

In(\dfrac{R_2}{R_1})= \dfrac{Ea}{R}(\dfrac{1}{T_1}- \dfrac{1}{T_2})

In(\dfrac{R_2}{R_1})= \dfrac{179.176}{8.314}(\dfrac{1}{362.8}- \dfrac{1}{386})

In (\dfrac{R_2}{R_1}) = 0.00357

\dfrac{R_2}{R_1}= e^{0.00357}

\dfrac{R_2}{R_1}= 1.0035

where ;

R_2 = \dfrac{1}7{}

R_1 = \dfrac{1}{t}

Now;

\dfrac{t}{7}= 1.0035

t = 7.0245 mins

Therefore; it will take  7.0245 mins for the same food to cook in an open pot of boiling water at an altitude of 10000 feet.

4 0
3 years ago
A cylinder of argon contains 50 L of Ar at 12.4 atm and 127°C . How many moles of argon are in the cylinder
joja [24]

Answer:

18.9 moles

Explanation:

We have the following data:

V = 50 L

P = 12.4 atm

T= 127°C + 273 = 400 K

R = 0.082 L.atm/K.mol (it is the gas constant)

We use the ideal gas equation to calculate the number of moles n of the gas:

PV = nRT

⇒ n = PV/RT = (12.4 atm x 50 L)/(0.082 L.atm/K.mol x 400 K) = 18.9 mol

6 0
2 years ago
What causes the appearance of lines in an emission spectrum?
Nikitich [7]

Answer:

key concepts and summary. When electrons move from a higher energy level to a lower one, photons are emitted, and an emission line can be seen in the spectrum. Absorption lines are seen when electrons absorb photons and move to higher energy levels

Explanation:

thank me later

pls be brainlest

3 0
2 years ago
Although Rutherford is credited with the famous gold foil scattering experiment, he didn’t actually conduct this experiment hims
DerKrebs [107]
If he was the primary scientist doing it as he did alot of the heavy lifting then yes its ok, but i also think how the others should also me at least mentioned. Or they could just not name the experiment by a person just so its not too biased
3 0
3 years ago
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