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Greeley [361]
3 years ago
7

Compared with the freezing-point depression of a 0.01 m c6h12o6 solution, the freezing-point depression of a 0.01 m nacl solutio

n is
Chemistry
1 answer:
agasfer [191]3 years ago
6 0

Answer:

Twice  as much.

Explanation:

That's because the freezing point depression depends on the total number of solute particles.

C₆H₁₂O₆(s) ⟶ C₆H₁₂O₆(aq)

0.01 mol of C₆H₁₂O₆ gives 0.01 mol of solute particles.

NaCl(s) ⟶ Na⁺(aq) + Cl⁻(aq)

1 mol of NaCl gives 0.01 mol of Na⁺(aq) and 0.01 mol of Cl⁻(aq).

That's 0.02 mol of particles, so the freezing point depression of 0.01 mol·L⁻¹ NaCl will be twice that of 0.01 mol·L⁻¹ C₆H₁₂O₆.

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The table shows data for two groups of plants one grown with fertilizer and the other without fetalizer was the mean height of t
Dmitrij [34]

Answer:

D is the correct answer.

Explanation:

I took the test on A pex. Everyone make sure you look at your answers. They are NOT always in the same order.

8 0
2 years ago
I need help with with number 1515. How many moles of NO2 are in a 37.058 L container at 101.28 kPa and -139.86
Andre45 [30]

Explanation:

To solve this question, we will use the Clayperon Equation:

P.V = n.R.T

where:

P = 101.28 kPa

1 atm = 101,325 Pa

x atm = 101,280 Pa

x = 1 atm

V = 37.058 L

n = we don't know

R = 0.082 atm.L/K.mol

T = -139.88 ºC = -139.88+273.15 = 133.27 K

1*37.058 = n*0.082*133.27

n = 0.29 moles

Answer: 0.29 moles

5 0
1 year ago
A corpse discovered in the desert at 5:00 PM was found to have larvae from the blow fly species Phormia regina that had just dev
horrorfan [7]

Answer:

12.19 hours prior to discovery was the corpse placed in the desert.

Explanation:

Heat required for fly eggs to develop into first instar larvae = Exposure\,time\times temperature

From the given,

Exposure time = 16 hrs

Temperature = 27 C

=16\times 27 = 427.2^{o}C

The day time exposed hours is 10 hrs.-(7.00am-5.00pm)

=10\times 37.8 = 378^{o}C

So, the extra heat  = 427.2-378= 49.2

The addition heat divided by average temperature.

Average temperature = 22.4 C

=\frac{49.2}{22.4}= 2.19 hrs

So, the total time exposed is night time hours to day time hours.

= 10+2.19 = 12.19 hrs.

Therefore, 12.19 hours prior to discovery was the corpse placed in the desert.

3 0
3 years ago
2.15 Liters of helium gas at a pressure of 58 kPa and a temperature of 25 degrees celsius, what is the new volume at stp?
Nastasia [14]
The ideal gas law is P1V1/T1 = P2V2/T2. STP means the temperature is 273 K and pressure is 101.3 kPa. According to this formula, the new volume V2=2.15*58*273/(298*101.3) = 1.13 L.
5 0
3 years ago
Which choice describes a population? A. all the gray squirrels that live in a forest B. a nonliving factor in an organism's envi
goldenfox [79]
I believe D. because all the organisms that live in a lake can be a bunch of living things that would be considered a population and not a species. I canceled out B. from the get go because, I don’t believe non living organisms can be in a population ?

Please get back to me if I’m right or wrong :)
5 0
3 years ago
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