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maw [93]
3 years ago
13

How is a pure substance different from a mixture? A.Pure substances are separated by physical means B. Mixtures cannot be separa

ted by physical means C. A pure substance is heterogeneous D. Mixtures are made up of more than one component.
Chemistry
1 answer:
zimovet [89]3 years ago
8 0

Answer:

d

Explanation:

mixtures are made up of more than one components

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PLEASE HELP WHAT IS THE CORRECT ANSWER (if possible let me know why)
ANTONII [103]

Answer:I’m not 100% sure but I think it’s the first answer.

Explanation:

The pressure would increase as the molecules are pushing out word more and more, leading to more space being taken and more pressure. But then again, I’m not sure. I’m just a freshman with below average grades.

8 0
3 years ago
The osmotic pressure of an aqueoussolution of urea at 300 K is
den301095 [7]

<u>Answer:</u> The freezing point of solution is -0.09°C

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=imRT

where,

\pi = osmotic pressure of the solution = 120 kPa

i = Van't hoff factor = 1 (for non-electrolytes)

m = concentration of solute in terms of molality = ?

R = Gas constant = 8.31\text{L kPa }mol^{-1}K^{-1}

T = temperature of the solution = 300 K

Putting values in above equation, we get:

120kPa=1\times m\times 8.31\text{ L kPa }mol^{-1}K^{-1}\times 300K\\\\m=0.05m

  • To calculate the depression in freezing point, we use the equation:

\Delta T=i\times K_f\times m

where,

i = Vant hoff factor = 1 (for non-electrolytes)

K_f = molal freezing point depression constant = 1.86°C/m

m = molality of solution = 0.05 m

Putting values in above equation, we get:

\Delta T=1\times 1.86^oC/m\times 0.05m\\\\\Delta T=0.09^oC

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.  

  • The equation used to calculate freezing point of solution is:

\Delta T=\text{freezing point of water}-\text{freezing point of solution}

where,

\Delta T = Depression in freezing point = 0.09 °C

Freezing point of water = 0°C

Freezing point of solution = ?

Putting values in above equation, we get:

0.09^oC=0^oC-\text{Freezing point of solution}\\\\\text{Freezing point of solution}=-0.09^oC

Hence, the freezing point of solution is -0.09°C

4 0
3 years ago
Im in a big time rush, PLEASE HELP ME!! Answer this correctly
xz_007 [3.2K]

Answer:

GG= Green Gg= Green Gg=yellow

Explanation:

Whenever there is a dominant allele it will ALWAYS show, the recessive alleles or lowercase letters show when they are paired together

3 0
3 years ago
A new metal alloy is found to have a specific heat capacity of 0.260 J/(g⋅∘C). First, 47 g of the new alloy is heated to 180. ∘C
Vlad1618 [11]

Answer:

The final temperature, at the equilibrium is 24.14 °C

Explanation:

Step 1: Data given

specific heat capacity of alloy = 0.260 J/(g°C)

MAss of alloy = 47 grams

Mass of water = 110 grams

Specific heat of water = 4.184 J/g°C

Initial temperature of water = 20.0 °C

Initial temperature of alloy = 180.0 °C

Step 2: Calculate the final temperature at equilibrium

Heat lost = heat gained

Qlost = -Qgained

Q(alloy) =- Q(water)

Q=m*c*ΔT

Q = m(alloy)*c(alloy)*ΔT(alloy) = -m(water) * c(water)* ΔT(water)

⇒with m(alloy) = the mass of alloy = 47.0 grams

⇒with c(alloy) = the specific heat of alloy = 0.260 J/g°C

⇒with ΔT(alloy) = the change of temperature = T2- T1 = T2 - 180 °C

⇒with m(water) = the mass of water = 110 grams

⇒with c(water) = the specific heat of water = 4.184 J/g°C

⇒with ΔT(water) = the change of temperature = T2 - 20.0°C

47.0*0.260 * (T2 - 180.0) = - 110 * 4.184 * (T2 - 20.0)

12.22(T2-180.0) = -460.24(T2- 20)

12.22T2 - 2199.6 = -460.24T2 + 9204.8

472.46T2 = 11404.4

T2 = 24.14 °C

The final temperature, at the equilibrium is 24.14 °C

8 0
3 years ago
What is the percent by mass of C in benzene (C6H6)? The molecular weight of carbon is 12.0107 g/mol and of hydrogen 1.00794 g/mo
pshichka [43]

Answer:

92.26% of C

Explanation:

To solve this problem we must assume we have 1 mole of benzene. The mole contains 6 moles of C and 6 moles of H. We have to convert these moles to grams in order to find the total mass and mass percent will be:

Mass atom / Total mass * 100

<em>Mass C: </em>6mol C * (12.0107g / mol) = 72.0642g

<em>Mass H: </em>6mol H * (1.00794g / mol) = 6.04764g

<em>total mass: </em>72.0642g + 6.04764g = 78.11184g

Mass percent of C will be:

72.0642g C / 78.11184g* 100

<h3>92.26% of C</h3>

3 0
3 years ago
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