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labwork [276]
3 years ago
9

During _______ materials move into the cell while during ______ materials move out of the cell.

Chemistry
1 answer:
Len [333]3 years ago
3 0
D. Endocytosis ; Exocytosis
You might be interested in
7. A 2.0 L container had 0.40 mol of He(g) and 0.60 mol of Ar(g) at 25°C.
Finger [1]

Answer:

a) Ek Ar > Ek He

b) v Ar < v He

c) If v Ar = 431 m/s ⇒ v He = 1710.44 m/s

d) Pt = 12.218 atm

e) P He = 4.887 atm and P Ar = 7.33 atm

Explanation:

container:

∴ V = 2.0 L

∴ n He = 0.4 mol

∴ n Ar = 0.6 mol

∴ T = 25°C ≅ 298 K

a) Internal energy (U) :

∴ U = Ek + Ep = kinetic energy + potential energy

∴ Ep: the potential interaction energy is neglected, assuming ideal gas mixture

⇒ U = Ek = N(1/2mv²)= 3/2 NKT

∴ N = nNo ....number of moleculas

∴ K = 1.380 E-23 J/K....Boltzmann's constant

∴ No = 6.022 E23 molec/mol....Avogadro's number

for He:

⇒ N = (0.4)(6.022 E23) = 2.4088 E23 molec

⇒ Ek = (3/2)(2.4088 E23)(1.380 E-23 J/K)(298) = 1485.892 J

for Ar:

⇒ N = (0.6)(6.022 E23) = 3.6132 E 23 molec

⇒ Ek = (3/2)(3.6132 E23)(1.380 E-23 J/K)(298) = 2228.838 J

** Ar gas has a greater average kinetic energy

b) He:

∴ N(1/2)mv² = (3/2)NKT

⇒ mv² = 3KT

⇒ v² = 3KT/m

⇒ v = √3KT/m

∴ m He = (0.4 mol)(4.0026 g/mol) = 1.601 g He = 1.601 E-3 Kg He

⇒ v = √(3(1.380 E-23)(298)/(1.601 E-3)) = 2.776 E-9 m/s He

Ar:

∴ m Ar = (0.6)(39.948 g/mol) = 23.969 g = 0.0239 Kg Ar

⇒ v = 6.99 E-10 m/s

** v Ar < v He

c) r = V Ar / v He = (6.99 E-10 m/s)/(2.776 E-9 m/s) = 0.252

∴ If v Ar = 431 m/s

⇒ v He = v Ar/0.252 = 431 m/s / 0.252 = 1710.44 m/s

d) Pt = ntRT / V

∴ nt = 0.4 + 0.6 = 1 mol

⇒ Pt = (1mol)(0.082 atm.L/K.mol)(298 K)/(2.00 L) = 12.218 atm

e) P He = nRT/V = (0.4)(0.082)(298)/2 = 4.8872 atm

⇒ P Ar = Pt - PHe = 12.218 - 4.8872 = 7.33 atm

3 0
3 years ago
When 2.5 mol of O2 are consumed in their reaction, ________ mol of CO2 are produced
Vika [28.1K]

The given question is incomplete. The complete question is:

The combustion of propane (C3H8) in the presence of excess oxygen yields CO_2 and H_2O

C_3H_8(g)+5O_2 (g)\rightarrow 3CO_2(g)+4H_2O (g)

When only 2.5 mol of O_2 are consumed in order to complete the reaction, ________ mol of CO_2 are produced.

Answer: Thus when 2.5 mol of O_2 are consumed in their reaction,  1.5 mol of CO_2 are produced

Explanation:

The balanced chemical equation is:

C_3H_8(g)+5O_2 (g)\rightarrow 3CO_2(g)+4H_2O (g)

According to stoichiometry :

5 moles of O_2 produce = 3 moles of  CO_2

Thus 2.5 moles of O_2 will produce = \frac{3}{5}\times 2.5=1.5 moles of  CO_2

Thus when 2.5 mol of O_2 are consumed in their reaction,  1.5 mol of CO_2 are produced

4 0
3 years ago
What is the value of k, at 25 degrees Celsius
Dimas [21]
<h3>Answer:</h3>

298.15 K

<h3>Explanation:</h3>

W e are supposed to calculate the Value of K at 25°C

Assuming the value of K represent K, the question wants us to convert  degree Celsius to Kelvin.

  • To convert degrees Celsius to kelvin scale, we use the relationship;
  • Kelvin (K) = Degrees Celsius + 273.15 ; 273.5 is a constant
  • That is, to convert temperature from °C to Kelvin we add a constant of 273.15 to the °C given.

In this case;

Temperature is 273.15 °c

Thus, to Kelvin scale temperature will be;

= 25°C + 273.15

= 298.15 K

Therefore, the value of K, at 25°C is 298.15 K

7 0
3 years ago
What is the value of R in the ideal gas law?
dybincka [34]
It’s option B 0.0821L.atm/mol.K
5 0
3 years ago
Can someone tell me how to do this steps by step<br>How many grams is there in 3.65 moles of CuSO4
Bad White [126]
1. you need a periodic table and find the atomic mass of Cu (copper), S (sulfur) and O (oxygen). The atomic mass is the number in the box that corresponds to the element and have several decimal places.

2. atomic mass of 
Cu = 63.546 
S = 32.065
O = 15.9994

3. Then according to the formula of the compound, you add as many time the atomic mass of each element as subindex in the formula and add all the values together to calculate the molecular mass of the compound in grams.

4. 63.546 g + 32.065 g + ( 4 x <span>15.9994) = 159.609 g

5.  this value </span><span>159.609 g is the mass in grams of one mol of CuSO4

6 the problem is asking not for the mass of one mole but the mass of 3.65 moles of CuSO4

7 then you have the multiply the value of one mol by the number of moles that the problem is asking you

8. </span><span>159.609 g x 3.65 = 582.571 g
</span>
9 the answer to the problem will be
"there are 582.571 g of CuSO4 in 3.65 moles of CuSO4"
 
6 0
3 years ago
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