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expeople1 [14]
3 years ago
11

Mushrooms, algae, and protists are all examples of _______________ cells.

Chemistry
2 answers:
Feliz [49]3 years ago
8 0
The answer is Eukaryotic cells.
hope I helped 
Triss [41]3 years ago
7 0
Hey you!

the answer is Eukaryotic 
hope it helped
You might be interested in
What elements are in NaC2HO4 and how many atoms are in each element​
JulijaS [17]

Answer:

Each molecule contains one atom of A and one atom of B. The reaction does not use all of the atoms to form compounds.

                 A + B ⟶ Product

Particles: 6      8              6

If six A atoms form six product molecules, each molecule can contain only one A atom.

The formula of the product is ABₙ.

If n = 1, we need six atoms of B.

If n = 2, we need 12 atoms of B. However, we have only eight atoms of B, so the formula of the product must be AB.

Thus, 6A + 6B ⟶ 6AB, with two B atoms left over.

Explanation:

Credit goes to @znk

Hope it helps you :))

7 0
3 years ago
An equilibrium mixture of PCl 5 ( g ) , PCl 3 ( g ) , and Cl 2 ( g ) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 To
antoniya [11.8K]

Answer: The new partial pressures of PCl_5,PCl_3\text{ and }Cl_2 when equilibrium is re-established are 223.4 torr, 6.82 torr and 26.4 torr respectively.

Explanation:

For the given chemical reaction:

PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)

The expression of K_p for above reaction follows:

K_p=\frac{P_{PCl_5}}{P_{PCl_3}\times P_{Cl_2}}         ........(1)

We are given:

P_{PCl_5}=217.0torr

P_{PCl_3}=13.2torr

P_{Cl_2}=13.2torr

Putting values in above equation, we get:

K_p=\frac{217.0}{13.2\times 13.2}\\\\K_p=1.24

Now we have to calculate the new partial pressure of Cl_2.

P_{PCl_5}+P_{PCl_3}+P_{Cl_2}=P_{Total}

217.0torr+13.2torr+P_{Cl_2}=263.0torr

P_{Cl_2}=32.8torr

The reaction is re-established and proceed to right direction by Le-Chatelier's principle to cancel the effect of addition of Cl_2.

Now, the equilibrium is shifting to the reactant side. The equation follows:

                       PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)

Initial:             13.2         32.8            217.0

At eqm:         13.2-x      32.8-x         217.0+x

Putting values in expression 1, we get:

1.24=\frac{(217.0+x)}{(13.2-x)(32.8-x)}\\\\x=40.4,6.38

Neglecting the 40.4 value of 'x'  because pressure can not be more than initial partial pressure.

Thus, the value of 'x' will be, 6.38 torr.

Now we have to calculate the new partial pressures after equilibrium is reestablished.

Partial pressure of PCl_5 = (217.0+x) = (217.0+6.38) = 223.4 torr

Partial pressure of PCl_3 = (13.2-x) = (13.2-6.38) = 6.82 torr

Partial pressure of Cl_2 = (32.8-x) = (32.8-6.38) = 26.4 torr

Hence, the new partial pressures of PCl_5,PCl_3\text{ and }Cl_2 when equilibrium is re-established are 223.4 torr, 6.82 torr and 26.4 torr respectively.

7 0
3 years ago
The reaction below will occur in a gaseous system at STP:
WARRIOR [948]

Answer:

V = 12.5 L

Explanation:

Given data:

Volume of NO = 15.0 L

Temperature and pressure = standard

Volume of nitrogen gas produced = ?

Solution:

Chemical equation:

6NO + 4NH₃    →     5N₂ + 6 H₂O

Number of moles of NO:

PV = nRT

n = PV/RT

n = 1 atm × 15.0 L / 0.0821 atm.L /mol.K × 273.15 K

n = 15.0 atm.L / 22.43 atm.L /mol

n = 0.67 mol

now we will compare the moles of No and nitrogen gas.

              NO           :         N₂

               6             :          5

              0.67         :         5/6×0.67 = 0.56

Volume of nitrogen gas:

 PV = nRT

1 atm × V = 0.56 mol ×  0.0821 atm.L /mol.K × 273.15 K

V = 12.5 atm.L / 1 atm

V = 12.5 L

4 0
3 years ago
Ort
ivanzaharov [21]

Answer:

1 ethanol is right answer

Explanation:

CH3- CH2-OH

4 0
3 years ago
Read 2 more answers
B<br><br> A. refraction <br> B. Diffraction<br> C. angle of reflection <br> D. angle of incidence
saw5 [17]

Answer:

B. Refraction

Explanation:

The angle of incidence equals the angle of refraction.

5 0
3 years ago
Read 2 more answers
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