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Finger [1]
2 years ago
9

6) Which of the following does not fall in the biotic component of the ecosystem?

Chemistry
1 answer:
zlopas [31]2 years ago
4 0

Answer:

6.grass

7.xylem

8.oval

9. poor growth of bones and teetn

You might be interested in
Which element would lose electrons during ionic bonding choose all that apply
yawa3891 [41]

Answer:

All of them!

Explanation:

Since Mg, Li, Ca, and Cs are all in groups 1 and 2 of the periodic table, they are alkali/alkaline earth metals and will all lose electrons during ionic bonding.

7 0
3 years ago
At 298 K, the osmotic pressure of a glucose solution (C6H12O6 (aq)) is 12.1 atm. Calculate the freezing point of the solution. T
Anarel [89]

<u>Answer:</u> The freezing point of solution is -0.974°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 = 12.1 atm

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

M = molarity of solute = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature of the solution = 298 K

Putting values in above equation, we get:

12.1atm=1\times M\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\M=\frac{12.1}{1\times 0.0821\times 298}=0.495M

This means that 0.495 moles of glucose is present in 1 L or 1000 mL of solution

  • To calculate the mass of solution, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of solution = 1.034 g/mL

Volume of solution = 1000 mL

Putting values in above equation, we get:

1.034g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.034g/mL\times 1000mL)=1034g

  • To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of glucose = 0.495 moles

Molar mass of glucose = 180.16 g/mol

Putting values in above equation, we get:

0.495mol=\frac{\text{Mass of glucose}}{180.16g/mol}\\\\\text{Mass of glucose}=(0.495mol\times 180.16g/mol)=89.18g

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

  • The equation used to calculate depression in freezing point follows:

\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}

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

\Delta T_f=iK_fm

Or,

\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Freezing point of pure solution = 0°C

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

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

m_{solute} = Given mass of solute (glucose) = 89.18 g

M_{solute} = Molar mass of solute (glucose) = 180.16  g/mol

W_{solvent} = Mass of solvent (water) = [1034 - 89.18] g = 944.82 g

Putting values in above equation, we get:

0-\text{Freezing point of solution}=1\times 1.86^oC/m\times \frac{89.18\times 1000}{180.16g/mol\times 944.82}\\\\\text{Freezing point of solution}=-0.974^oC

Hence, the freezing point of solution is -0.974°C

8 0
3 years ago
For the equilibrium PCl5(g) PCl3(g) + Cl2(g), Kc = 2.0 × 101 at 240°C. If pure PCl5 is placed in a 1.00-L container and allowed
Cloud [144]

Answer:

the equilibrium concentration of [PCl₅] is 3.64*10⁻³ M

Explanation:

for the reaction

PCl₅(g) → PCl₃(g) + Cl₂(g)

where

Kc= [PCl₃]*[Cl₂]/[PCl₅] = 2.0*10¹ M = 20 M

and [A] denote concentrations of A

if initially the mixture is pure PCl₅ , then it will dissociate according to the reaction and since always one mole of PCl₃(g) is generated with one mole of Cl₂(g) , the total number of moles of both at the end is the same → they have the same concentration → [PCl₃(g)] = [Cl₂]=0.27 M

therefore

Kc= [PCl₃]*[Cl₂]/[PCl₅] = 0.27 M* 0.27 M /[PCl₅] = 20 M

[PCl₅]  =  0.27 M* 0.27 M / 20 M = 3.64*10⁻³ M

[PCl₅]  = 3.64*10⁻³ M

the equilibrium concentration of [PCl₅] is 3.64*10⁻³ M

6 0
3 years ago
What is the coefficient for calcium oxide? CaO(s) + CO2(g) → CaCO3(s)
VARVARA [1.3K]

Answer:

The coefficient is 1

Explanation:

CaO(s) + CO2(g) -> CaCO3(s)

In the balanced equation, the coefficient for CaO is 1

The coefficient represents the number of moles of a compound in the stoichiometry of the reaction

4 0
3 years ago
Which 3 elements had complete out shells? (you will probably need to use the periodic table)
wlad13 [49]
The inert gases like neon,argon ,krypton have complete shells
3 0
3 years ago
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