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kondor19780726 [428]

3 years ago

14

3. Which of the following is NOT an abiotic factor in an ecosystem?

2 answers:

ozzi3 years ago

7 0

It is Bacteria........

sasho [114]3 years ago

6 0

The answer is Bacteria. Bacteria is biotic.

You might be interested in

Acids reacts with base and produces salt and water represent it in the form of chemical equation

tatiyna

Answer:

HCl+ NaoH- Nacl+H2O

Explanation:

base react with acid

3 0

3 years ago

Consider the reaction of CaCN2 and water to produce CaCO3 and NH3 according to the reaction CaCN2 + 3H2O → CaCO3 + 2 NH3 . How m

adelina 88 [10]

Answer:

56 g. Option 3.

Explanation:

The reaction is: CaCN₂ + 3H₂O → CaCO₃ + 2 NH₃

1 mol of calcium cianide reacts with 3 moles of water in order to produce 1 mol of calcium carbonate and 2 moles of ammonia

We have the mass of each reactant, so let's convert the mass to moles:

45 g. 1mol / 80.08 g = 0.562 moles of cianide

45 g. 1mol / 18 g = 2.5 moles of water

The cianide is the limiting reactant:

3 moles of water need 1 mol of cianide to react

Then, 2.5 moles of water will need (2.5 . 1)/ 3 = 0.833 moles

As we have 0.562 moles of CN⁻ we don't have enough

We can work now, on the reaction:

Ratio is 1:1. Therefore 0.562 moles of cianide will produce 0.562 moles of carbonate

Let's convert the mass to moles to find the answer:

0.562 mol . 100.08 g / 1 mol = 56.2 g

8 0

2 years ago

One sphere has a radions of 181 cm, another has a radius of 5.01 cm. What is the difference in volume (in cubic centimeters) bet

PSYCHO15rus [73]

Answer:

$2.4*10^{7}cm^{3}$

Explanation:

First you should calculate the volume of a big sphere,so:

$V_{big}=\frac{4}{3}\pi r^{3}$

$V_{big}=\frac{4}{3}\pi (181cm)^{3}$

$V_{big}=2.4*10^{7}cm^{3}$

Then you calculate the volume of a small spehre, so:

$V_{small}=\frac{4}{3}\pi r^{3}$

$V_{small}=\frac{4}{3}\pi (5.01cm)^{3}$

$V_{small}=5.3*10^{2}cm^{3}$

Finally you subtract the two quantities:

$V_{big}-V_{small}=2.4*10^{7}cm^{3}-5.3*10^{2}cm^{3}$

$V_{big}-V_{small}=2.4*10^{7}cm^{3}$

5 0

3 years ago

1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.

solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2) The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3)The given statement is false.

Explanation:

1)

Solubility of lead chloride = $S=3.1\times 10^-2M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

S 2S

The solubility product of the lead(II) chloride = $K_{sp}$

$K_{sp}=[Pb^{2+}][Cl^-]^2$

$K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}$

The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2)

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion = $1\timed 0.000010 M=0.000010 M$

Solubility of aluminium hydroxide in aluminum nitrate solution = $S$

$Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)$

S 3S

The solubility product of the aluminium nitrate = $K_{sp}=1.0\times 10^{-33}$

$K_{sp}=[Al^{3+}][OH^-]^3$

$1.0\times 10^{-33}=(0.000010+S)\times (3S)^3$

$S=1.6\times 10^{-10} M$

The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3.

$Molarity=\frac{Moles}{Volume (L)}$

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = $\frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol$

Volume of the solution = 0.250 L

$[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M$

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

$[Cl^-]=[NaCl]=0.00024 M$

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

$n=0.12 M]\times 0.250 L=0.030 mol$

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

$[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

Solubility of lead(II) chloride = $K_{sp}=1.2\times 10^{-4}$

Ionic product of the lead chloride in solution :

$Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}$

$Q_i$ ( no precipitation)

The given statement is false.

3 0

3 years ago

Lye is a common ingredient in drain cleaning products. It is made of one atom of sodium, one atom of oxygen, and one atom of hyd

il63 [147K]

The answer would be NaOH

8 0

3 years ago