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natita [175]
3 years ago
7

Explain in your own words the process using the following key words: carbon-di-oxide, water, sugar, oxygen, thylakoid, chloropla

st, atp, photosystem, Calvin cycle, chlorophyll, and sunlight.
Chemistry
1 answer:
AleksAgata [21]3 years ago
5 0

Answer:

The Photosynthesis process

Explanation:

Plants, algae, and some other organisms can transform the sunlight energy into chemical energy. The photosynthesis process occur thanks to the chloroplasts. The chloroplast is an organelle found in all green plants. Inside of the chloroplast you can find the thylakoids which are arranged in stacks named grana, they have membranes with chloropyll a photosynthetic pigment, also you can find the photosystems, they are functional and structural units of protein complexes. The thylakoids capture the light and allow the reactions to transform CO2. The set of reactions that occurs in the chloroplasts are known as the Calvin cycle.

The general equation of photosynthesis is:

6 CO_{2}                 + 6 H_{2} O + Energy  -> C_{6} H_{12} 12O_{6}  +  6 O_{2}

6 CO2                + 6 H2O + Energy  -> C6H12O6         +  6 O2  

Carbon Dioxide  + water  +  Light   -> Glucose (sugar) + Oxygen

After, this glucose is transformed into pyruvate, and it allowed the release of denosine triphosphate (ATP) by cellular respiration. The ATP is an organic chemical that is requires for the cell to perform any process (any kind or work).

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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)

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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)

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Explanation:

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