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3 years ago

Which statement regarding the Calvin cycle is false?

Chemistry

1 answer:

scoundrel [369]3 years ago

3 0

Answer:

The statement "Six turns of the cycle are required for every glucose molecule later produced in non–Calvin cycle reactions" is incorrect. It really looks not well-worded.

Explanation:

It is incorrect because Six turns of the cycle are required for every glucose molecule produced in Calvin cycle reactions, no in non-Calvin cycle reactions. This process includes the fixation of 6 molecules of carbon dioxide to produce 1 Glucose (seen as the addition of the two Phosphoglyceraldehide molecules (PGAL). Moreover, the other statements in the questions are correct:

ATP is required during carbon fixation.

The most intensive energy phase is reduction and sugar production.

Twelve NADPH are required for every six CO2 fixed.

NADPH is required for reduction and sugar production.

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A saturated solution of Pb(IO3)2 in pure water has a lead ion concentration of 5.0 x 10-5 Molar. What is the Ksp value of Pb(IO3

Orlov [11]

Answer:

Option (E) is correct

Explanation:

Solubility equilibrium of $Pb(IO_{3})_{2}$ is given as follows-

$Pb(IO_{3})_{2}\rightleftharpoons Pb^{2+}+2IO_{3}^{-}$

Hence, if solubility of $Pb(IO_{3})_{2}$ is S (M) then-

$[Pb^{2+}]=S(M)$ and $[IO_{3}^{-}]=2S(M)$

Where species under third bracket represent equilibrium concentrations

So, solubility product of $Pb(IO_{3})_{2}$ , $K_{sp}=[Pb^{2+}][IO_{3}^{-}]^{2}$

Here, $[Pb^{2+}]=S(M)=5.0\times 10^{-5}M$

So, $[IO_{3}^{-}]=2S(M)=(2\times 5.0\times 10^{-5})M=1.0\times 10^{-4}M$

So, $K_{sp}=(5.0\times 10^{-5})\times (1.0\times 10^{-4})^{2}=5.0\times 10^{-13}$

Hence option (E) is correct

7 0

3 years ago

You might need:

nikitadnepr [17]

Answer:

Mass = 14.3 g

Explanation:

Given data:

Mass of Mg(OH)₂ = 16.0 g

Mass of HCl = 11.0 g

Mass of MgCl₂ = ?

Solution:

Chemical equation:

Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O

Number of moles of Mg(OH)₂ :

Number of moles = mass/ molar mass

Number of moles = 16.0 g/ 58.3 g/mol

Number of moles = 0.274 mol

Number of moles of HCl :

Number of moles = mass/ molar mass

Number of moles = 11.0 g/ 36.5 g/mol

Number of moles = 0.301 mol

Now we will compare the moles of Mg(OH)₂ and HCl with MgCl₂.

Mg(OH)₂ : MgCl₂

1 : 1

0.274 : 0.274

HCl : MgCl₂

2 : 1

0.301 : 1/2×0.301 = 0.150

The number of moles of MgCl₂ produced by HCl are less so it will limiting reactant.

Mass of MgCl₂:

Mass = number of moles × molar mass

Mass = 0.150 × 95 g/mol

Mass = 14.3 g

8 0

3 years ago

Need help with balancing equations!

shusha [124]

Answer:

Fe2O3 + 3CO → 2Fe + 3CO2

Explanation:

the numbers in front are the numbers you need

4 0

2 years ago

Read 2 more answers

A dilute solution of bromine in carbon tetrachloride behaves as an ideal-dilute solution. The vapour pressure of pure CCl4 is 33

ANEK [815]

Explanation:

The given data is as follows.

Vapour pressure of pure $CCl_{4}$ = 33.85 Torr

Temperature = 298 K

Mole fraction of $Br_{2}$ = 122.36 torr

Therefore, calculate the vapor pressure of $Br_{2}$ as follows.

Vapour pressure of $Br_{2}$ = mole fraction of $Br_{2}$ x K of $Br_{2}$

= 0.050 x 122.36 Torr

= 6.118 Torr

So, vapor pressure of $Br_{2}$ is 6.118 Torr .

Now, calculate the vapor pressure of carbon tetrachloride as follows.

Vapour pressure of $CCl_{4}$ = mole fraction of $CCl_{4}$ x Pressure of $CCl_{4}$

= (1 - 0.050) × 33.85 Torr

= 32.1575 Torr

So, vapor pressure of $CCl_{4}$ is 32.1575 Torr .

Hence, the total pressure will be as follows.

= 6.118 Torr + 32.1575 Torr

= 38.2755 Torr

Therefore, composition of $CCl_{4}$ = $\frac{32.1575 Torr}{38.2755 Torr}$

= 0.8405

Composition of $CCl_{4}$ is 0.8405 .

And, composition of $Br_{2}$ = $\frac{6.118 Torr}{38.2755 Torr}$

= 0.1598

Composition of $Br_{2}$ is 0.1598 .

6 0

3 years ago

Plz help (it’s a picture)

Feliz [49]

Umm I’ll figure it out rn! Will come back in 1 min

7 0

3 years ago