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2 years ago

11

Lactic acid and energy is produced in muscle cells during

1 answer:

mr Goodwill [35]2 years ago

7 0

It's C anaerobic respiration

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In which of these systems is the entropy decreasing?

Delicious77 [7]

"A <span>gas condensing to a liquid" is the one system among the following choices given in the question where the entropy is decreasing. The correct option among all the options that are given in the question is the fourth option or option "D". I hope that this is the answer that has come to your desired help.</span>

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3 years ago

Hydrazine (N₂H₄), a rocket fuel, reacts with oxygen to form nitrogen gas and water vapor. The reaction is represented with the e

Vinvika [58]

At STP one mol weighs 22.4L

Moles of O_2

4.2/22.4
2.1/11.2
0.19mol

1 mol.O_2 can create 2mol water

moles of water

2(0.19)
0.38mol

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8 0

1 year ago

Read 2 more answers

Using the standard reduction potentials, Pb 2+(aq) + 2e– => Pb(s), E° = –0.13 V Fe 2+(aq) + 2e– => Fe(s), E° = –0.44 V Zn

mojhsa [17]

Answer:

Pb(s), Fe(s) and Zn(s) will reduce $Mn^{3+}$ to $Mn^{2+}$

Fe(s) and Zn(s) will reduce $Cr^{3+}$ to $Cr$

Explanation:

Standard reduction potential denotes ability to consume electrons from another species.

Hence, higher the standard reduction potential, higher will be the ability to oxidize another species.

Metal with $E_{red}^{0}$ value lower than 1.51 V will donate electron to $Mn^{3+}$ and thus reduces $Mn^{3+}$ to $Mn^{2+}$ .

So, Pb(s), Fe(s) and Zn(s) will reduce $Mn^{3+}$ to $Mn^{2+}$ .

Metal with $E_{red}^{0}$ value lower than -0.40 V will donate electron to $Cr^{3+}$ and thus reduces $Cr^{3+}$ to $Cr$ .

So, Fe(s) and Zn(s) will reduce $Cr^{3+}$ to $Cr$ .

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2 years ago

How to make a sugar cube dissolve more quickly in water?

Vesna [10]

Heat the water up a little

3 0

2 years ago

Consider the reaction of diboron trioxide with carbon and chlorine. B2O3 (s) + 3C (s) + 3Cl2 (g) 2BCl3 (g) + 3CO (g) Determine t

Sholpan [36]

Answer:

Limiting reactant = B2O3

Amount of BCl3 formed = 468 g

Explanation:

The given reaction is:

$B2O3 (s) + 3C (s) + 3Cl2 (g) \rightarrow 2BCl3 (g) + 3CO (g)$

In order to identify the limiting reagent calculate the moles of B2O3, C and Cl2. The reagent with the lowest moles is the limiting reactant

$Moles(B2O3)=\frac{Mass(B2O3)}{Mol.wt(B2O3)}=\frac{139g}{69.6g/mol}=1.997moles$

$Moles(C)=\frac{Mass(C)}{At.wt(C)}=\frac{87.8g}{12g/mol}=7.317moles$

$Moles(Cl2)=\frac{Mass(Cl2)}{Mol.wt(Cl2)}=\frac{650g}{70.9g/mol}=9.168moles$

Since the moles of B2O3 < C < Cl2, the limiting reactant is B2O3

Based on the reaction stoichiometry:

1 mole of B2O3 produces 2 moles of BCl3

Hence, the number of moles of BCl3 produced under the experimental conditions = 2*1.997=3.994 moles

$Mass(BCl3)= Moles* Mol.wt = 3.994 moles*117.17g/mol = 468 g$

8 0

2 years ago