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viva [34]
3 years ago
5

Materials are transported within a single celled organism by the

Chemistry
1 answer:
CaHeK987 [17]3 years ago
6 0
The cytoplasm transports materials within a single celled organism. that is the answer.
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Calculate the cell potential E at 25°C for the reaction 2 Al(s) + 3 Fe2+(aq) → 2 Al3+(aq) + 3 Fe(s) given that [Fe 2+] = 0.020 M
Elodia [21]

Answer:

1.18 V

Explanation:

The given cell is:

Al(s)/Al^{3+}(0.10M)||Fe^{2+}(0.020M)/Fe(s)

Half reactions for the given cell follows:

Oxidation half reaction: Al(s)\rightarrow Al^{3+}(0.10M)+2e^-;E^o_{Al^{3+}/Al}=-1.66V

Reduction half reaction: Fe^{2+}(0.020M)+2e^-\rightarrow Fe(s);E^o_{Fe^{2+}/Fe}=-0.45V

Multiply Oxidation half reaction by 2 and Reduction half reaction by 3

Net reaction: 2Al(s)+3Fe^{2+}(0.020M)\rightarrow 2Al^{3+}(0.10M)+3Fe(s)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=-0.45-(-1.66)=1.21V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Al^{3+}]^2}{[Fe^{2+}]^3}

where,

E_{cell} = electrode potential of the cell = ?V

E^o_{cell} = standard electrode potential of the cell = +1.21 V

n = number of electrons exchanged = 6

Putting values in above equation, we get:

E_{cell}=1.21-\frac{0.059}{6}\times \log(\frac{0.10^2}{0.020^3})\\\\E_{cell}=1.18V

5 0
4 years ago
Avogadro counted the number of __ in carbon 12 gas
Burka [1]

Answer:

(atoms.............) is the homogeneous mixture

5 0
3 years ago
Read 2 more answers
A three dimensional circular shape
Bond [772]
A three dimensional circular shape would be a sphere :)
4 0
4 years ago
Read 2 more answers
What is the empirical formula of a molecule containing 65.5% carbon, 5.5% hydrogen, and 29.0% oxygen?
rusak2 [61]

Answer:

The empirical formula is C3H3O

Explanation:

Step 1: Data given

Suppose the mass of the molecule = 100 grams

The molecule contains:

65.5 % Carbon = 65.5 grams

5.5 % Hydrogen = 5.5 grams

29.0% Oxygen = 29.0 grams

Molar mass of C = 12 g/mol

Molar mass of H = 1.01 g/mol

Molar mass of O = 16 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles C = 65.5 grams / 12 g/mol = 5.46 moles

Moles H = 5.5 / 1.01 g/mol = 5.45 moles

Moles O = 29.0 grams / 16 g/mol = 1.8125 moles

Step 3: Calculate mol ratio

We divide by the smallest amount of moles

C: 5.46 / 1.8125 = 3

H = 5.45 / 1.8125 = 3

O = 1.8125/1.8125 = 1

The empirical formula is C3H3O

8 0
3 years ago
I need a bit of help on this I kinda forgot about this lesson
Softa [21]
A is correct because the fruit salad can be easily separated and the puch cannot so it is a solution and the fruit salad is a mixture.
3 0
4 years ago
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