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3 years ago

What is 2\5 please let me know what the answer

Chemistry

1 answer:

mariarad [96]3 years ago

4 0

Answer: 0.4

Explanation: The easiest way to do this is to get a denominator of 10. To do this, multiply each number in 2/5 by 2. You will get the fraction 4/10. With the denominator of 10, the answer would be 0.4 by putting the decimal in front of the number.

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A chemist designs a galvanic cell that uses these two half-reactions:

Nonamiya [84]

Answer :

(a) Reaction at anode (oxidation) : $4Fe^{2+}\rightarrow 4Fe^{3+}+4e^-$

(b) Reaction at cathode (reduction) : $O_2+4H^++4e^-\rightarrow 2H_2O$

(c) $O_2+4H^++4Fe^{2+}\rightarrow 2H_2O+4Fe^{3+}$

(d) Yes, we have have enough information to calculate the cell voltage under standard conditions.

Explanation :

The half reaction will be:

Reaction at anode (oxidation) : $Fe^{2+}\rightarrow Fe^{3+}+e^-$ $E^0_{anode}=+0.771V$

Reaction at cathode (reduction) : $O_2+4H^++4e^-\rightarrow 2H_2O$ $E^0_{cathode}=+1.23V$

To balance the electrons we are multiplying oxidation reaction by 4 and then adding both the reaction, we get:

Part (a):

Reaction at anode (oxidation) : $4Fe^{2+}\rightarrow 4Fe^{3+}+4e^-$ $E^0_{anode}=+0.771V$

Part (b):

Reaction at cathode (reduction) : $O_2+4H^++4e^-\rightarrow 2H_2O$ $E^0_{cathode}=+1.23V$

Part (c):

The balanced cell reaction will be,

$O_2+4H^++4Fe^{2+}\rightarrow 2H_2O+4Fe^{3+}$

Part (d):

Now we have to calculate the standard electrode potential of the cell.

$E^o=E^o_{cathode}-E^o_{anode}$

$E^o=(1.23V)-(0.771V)=+0.459V$

For a reaction to be spontaneous, the standard electrode potential must be positive.

So, we have have enough information to calculate the cell voltage under standard conditions.

4 0

3 years ago

What is the rock that has no exposure to chemical or mechanical weathering known as?

den301095 [7]

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3 years ago

Who earned a Nobel Peace Prize for their study of the photoelectric effect?

riadik2000 [5.3K]

ALBERT ENSTIEN !!!!!

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4 years ago

The room temperature electrical conductivity of a semiconductor specimen is 2.8 x 104 (?-m)-1. The electron concentration is kno

Julli [10]

Answer:

7.43 × 10²⁴ m⁻³

Explanation:

Data provided in the question:

Conductivity of a semiconductor specimen, σ = 2.8 × 10⁴ (Ω-m)⁻¹

Electron concentration, n = 2.9 × 10²² m⁻³

Electron mobility, $\mu_n$ = 0.14 m²/V-s

Hole mobility, $\mu_p$ = 0.023 m²/V-s

Now,

σ = $nq\mu_n+pq\mu_p$

σ = $q(n\mu_n+p\mu_p)$

here,

q is the charge on electron = 1.6 × 10⁻¹⁹ C

p is the hole density

thus,

2.8 × 10⁴ = 1.6 × 10⁻¹⁹( 2.9 × 10²² × 0.14 + p × 0.023 )

1.75 × 10²³ = 0.406 × 10²² + 0.023p

17.094 × 10²² = 0.023p

p = 743.217 × 10²²

p = 7.43 × 10²⁴ m⁻³

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4 years ago

Is Cytoplasm found in animal cells,plant cells or in both?

nataly862011 [7]

Answer:

cytoplasm is found in both plant and animal cell.

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3 years ago

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What is 2\5 please let me know what the answer​

What is 2\5 please let me know what the answer