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Montano1993 [528]
3 years ago
13

How many meters are there in 23.1 miles?

Chemistry
1 answer:
Hunter-Best [27]3 years ago
4 0

Answer:

3.7 ×10⁴ m

Explanation:

We know that,

1 miles = 1609.34 meter

We need to find how many meters are there in 23.1 miles.

23 miles = 37175.85 m

or

23 miles = 3.7 ×10⁴ m

So, there are 3.7 ×10⁴ m in 23.1 miles.

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IT will be easy to know how chemical reactions occur
It will helps to know the fundamental particles of an atom
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The light reactions of photosynthesis use _____ and produce _____. The light reactions of photosynthesis use _____ and produce _
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<h2>Input = NADP^+, water and Output = NADPH + O_2</h2>

Explanation:

The light reactions of photosynthesis use water and produce Oxygen, NADPH.

The equation for photosynthesis :

6 CO_2 + 6 H_2O → C_6H_12O_6 + 6 O_2

The process of photosynthesis in two stages -

  • The first stage is called the light reaction in which the light energy from the sun is captured and converted into chemical energy stored in the form of ATP and NADPH
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For a light reaction -

Net Input is of, NADP^+, Light, Water, ADP

Net Output is of, ATP, NADPH, O_2

8 0
3 years ago
Hydrogen react ls with oxygen at room temperature to form
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Answer:

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3 years ago
What is the answer to the net ionic equation FeO(s)+2HClO4(aq)--&gt; Fe(ClO4)2 (aq)+ H2O(l)
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8 0
3 years ago
Read 2 more answers
Calculate the standard reaction Gibbs free energy for the following cell reactions: (a) 2 Ce41(aq) 1 3 I2(aq) S 2 Ce31(aq) 1 I32
Law Incorporation [45]

<u>Answer:</u>

<u>For a:</u> The standard Gibbs free energy of the reaction is -347.4 kJ

<u>For b:</u> The standard Gibbs free energy of the reaction is 746.91 kJ

<u>Explanation:</u>

Relationship between standard Gibbs free energy and standard electrode potential follows:

\Delta G^o=-nFE^o_{cell}           ............(1)

  • <u>For a:</u>

The given chemical equation follows:

2Ce^{4+}(aq.)+3I^{-}(aq.)\rightarrow 2Ce^{3+}(aq.)+I_3^-(aq.)

<u>Oxidation half reaction:</u>   Ce^{4+}(aq.)\rightarrow Ce^{3+}(aq.)+e^-       ( × 2)

<u>Reduction half reaction:</u>   3I^_(aq.)+2e^-\rightarrow I_3^-(aq.)

We are given:

n=2\\E^o_{cell}=+1.08V\\F=96500

Putting values in equation 1, we get:

\Delta G^o=-2\times 96500\times (+1.80)=-347,400J=-347.4kJ

Hence, the standard Gibbs free energy of the reaction is -347.4 kJ

  • <u>For b:</u>

The given chemical equation follows:

6Fe^{3+}(aq.)+2Cr^{3+}+7H_2O(l)(aq.)\rightarrow 6Fe^{2+}(aq.)+Cr_2O_7^{2-}(aq.)+14H^+(aq.)

<u>Oxidation half reaction:</u>   Fe^{3+}(aq.)\rightarrow Fe^{2+}(aq.)+e^-       ( × 6)

<u>Reduction half reaction:</u>   2Cr^{2+}(aq.)+7H_2O(l)+6e^-\rightarrow Cr_2O_7^{2-}(aq.)+14H^+(aq.)

We are given:

n=6\\E^o_{cell}=-1.29V\\F=96500

Putting values in equation 1, we get:

\Delta G^o=-6\times 96500\times (-1.29)=746,910J=746.91kJ

Hence, the standard Gibbs free energy of the reaction is 746.91 kJ

7 0
3 years ago
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