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3 years ago

Methane and water have different chemical properties because

Chemistry

2 answers:

Shtirlitz [24]3 years ago

5 0

C. They are in different states of matter

Explanation:
Water is a liquid while methane is a gas

jasenka [17]3 years ago

3 0

Answer:

Explanation:

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Why do we use commas after transition words?

ra1l [238]

Answer:

C - commas indicate a natural pause in the sentence

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2 years ago

science-what kind of input #o you need to provide to make the pen work. Describe the process and out put that result from the in

seropon [69]

Answer:

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Explanation:I looked it UP

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2 years ago

If the reduction reaction has a reduction potential of 0.1 V, and the oxidation reaction has a reduction potential of -0.4V, and

aleksley [76]

Answer : The value of ΔG expressed in terms of F is, -1 F

Explanation :

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

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

or,

$E^o=E^o_{reduction}-E^o_{oxidation}$

$E^o=(0.1V)-(-0.4V)=+0.5V$

Now we have to calculate the standard cell potential.

Formula used :

$\Delta G^o=-nFE^o$

where,

$\Delta G^o$ = Gibbs free energy = ?

n = number of electrons = 2

F = Faraday constant

$E^o$ = standard e.m.f of cell = +0.5 V

Now put all the given values in this formula, we get the Gibbs free energy.

$\Delta G^o=-(2\times F\times 0.5)$

$\Delta G^o=-1F$

Therefore, the value of ΔG expressed in terms of F is, -1 F

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3 years ago

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aliina [53]

Answer:

<h2>The answer you are looking for is (B)</h2>

Explanation:

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8 0

2 years ago

Read 2 more answers

A 50.0 g sample of scandium, sc, is heated by exposure to 1.50 x 10 3 j. The temperature of the sc is raised by 61.1 o

statuscvo [17]

Given mass of Scandium = 50.0 g

Increase in temperature of the metal when heated = $61.1^{0}C$

Heat absorbed by Scandium = $1.50*10^{3}J$

The equation showing the relationship between heat, mass, specific heat and temperature change:

$Q = m C (deltaT)$

Where Q is heat = $1.50*10^{3}J$

m is mass = 50.0 g

ΔT = $61.1^{0}C$

On plugging in the values and solving for C(specific heat) we get,

$1.50*10^{3}J$ =50.0g(C)( $61.1^{0}C$ )

C = 0.491 $\frac{J}{g^{0}C }$

Specific heat of the metal = 0.491 $\frac{J}{g^{0}C }$

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3 years ago