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3 years ago

The question is below it was just easier to take a screenshot.

Chemistry

1 answer:

photoshop1234 [79]3 years ago

4 0

In this situation we need to balance the two half cell equations to form a proper reaction. This means reversing one of the two half-cell equations and balancing.

In this case we do not need to actually worry about balancing because the electrical potentials for each half reaction remain unchanged regardless of reactant and product quantities. What does change however, is the sign of the electrical potential. As we are reversing one half-cell reaction, it's electrical potential will be reversed aswell. Then the overall cell potential is the sum of these two potentials.

The anode reaction is Fe3+ + e -> Fe2+ as it requires electrons as reactants and as a positive electrical potential, meaning this reaction favours the products. The cathode reaction is therefore manganese.

So the overall electrical potential Ecell = Ecathode - Eanode
Ecell = -1.18V - 0.77 = -1.95V

(remember we switch the polarity of the electrical potential for the reaction that we reverse, in this case the anode reaction involving iron).

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3 years ago

A hot lump of 39.9 g of iron at an initial temperature of 78.1 °C is placed in 50.0 mL H 2 O initially at 25.0 °C and allowed to

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Answer : The final temperature of the mixture is $29.6^oC$

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In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of iron = $0.499J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of iron = 39.9 g

$m_2$ = mass of water = $Density\times Volume=1g/mL\times 50.0mL=50.0g$

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of iron = $78.1^oC$

$T_2$ = initial temperature of water = $25.0^oC$

Now put all the given values in the above formula, we get

$(39.9g)\times (0.499J/g^oC)\times (T_f-78.1)^oC=-(50.0g)\times 4.18J/g^oC\times (T_f-25.0)^oC$

$T_f=29.6^oC$

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4 years ago

If the graph of y = a x 2 + b x + c does not intersect the x -axis, then what is true about the roots? Both are real roots. Both

Reptile [31]

Answer:

Both roots are imaginary roots.

Explanation:

Consider these things:

If we try to solve x²+1 = 0, notice that we aren't able to solve the equation in Real Number system because there are no negative outputs for quadratic function.

Remember that quadratic function has range greater or equal to the max-min value.

x-axis plane represents the solutions of that equation. If a graph intersects x-axis plane then it has a solution.

While a graph that doesn't have any intersects on x-plane, it means that the equation for that graph doesn't have real solutions but imaginary solutions.

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