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3 years ago

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Balance the redox reaction equation (occurring in acidic solution) and choose the correct coefficients for each reactant and pro

duct. __MnO4−(aq)+Ag(s)+H+(aq)→Mn2+(aq)+Ag+(aq)+__H2O(l)

1 answer:

dlinn [17]3 years ago

5 0

Explanation:

A balanced redox chemical equation is an equation that contains same number of atoms on both reactant and product side. Also, it contains same total charge on reactant and product side.

For example, $MnO^{-}_{4}(aq) + Ag(s) + H^{+}(aq) \rightarrow Mn^{2+}(aq) + Ag^{+}(aq) + H_{2}O(l)$

Number of atoms on reactant side are as follows.

Mn = 1

O = 4

Ag = 1

H = 1

Number of atoms on product side are as follows.

Mn = 1

O = 1

Ag = 1

H = 2

Therefore, to balance the equation multiply $H^{+}(aq)$ on reactant side by 8 and multiply $H_{2}O$ on product side by 4.

$MnO^{-}_{4}(aq) + Ag(s) + 8H^{+}(aq) \rightarrow Mn^{2+}(aq) + Ag^{+}(aq) + 4H_{2}O(l)$

Now, total charge on reactant side is +7 but total charge on product side is +2. Therefore, to balance the charges we multiply Ag(s) on reactant side by 5 and $Ag^{+}$ on product side by 5.

Therefore, completely balanced redox reaction equation will be as follows.

$MnO^{-}_{4}(aq) + 5Ag(s) + 8H^{+}(aq) \rightarrow Mn^{2+}(aq) + 5Ag^{+}(aq) + 4H_{2}O(l)$

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If a ball rolling down a hill is half way between the top and bottom, how much potential energy does the ball have compared to k

madreJ [45]

Answer:

The gravitational potential energy and kinetic energy of this ball should be equal (assuming that there is no energy loss due to friction.)

Explanation:

The ball loses gravitational potential energy as it rolls down the hill. At the same time, the speed of the ball increases, such that the ball gains kinetic energy.

If there is no friction on this ball (and that the ball did not deshape,) all the gravitational potential energy that this ball lost would be converted to kinetic energy.

If the gravitational field strength $g$ is constant throughout, the gravitational potential energy of an object in that gravitational field would be proportional to its height.

If $m$ denote the mass of this ball, the gravitational potential energy ( $\rm GPE$ ) of this ball at height $h$ would be ${\rm GPE} = (m \cdot g) \cdot h$ , which is proportional to $h\!$ .

The value of $g$ near the surface of the earth is indeed approximately constant (typically $g \approx 9.8\; \rm m \cdot s^{-2}$ .)

At halfway between the top and bottom of this hill, the height of this ball would be $(1/2)$ of its initial value (the value when the ball was at the top of the hill.) Because the $\rm GPE$ of this ball is proportional to its height, at halfway down the hill, the $\rm GPE\!$ of this ball would also be $(1/2)\!$ its initial value.

However, if there was no friction on this ball (and that the ball did not deshape,) that $(1/2)$ of the initial $\rm GPE\!$ of this ball was not lost. Rather, these $(1/2)\!$ of the initial $\rm GPE$ would have been converted to the kinetic energy ( $\rm KE$ ) of this ball.

Hence, when the ball is halfway down the hill:

$\displaystyle \text{GPE halfway down the hill} = \frac{1}{2}\, \text{Initial GPE}$ .

$\begin{aligned}& \text{KE halfway down the hill}\\ &= \text{Initial GPE} - \text{GPE halfway down the hill}\\ &= \text{Initial GPE} - \frac{1}{2}\, \text{initial GPE}\\ &= \frac{1}{2}\, \text{Initial GPE}\end{aligned}$ .

Therefore:

$\begin{aligned}& \text{GPE halfway down the hill} \\ &= \frac{1}{2}\, \text{Initial GPE} \\ &= \text{KE halfway down the hill}\end{aligned}$ .

In other words, under these assumptions, when this ball is halfway down the hill, the gravitational potential energy and the kinetic energy of this ball would be equal.

3 0

3 years ago

Help Tell me the answer.

Paladinen [302]

Answer:

it's easy u just have to put them in a calculator the way they are it will give you your answer atleast I think so hope this helps

5 0

3 years ago

Calculate the molarity when 135 g of KCl is dissolved to make a 2.50 L solution. Round to two significant digits.

inysia [295]

<h2>0.72 mol/L</h2>

Explanation:

$\textrm{Molarity of a solution}=\frac{\textrm{Number of moles of solvent}}{\textrm{Volume of solution in Liters}}$

We are given a $2.50L$ solution which contains $135g$ of $KCl$ .

To calculate number of moles of $KCl$ present, we need to find it's molar mass from charts. Molar Mass of $KCl$ is known to be $74.55\frac{g}{Mol}$ .

Number of moles of $KCl$ present = $\frac{\textrm{Given weight}}{\textrm{Molar Mass}}\textrm{ = }\frac{135g}{74.55\frac{g}{Mol}}\textrm{ = }1.81087mol$

Molarity = $\frac{1.81087mol}{2.50L}=0.7243\frac{mol}{L}$

∴ Molarity of given $KCl$ solution = $0.72\frac{mol}{L}$

7 0

3 years ago

A solution is 40 cetic acid by mass. the density of this solution is 1.049 g/ml. Calculate the mass of pure acetic acid in 220 m

taurus [48]

The mass of pure acetic acid in 220 ml of the given solution at 20°C is 92.311 g

<h3>What is Acetic acid?</h3>

Acetic acid is a type of carboxylic acid and also known as ethanoic acid

Its formula is CH₃COOH.

It is an organic compound and is a colorless liquid

It is mostly used in the production of vinegar

40 % acetic acid by mass means,

40 g of acetic acid is dissolved in 100 g of solution.

The density of solution at 20°C,

$\rho = 1.049 g/ml$

We know,

$\rho = \frac{m}{V}$

$V = \frac{m}{\rho}$

The volume of the solution, $V = \frac{100}{1.049}$ = 95.33 ml

95.33 ml of solution contains 40 g of pure Acetic acid

220 ml of solution contains $\frac{40 \times 220}{95.33}$ = 92.311 g of pure Acetic acid

Thus, the mass of pure Acetic acid in 220 ml of solution at 20°C is 92.311 g

Learn more about acetic acid:

brainly.com/question/24304533

#SPJ4

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2 years ago

Which of the following summertime activities consists solely of physical change?

Evgen [1.6K]

A: making s sandcastle. This is because water and sand is only a mixture, so they do not react with each other. All the rest include chemical reactions!

5 0

4 years ago