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DENIUS [597]
3 years ago
11

Which of the following will reduce copper? zinc mercury fluorine chlorine

Chemistry
2 answers:
garri49 [273]3 years ago
6 0

<u>Answer:</u> The correct answer is Zinc.

<u>Explanation:</u>

Reactivity is defined as the tendency of an element to loose or gain electrons. Metals which loose electrons easily will be considered as more reactive metal.

The reactivity of elements will be judged on the basis of reactivity series. The elements which lie above in the series are more reactive than the elements which lie below in the series. The metals which lie above in the series will easily reduce the metals which lie below in the series.

Copper is the metal which lie above mercury and below zinc.

Thus, zinc is more reactive metal than copper and hence will easily reduce copper from its reaction.

Hence, the correct answer is Zinc.

laiz [17]3 years ago
4 0
Zinc because the only metals that would be able to reduce copper ions in solution would be hydrogen, lead, tin, nickel, iron, zinc, aluminum, Magnesium, sodium, calcium, potassium, and lithium. and according to your answer choices Zinc is the answer.
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(04.02 LC) Based on the cell theory, which of the following is true?
prohojiy [21]

Answer:

c

Explanation:

there are billions of cells in living things. We are made out of millions of cells

4 0
2 years ago
How much energy is released by the decay of 3 grams of 230Th in the following reaction 230 Th - 226Ra + "He (230 Th = 229.9837 g
Serhud [2]

<u>Answer:</u> The energy released for the decay of 3 grams of 230-Thorium is 2.728\times 10^{-15}J

<u>Explanation:</u>

First we have to calculate the mass defect (\Delta m).

The equation for the alpha decay of thorium nucleus follows:

_{90}^{230}\textrm{Th}\rightarrow _{88}^{226}\textrm{Ra}+_2^{4}\textrm{He}

To calculate the mass defect, we use the equation:

Mass defect = Sum of mass of product - Sum of mass of reactant

\Delta m=(m_{Ra}+m_{He})-(m_{Th})

\Delta m=(225.9771+4.008)-(229.9837)=1.4\times 10^{-3}amu=2.324\times 10^{-30}kg

(Conversion factor: 1amu=1.66\times 10^{-27}kg )

To calculate the energy released, we use Einstein equation, which is:

E=\Delta mc^2

E=(2.324\times 10^{-30}kg)\times (3\times 10^8m/s)^2

E=2.0916\times 10^{-13}J

The energy released for 230 grams of decay of thorium is 2.0916\times 10^{-13}J

We need to calculate the energy released for the decay of 3 grams of thorium. By applying unitary method, we get:

As, 230 grams of Th release energy of = 2.0916\times 10^{-13}J

Then, 3 grams of Th will release energy of = \frac{2.0916\times 10^{-13}}{230}\times 3=2.728\times 10^{-15}J

Hence, the energy released for the decay of 3 grams of 230-Thorium is 2.728\times 10^{-15}J

5 0
3 years ago
What is the speed of sound in dry air at 20°C?
Lubov Fominskaja [6]
343 meters per second
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3 years ago
CO2<br> +<br> NaOH<br> &lt;=&gt;<br> NaHCO3<br> Balance
NeTakaya

Answer:

Explanation:

It is balaned

Left side

1 Na

1 C

3 Os

1 H

=========

Right side

1 Na

1 H

1 C

3 Os

5 0
3 years ago
Read 2 more answers
Equation balancing
meriva

<u>Answer:</u>

<u>For a:</u> The balanced equation is 2S(s)+3O_2(g)\rightarrow 2SO_3(g)

<u>For c:</u> The balanced equation is 2NaOH(s)+H_2SO_4(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)

<u>Explanation:</u>

A balanced chemical equation is one where all the individual atoms are equal on both sides of the reaction. It follows the law of conservation of mass.

  • <u>For (a):</u>

The given unbalanced equation follows:

S(s)+O_2(g)\rightarrow SO_3(g)

To balance the equation, we must balance the atoms by adding 2 infront of both S(s) and SO_3 and 3 in front of O_2

For the balanced chemical equation:

2S(s)+3O_2(g)\rightarrow 2SO_3(g)

  • <u>For (b):</u>

The given balanced equation follows:

2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3(s)

The given equation is already balanced.

  • <u>For (c):</u>

The given unbalanced equation follows:

2NaOH(s)+H_2SO_4(aq)\rightarrow Na_2SO_4(aq)+H_2O(l)

To balance the equation, we must balance the atoms by adding 2 infront of H_2O(l)

For the balanced chemical equation:

2NaOH(s)+H_2SO_4(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)

  • <u>For (d):</u>

The given balanced equation follows:

C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)

The given equation is already balanced.

4 0
3 years ago
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