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adell [148]
2 years ago
7

How do I find the number of electrons in an Isotope? For example, I know that in the image attached the isotope has 45 protons a

nd 58 neutrons, but I am unsure as to how to find the number of electrons in the isotope.

Chemistry
1 answer:
vova2212 [387]2 years ago
7 0

Answer:

The answer to your question is: 45 electrons

Explanation:

Isotopes are molecules that have the same number of protons but the number of neutrons is different.

Also, in a neutral atom, the number of protons and electrons is the same.

In the example given, we can see that the atom is neutral so, the number of protons = 45 and the number of electrons = 45.

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Describe the oxygen/carbon dioxide cycle using plants and animals.
Vikentia [17]

Answer:

The process of photosynthesis in plants releases oxygen into the atmosphere. Respiration by plants and animals, as they use the energy stored in food, and the process of decomposition of dead organisms, releases carbon dioxide into the atmosphere. All three work together to maintain the carbon dioxide-oxygen cycle.

5 0
2 years ago
A block of metal has a width of 3.2 cm , a length of 17.1 cm , and height of 4.6 cm . Its mass is 1.4 kg . Calculate the density
dsp73

The Density of  the metal  is 5.6 g/cm³

<h3>What is the density of a substance?</h3>

The density of a substance is the ratio of the mass and the volume of the substance.

  • Density = mass/volume


The density of the metal is calculated as follows:

mass of metal = 1.4 kg = 1400 g

volume of metal = 3.2 * 17.1 * 4.6 = 251.712 cm³

Density of metal = 1400 g/251.712 cm³

Density of the metal  = 5.6 g/cm³

Therefore, the density of the metal is obtained from the mass and the volume of the metal.

Learn more about density at: brainly.com/question/1354972

#SPJ1

3 0
1 year ago
Using the periodic table, choose the more reactive nonmetal.<br> Br or As
raketka [301]

Reactivity of non-metals depend on their ability to gain electrons. So, smaller is the size of a non-metal more readily it will attract electrons because then nucleus will be more closer to valence shell. ... Hence, Br is the non-metal which will be more reactive than At.

8 0
2 years ago
Read 2 more answers
Determine the free energy(ΔG) from the standard cell potential (Ecell0 ) for the reaction:2ClO2-(aq)+Cl2(g)→2ClO2(g)+ 2Cl-(aq)wh
Dima020 [189]

<u>Answer:</u> The \Delta G^o for the given reaction is -7.84\times 10^4J

<u>Explanation:</u>

For the given chemical reaction:

2ClO_2^-(aq.)+Cl_2(g)\rightarrow 2ClO_2(g)+2Cl^-(aq.)

Half reactions for the given cell follows:

<u>Oxidation half reaction:</u> ClO_2^-\rightarrow ClO_2+e^-;E^o_{ClO_2^-/ClO_2}=0.954V  ( × 2)

<u>Reduction half reaction:</u> Cl_2+2e^-\rightarrow 2Cl(g);E^o_{Cl_2/2Cl^-}=1.36V

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=1.36-(0.954)=0.406V

To calculate standard Gibbs free energy, we use the equation:

\Delta G^o=-nFE^o_{cell}

Where,

n = number of electrons transferred = 2

F = Faradays constant = 96500 C

E^o_{cell} = standard cell potential = 0.406 V

Putting values in above equation, we get:

\Delta G^o=-2\times 96500\times 0.406=-78358J=-7.84\times 10^4J

Hence, the \Delta G^o for the given reaction is -7.84\times 10^4J

4 0
3 years ago
The solubility of o2 at 20c is 1.38 x10^-3. the partial presure of o2 in the air at sea level is 0.27 atm. using henery;s law, c
netineya [11]

<u>Answer:</u> The solubility of oxygen at 682 torr is 4.58\times 10^{-3}M

<u>Explanation:</u>

To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_{A}=K_H\times p_{A}

Or,

\frac{C_{1}}{C_{2}}=\frac{p_{1}}{p_2}

where,

C_1\text{ and }p_1 are the initial concentration and partial pressure of oxygen gas

C_2\text{ and }p_2 are the final concentration and partial pressure of oxygen gas

We are given:

Conversion factor used:  1 atm = 760 torr

C_1=1.38\times 10^{-3}M\\p_1=0.27atm\\C_2=?\\p_2=682torr=0.897atm

Putting values in above equation, we get:

\frac{1.38\times 10^{-3}}{C_2}=\frac{0.27atm}{0.897atm}\\\\C_2=\frac{1.38\times 10^{-3}\times 0.897atm}{0.27atm}=4.58\times 10^{-3}M

Hence, the solubility of oxygen gas at 628 torr is 4.58\times 10^{-3}M

4 0
2 years ago
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