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Ipatiy [6.2K]
3 years ago
12

An atom of the element iron has an atomic number of 26 and an atomic mass number of 56. if it is neutral, how many protons, neut

rons, and electrons does it have?
Chemistry
2 answers:
allsm [11]3 years ago
6 0
<span> 26 protons, 30 neutrons, 26 electrons</span>
asambeis [7]3 years ago
6 0

Answer:

26 protons, 26 electrons, 30 neutrons

Explanation:

Let P = protons, N = neutrons, E = electrons

The mass number of an atom is the sum of protons and neutrons in the nucleus of the atom.

The number of neutrons is the difference between the mass and atomic numbers of the atom

Therefore, 56 = P + N

where N = 56 - 26

N = 30

That means the number of neutrons is 30

To find P, 56 = P + N, N being 30

56 is now P + 30

56 = P + 30

56 - 30 = P

P = 26

Now, for an atom in its neutral state, the number of electrons is equal to the number of protons

Therefore, E is also 26

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Answer:

Approximately 0.291\; \rm M (rounded to two significant figures.)

Explanation:

The unit of concentration \rm M is the same as \rm mol \cdot L^{-1} (moles per liter.) On the other hand, the volume of both the \rm NaOH solution and the original \rm HCl solution here are in milliliters. Convert these two volumes to liters:

  • V(\mathrm{NaOH}) = 18.2\; \rm mL = 18.2 \times 10^{-3}\; \rm L = 0.0182\; \rm L.
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Calculate the number of moles of \rm NaOH in that 0.0182\; \rm L of 0.160\; \rm M solution:

\begin{aligned} n(\mathrm{NaOH}) &= c(\mathrm{NaOH})\cdot V(\mathrm{NaOH})\\ &= 0.160\; \rm mol \cdot L^{-1} \times 0.0182\; \rm L \approx 0.00291\; \rm mol\end{aligned}.

\rm HCl reacts with \rm NaOH at a one-to-one ratio:

\rm HCl\; (aq) + NaOH\; (aq) \to NaCl\; (aq) + H_2O\; (l).

Coefficient ratio:

\displaystyle \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} = 1.

In other words, one mole of \rm NaOH would neutralize exactly one mole of \rm HCl. In this titration, 0.291\; \rm mol of \rm NaOH\! was required. Therefore, the same amount of \rm HC should be present in the original solution:

\begin{aligned}&n(\text{$\mathrm{HCl}$, original})\\ &= n(\mathrm{NaOH})\cdot \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} \\ &\approx 0.00291\; \rm mol \times 1 = 0.00291\; \rm mol\end{aligned}.

Calculate the concentration of the original \rm HCl solution:

\displaystyle c(\text{$\mathrm{HCl}$, original}) = \frac{n(\text{$\mathrm{HCl}$, original})}{V(\text{$\mathrm{HCl}$, original})} \approx \frac{0.00291\; \rm mol}{0.0100\; \rm L} \approx 0.291\; \rm M.

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3 years ago
Does ice particles have high interparticle force of attraction? Justify your answer.<br> please and
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Answer:

¿Las partículas de hielo tienen una alta fuerza de atracción entre partículas? Justifica tu respuesta

Explanation:

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Question: In experiment your teacher has provided you with a sample of sandy, salty, ocean water and
liq [111]

Answer:

Explanation:

Homogeneous mixture is a mixture in which the components of the mixture are in the same proportion throughout any sample extracted from the mixture while an heterogeneous mixture is a mixture in which the components of the mixture differ in term of proportion when different samples of the mixture are extracted and compared.

For example, a sandy water will have some parts (usually the bottom) of the mixture with more sand than other parts of the mixture, hence, it (sandy water) is a heterogeneous mixture. While salty and ocean water has it's salt dissolved in the same proportion throughout the water in the mixture, hence salty and/or ocean water is a homogeneous mixture.

Sandy water can be separated by filtration (i.e using a filter paper to separate the sand from the water when the mixture is poured over a filter paper) while salty and ocean water can be separated by distillation (i.e boiling of the mixture so the water molecules can boil and move through a tube as gas or steam into another container where they are cooled and converted back to liquid or water while leaving the solid salt component of the mixture in the boiling tube).

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3 years ago
A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0-L vessel at 300 K. The following equilibriu
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Answer:

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[N2] =    0.019 M

[H2O] =  0.057 M

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i  mol            0.10                   0.050                                             0.10

c mol            -0.038                -0.038                +0019                +0.038                                                

e mol            0.062                 0.012                  00.019               0.057

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[NO] =   0.062 M

[H2] =    0.012 M

[N2] =    0.019 M

[H2O] =  0.057 M

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