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BlackZzzverrR [31]
3 years ago
12

Show bond formation in magnesium chloride​

Chemistry
1 answer:
Nitella [24]3 years ago
7 0

mg has a 2 in its valence shell

it will become mgcl2

. . . .

: Cl -------- Mg ------ Cl :

. . . .

You might be interested in
One liter of oxygen gas at standard temperature and pressure has a mass of 1.43 g. The same volume of hydrogen gas under these c
Alchen [17]

Answer:

Indeed, the two samples should contain about the same number of gas particles. However, the molar mass of \rm O_2\; (g) is larger than that of \rm H_2\; (g) (by a factor of about 16.) Therefore, the mass of the \rm O_2\; (g) sample is significantly larger than that of the \rm H_2\; (g) sample.

Explanation:

The \rm O_2\; (g) and the \rm H_2\; (g) sample here are under the same pressure and temperature, and have the same volume. Indeed, if both gases are ideal, then by Avogadro's Law, the two samples would contain the same number of gas particles (\rm O_2\; (g) and \rm H_2\; (g) molecules, respectively.) That is:

n(\mathrm{O_2}) = n(\mathrm{H}_2).

Note that the mass of a gas m is different from the number of gas particles n in it. In particular, if all particles in this gas have a molar mass of M, then:

m = n \cdot M.

In other words,

  • m(\mathrm{O_2}) = n(\mathrm{O_2}) \cdot M(\mathrm{O_2}).
  • m(\mathrm{H_2}) = n(\mathrm{H_2}) \cdot M(\mathrm{H_2}).

The ratio between the mass of the \rm O_2\; (g) and that of the \rm H_2\; (g) sample would be:

\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})}\end{aligned}.

Since n(\mathrm{O_2}) = n(\mathrm{H}_2) by Avogadro's Law:

\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})} = \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}\end{aligned}.

Look up relative atomic mass data on a modern periodic table:

  • \rm O: 15.999.
  • \rm H: 1.008.

Therefore:

  • M(\mathrm{O_2}) = 2 \times 15.999 \approx 31.998\; \rm g \cdot mol^{-1}.
  • M(\mathrm{H_2}) = 2 \times 1.008 \approx 2.016\; \rm g \cdot mol^{-1}.

Verify whether \begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})}= \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}\end{aligned}:

  • Left-hand side: \displaystyle \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})}= \frac{1.43\; \rm g}{0.089\; \rm g} \approx 16.1.
  • Right-hand side: \displaystyle \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}= \frac{31.998\; \rm g \cdot mol^{-1}}{2.016\; \rm g \cdot mol^{-1}} \approx 15.9.

Note that the mass of the \rm H_2\; (g) sample comes with only two significant figures. The two sides of this equations would indeed be equal if both values are rounded to two significant figures.

7 0
3 years ago
Name two hormones produced by the endocrine system and describe how they work with other organ systems to maintain homeostasis
Annette [7]
The 2 hormones are insulin & glucagon.

A hormone will only act on a part of the body it 'fits'. A hormone can be thought of as a key, and its target site ( i.e an organ) has specially shaped locks on the cell walls. 
If the hormone fits, then it will work.

The hormone can set off a cascade of other singling pathways in the cell to cause an immediate effect ( for instance, insulin signaling leads to a rapid uptake of glucose in muscle cells)

The endocrine system is a tightly regulated system that keeps the hormones and their effects at just the right level. One way this is achieved is through ' feedback loops'. The release of hormones is regulated by other hormones, proteins or neuronal signals.

The released hormone then has its effect on other organs. This effect on the organ feeds back to the original signal to control any further hormone release.


btw- found all this info @ the Better Health channel, an australian government health website , so if your still confused by my answer, check out this website

www.betterhealth.vic.gov.au/health/conditionsandtreatments/hormonal-endocrine-system

8 0
3 years ago
Read 2 more answers
A chemistry teacher adds 50.0 mL of 1.50 M H2SO4 solution to 200 mL of water. What is the concentration of the final solution? A
anygoal [31]

this is a dilution equation where 50.0 mL of 1.50 M H₂SO₄ is taken and added to 200 mL of water.

c1v1 = c2v2

where c1 is concentration and v1 is volume of the concentrated solution

and c2 is concentration and v2 is volume of the diluted solution to be prepared

50.0 mL of 1.50 M H₂SO₄ is added to 200 mL of water so the final solution volume is - 200 + 50.0 = 250 mL

substituting these values in the formula

1.50 M x 50.0 mL = C x 250 mL

C = 0.300 M

concentration of the final solution is A) 0.300 M

4 0
3 years ago
Read 2 more answers
Who was the scientist responsible for the discovery of the keystone species
sveticcg [70]

Answer:

Ecologist Robert Paine

Explanation:

:)

3 0
3 years ago
An ionic bond can be formed when one or more electrons are 1) equally shared by two atoms 2) unequally shared by two atoms 3) tr
Lyrx [107]

Answer:

4) transferred from the valence shell of one atom to the valence shell of another atom

Explanation:

Electrons are located outside of the nucleus which contains the protons and the neutrons.

For bonds to form, valence electrons located in the outermost shell electrons are involved. These are the valence electrons. These outer shell electrons can be shared or transferred between two combining atoms to form stable atoms.

In ionic bonds, the electrons are transferred from one specie to another. The atom that loses the electrons becomes positively charged and the receiving atom becomes negatively charged. This is the crux of ionic bonds.

6 0
3 years ago
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