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GREYUIT [131]
4 years ago
15

The sphere is the layer of the earth that contains a mixture of gases

Chemistry
2 answers:
jasenka [17]4 years ago
6 0
Omoze layer I know this

konstantin123 [22]4 years ago
6 0
No... I believe that the part of Earth that contains mixtures of gases is the atmosphere, along with its layers.
You might be interested in
Excess magnesium reacts with 165.0 grams of hydrochloric acid in a single displacement reaction.
JulsSmile [24]

Answer:

The volume of hydrogen gas produced will be approximately 50.7 liters under STP.

Explanation:

Relative atomic mass data from a modern periodic table:

  • H: 1.008;
  • Cl: 35.45.

Magnesium is a reactive metal. It reacts with hydrochloric acid to produce

  • Hydrogen gas \rm H_2, and
  • Magnesium chloride, which is a salt.

The chemical equation will be something like

\rm ?\;Mg\;(s) + ?\;HCl \;(aq)\to ?\;H_2 \;(g)+ [\text{Formula of the Salt}],

where the coefficients and the formula of the salt are to be found.

To determine the number of moles of \rm H_2 that will be produced, first find the formula of the salt, magnesium chloride.

Magnesium is a group 2 metal. The oxidation state of magnesium in compounds tends to be +2.

On the other hand, the charge on each chloride ion is -1. Each magnesium ion needs to pair up with two chloride ions for the charge to balance in the salt, magnesium chloride. The formula for the salt will be \rm MgCl_2.

\rm ?\;Mg\;(s) + ?\;HCl\;(aq) \to ?\;H_2 \;(g)+ ?\;MgCl_2\;(aq).

Balance the equation. \rm MgCl_2 contains the largest number of atoms among all species in this reaction. Start by setting its coefficient to 1.

\rm ?\;Mg\;(s) + ?\;HCl\;(aq) \to ?\;H_2 \;(g)+ {\bf 1\;MgCl_2}\;(aq).

The number of \rm Mg and \rm Cl atoms shall be the same on both sides. Therefore

\rm {\bf 1\;Mg}\;(s) + {\bf 2\;HCl}\;(aq) \to ?\;H_2 \;(g)+ {1\;\underset{\wedge}{Mg}\underset{\wedge}{Cl_2}}\;(aq).

The number of \rm H atoms shall also conserve. Hence the equation:

\rm {1\;Mg}\;(s) + {2\;\underset{\wedge}{H}Cl}\;(aq) \to {\bf 1\;H_2 \;(g)}+ {1\;MgCl_2}\;(aq).

How many moles of HCl are available?

M(\rm HCl) = 1.008 + 35.45 = 36.458\;g\cdot mol^{-1}.

\displaystyle n({\rm HCl}) = \frac{m(\text{HCl})}{M(\text{HCl})} = \rm \frac{165.0\;g}{36.458\;g\cdot mol^{-1}} = 4.52576\;mol.

How many moles of Hydrogen gas will be produced?

Refer to the balanced chemical equation, the coefficient in front of \rm HCl is 2 while the coefficient in front of \rm H_2 is 1. In other words, it will take two moles of \rm HCl to produce one mole of \rm H_2. \rm 4.52576\;mol of \rm HCl will produce only one half as much \rm H_2.

Alternatively, consider the ratio between the coefficient in front of \rm H_2 and \rm HCl is:

\displaystyle \frac{n(\text{H}_2)}{n(\text{HCl})} = \frac{1}{2}.

\displaystyle n(\text{H}_2) = n(\text{HCl})\cdot \frac{n(\text{H}_2)}{n(\text{HCl})} = \frac{1}{2}\;n(\text{HCl}) = \rm \frac{1}{2}\times 4.52576\;mol = 2.26288\;mol.

What will be the volume of that many hydrogen gas?

One mole of an ideal gas occupies a volume of 22.4 liters under STP (where the pressure is 1 atm.) On certain textbook where STP is defined as \rm 1.00\times 10^{5}\;Pa, that volume will be 22.7 liters.

V(\text{H}_2) = \rm 2.26288\;mol\times 22.4\;L\cdot mol^{-1} = 50.69\; L, or

V(\text{H}_2) = \rm 2.26288\;mol\times 22.7\;L\cdot mol^{-1} = 51.37\; L.

The value "165.0 grams" from the question comes with four significant figures. Keep more significant figures than that in calculations. Round the final result to four significant figures.

5 0
3 years ago
Which layer of the atmosphere contains a substance that was created from a product of living things and that protects living thi
aev [14]
<h2>Answer : Option B) Ozone</h2><h3>Explanation :</h3>

The layer of atmosphere which contains a substance(Ozone) that was created from a product of living thins (oxygen) and protects living things (from harmful UV rays).

Oxygen in the atmosphere reacts in presence of UV rays and forms Ozone.

This ozone forms a protective layer around the earth and protects it from harmful UV rays.

8 0
3 years ago
Read 2 more answers
NEED HELP ASAP NOT DIFFICULT
Burka [1]
Answer- 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.

Given - Number of moles of Al(NO3)3 - 4 moles
Number of moles of NaCl - 9 moles
Find - Maximum amount of AlCl3 produced during the reaction.
Solution - The complete reaction is - Al(NO3)3 + 3NaCl --> 3NaNO3 + AlCl3
To find the maximum amount of AlCl3 produced during the reaction, we need to find the limiting reagent.
Mole ratio Al(NO3)3 - 4/1 - 4
Mole ratio NaCl - 9/3 - 3
Thus, NaCl is the limiting reagent in the reaction.
Now, 3 moles of NaCl produces 1 mole of AlCl3
9 moles of NaCl will produce - 1/3*9 - 3 moles.
Weight of AlCl3 - 3*133.34 - 400 grams
Thus, 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
5 0
3 years ago
Read 2 more answers
6 Apply fill in the genotypes and phenotypes of the
Nuetrik [128]
The scientific geographic name is usually a great election
6 0
3 years ago
2. Sitting on a bench top are several samples: lithium metal (d = 0.53 g/mL), gold (d = 19.3 g/mL), aluminum (d = 2.70 g/mL), an
Snezhnost [94]

Answer:

The sample of lithium occupies the largest volume.

Explanation:

Given the densities for the four elements, we have the expression d=\frac{m}{V} that shows the relationship between mass and Volume to express the density of an element.

For each element we have:

d_{lithium}=\frac{m_{lithium}}{V_{lithium}}=0.53g/mL

d_{gold}=\frac{m_{gold}}{V_{gold}}=19.3g/mL

d_{aluminum}=\frac{m_{aluminum}}{V_{aluminum}}=2.70g/mL

d_{lead}=\frac{m_{lead}}{V_{lead}}=11.3g/mL

The problem says that all the samples have the same mass, so:

m_{lithium}=m_{gold}=m_{aluminum}=m_{lead}=m

it means that m is a constant

Now, solving for the Volume in each element and with m as a constant, we have:

V_{lithium}=\frac{m}{d_{lithium}}

V_{lithium}=\frac{1}{0.53\frac{g}{mL}} *m

V_{lithium}=1.88\frac{mL}{g}*m

V_{gold}=\frac{m}{d_{gold}}

V_{gold}=\frac{1}{19.3\frac{g}{mL}} *m

V_{gold}=5.18*10^{-2}\frac{mL}{g}*m

V_{aluminum}=\frac{m}{d_{aluminum}}

V_{aluminum}=\frac{1}{2.70\frac{g}{mL}} *m

V_{aluminum}=3.70*10^{-1}\frac{mL}{g}*m

V_{lead}=\frac{m}{d_{lead}}

V_{lead}=\frac{1}{11.3\frac{g}{mL}} *m

V_{lead}=8.85*10^{-2}\frac{mL}{g}*m

If we assume m = 1g, we find that:

V_{lithium}=1.88mL

V_{gold}=5.18*10^{-2}mL

V_{aluminum}=3.70*10^{-1}mL

V_{lead}=8.85*10^{-2}mL

So we can see that the sample of lithium occupies the largest volume with 1.88mL

Note that m only can take positive values, so if you change the value of m, always will be the lithium which occupies the largest volume.

4 0
3 years ago
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