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2 years ago

14

Three students are labeling organisms in their diorama. Zola wants to label the algae and mushrooms as photosynthetic. Saffi wan

ts to label plants and algae as photosynthetic. Ahmed wants to label plants and all bacteria as photosynthetic. Who is correct?
-Saffi and Ahmed
-Zola only
-Saffi only
-Zola and Ahmed

1 answer:

Nookie1986 [14]2 years ago

3 0

Answer:

Saffi only

Explanation:

I just took the test and that was the correct answer :)

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Lead will float in water.

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Excess magnesium reacts with 165.0 grams of hydrochloric acid in a single displacement reaction.

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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.

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how many molecules of sulfuric acid are in a spherical raindrop of diameter 6.0 mm if the acid rain has a concentration of 4.4 *

Vitek1552 [10]

Answer:

The number of moles =

$Moles=4.97\times 10^{-8}$

The number of molecules =

$Molecules = 2.99\times 10^{16}$

Explanation:

Volume of the sphere is given by :

$V=\frac{4}{3}\pi r^{3}$

here, r = radius of the sphere

$radius=\frac{diameter}{2}$

$radius=\frac{6.0}{2}$

Radius = 3 mm

r = 3 mm

1 mm = 0.01 dm (1 millimeter = 0.001 decimeter)

3 mm = 3 x 0.01 dm = 0.03 dm

r = 0.03 dm

<em>("volume must be in dm^3 , this is the reason radius is changed into dm"</em>

<em>"this is done because 1 dm^3 = 1 liter and concentration is always measured in liters")</em>

$V=\frac{4}{3}\pi 0.03^{3}$

$V=\frac{4}{3}\pi 2.7\times 10^{-5}$

$V=1.13\times 10^{-4}dm^{3}$

$V=1.13\times 10^{-4}L$ (1 L = 1 dm3)

Now, concentration "C"=

$C=4.4\times 10^{-4}moles/liter$

The concentration is given by the formula :

$C=\frac{moles}{Volume(L)}$

This is also written as,

$Moles = C\times Volume$

$Moles=1.13\times 10^{-4}\times 4.4\times 10^{-4}$

$Moles=4.97\times 10^{-8}$ moles

One mole of the substance contain "Na"(= Avogadro number of molecules)

So, "n" mole of substance contain =( n x Na )

$N_{a}=6.022\times 10^{23}$

Molecules =

$Molecule=4.97\times 10^{-8}\times 6.022\times 10^{23}$

$Molecules = 2.99\times 10^{16}$ molecules

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