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3 years ago

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Chlorophyll a is one of the green pigments found in plants. Chlorophyll a has the molecular formula C55H72MgN4O5. How many atoms

are in this molecule?

1 answer:

algol133 years ago

8 0

The number of atoms in a molecule can be calculated by adding all of the number of elements from the chemical formula. For chlorophyll, the chemical formula would be <span>C55H72MgN4O5 adding all the elements we have 137 atoms. Hope this answers the question.</span>

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What is the concentration (in M) of a sample of the unknown dye with an absorbance of 0.25 at 542 nm?

Sladkaya [172]

Answer:

see explanation below

Explanation:

Question is incomplete, so in picture 1, you have a sample of this question with the missing data.

Now, in general terms, the absorbance of a substance can be calculated using the beer's law which is the following:

A = εlc

Where:

ε: molar absortivity

l: distance of the light in solution

c: concentration of solution

However, in this case, we have a plot line and a equation for this plot, so all we have to do is replace the given data into the equation and solve for x, which is the concentration.

the equation according to the plot is:

A = 15200c - 0.018

So solving for C for an absorbance of 0.25 is:

0.25 = 15200c - 0.018

0.25 + 0.018 = 15200c

0.268 = 15200c

c = 0.268/15200

c = 1.76x10⁻⁵ M

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3 years ago

Can anyone check my work and see if it is correct? If not, may someone help me?

shusha [124]

It looks all correct to me, great job!

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3 years ago

What are the monomers of bakelite

topjm [15]

Answer:

Bakelite is a polymer made up of the monomers phenol and formaldehyde. This phenol-formaldehyde resin is a thermosetting polymer.

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Which of these oxides will likely form a colored solution when dissolved in water?

Alik [6]

Mn₂O

Explanation:

The oxide that will most likely form colored solutions is Mn₂O.

This is because most transition metals form colored compounds. Manganese is a transition metal belonging to the d-block on the periodic table.

Other examples of transition metals are scandium, titanium, iron, copper, cobalt, nickel, zinc
They belong to the d-block on the periodic table.
They have variable oxidation states.
Most of their solutions are always colored.

Learn more:

Periodic table brainly.com/question/8543126

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3 years ago

20cm of 0.09M solution of H2SO4. requires 30cm of NaOH for complete neutralization. Calculate the

kirill115 [55]

Answer:

Choice A: approximately $0.12\; \rm M$ .

Explanation:

Note that the unit of concentration, $\rm M$ , typically refers to moles per liter (that is: $1\; \rm M = 1\; \rm mol\cdot L^{-1}$ .)

On the other hand, the volume of the two solutions in this question are apparently given in $\rm cm^3$ , which is the same as $\rm mL$ (that is: $1\; \rm cm^{3} = 1\; \rm mL$ .) Convert the unit of volume to liters:

$V(\mathrm{H_2SO_4}) = 20\; \rm cm^{3} = 20 \times 10^{-3}\; \rm L = 0.02\; \rm L$ .
$V(\mathrm{NaOH}) = 30\; \rm cm^{3} = 30 \times 10^{-3}\; \rm L = 0.03\; \rm L$ .

Calculate the number of moles of $\rm H_2SO_4$ formula units in that $0.02\; \rm L$ of the $0.09\; \rm M$ solution:

$\begin{aligned}n(\mathrm{H_2SO_4}) &= c(\mathrm{H_2SO_4}) \cdot V(\mathrm{H_2SO_4})\\ &= 0.02 \; \rm L \times 0.09 \; \rm mol\cdot L^{-1} = 0.0018\; \rm mol \end{aligned}$ .

Note that $\rm H_2SO_4$ (sulfuric acid) is a diprotic acid. When one mole of $\rm H_2SO_4$ completely dissolves in water, two moles of $\rm H^{+}$ ions will be released.

On the other hand, $\rm NaOH$ (sodium hydroxide) is a monoprotic base. When one mole of $\rm NaOH$ formula units completely dissolve in water, only one mole of $\rm OH^{-}$ ions will be released.

$\rm H^{+}$ ions and $\rm OH^{-}$ ions neutralize each other at a one-to-one ratio. Therefore, when one mole of the diprotic acid $\rm H_2SO_4$ dissolves in water completely, it will take two moles of $\rm OH^{-}$ to neutralize that two moles of $\rm H^{+}$ produced. On the other hand, two moles formula units of the monoprotic base $\rm NaOH$ will be required to produce that two moles of $\rm OH^{-}$ . Therefore, $\rm NaOH$ and $\rm H_2SO_4$ formula units would neutralize each other at a two-to-one ratio.

$\rm H_2SO_4 + 2\; NaOH \to Na_2SO_4 + 2\; H_2O$ .

$\displaystyle \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})} = \frac{2}{1} = 2$ .

Previous calculations show that $0.0018\; \rm mol$ of $\rm H_2SO_4$ was produced. Calculate the number of moles of $\rm NaOH$ formula units required to neutralize that

$\begin{aligned}n(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})}\cdot n(\mathrm{H_2SO_4}) \\&= 2 \times 0.0018\; \rm mol = 0.0036\; \rm mol\end{aligned}$ .

Calculate the concentration of a $0.03\; \rm L$ solution that contains exactly $0.0036\; \rm mol$ of $\rm NaOH$ formula units:

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} = \frac{0.0036\; \rm mol}{0.03\; \rm L} = 0.12\; \rm mol \cdot L^{-1}\end{aligned}$ .

3 0

3 years ago