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2 years ago

3.00 moles of NO2 have a mass of

1 answer:

6 0

When solving for the mass of a compound when you’re given the number of moles present, you need to know the molar mass (how many grams there are in a mole of that compound).

In this case, we can make the following equation:

3.00(14.01 + 2(16.00))

In Nitrogen, the molar mass is 14.01 grams per mole, and Oxygen is 16.01 grams per mole.

However, because there are 2 oxygen atoms present per molecule, you must multiply it by two in order to solve for the molar mass.

Anyway:
3.00(14.01 + 2(16.00))

Simplify:
3.00(14.01 + 32.00)
3.00(46.01)

Multiply:
3.00(46.01) = 138.03

3.00 moles of NO2 has a mass of 138.03 grams.
-T.B.

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Design a synthesis of 2-ethyl-2-hexenoic acid from alcohols of four carbons or fewer.

Studentka2010 [4]

Answer:

Enolate Alkylation

The anions from ketones, called enolates, can act as a nucleophile in SN2 type reactions. Overall an α hydrogen is replaced with an alkyl group and a new carbon-carbon bond is formed. These alkylations are affected by the same limitations as SN2 reactions previously discussed. A good leaving group, chloride, bromide, iodide or tosylate, should be used. Also, secondary and tertiary leaving groups should not be used because of poor reactivity and possible competition with elimination reactions. Lastly, it is important to use a strong base, such as LDA or sodium amide, for preparing the enolate from the ketone. Using a weaker base such as hydroxide or an alkoxide leaves the possibility of multiple alkylations occurring, and competing SN2 reactions with the base.

Explanation:

Design is illustrated in the attached document

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2 years ago

Franklin was performing an experiment by combining hydrochloric acid and sodium hydroxide. He measured the mass of his reactant

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Answer:

Explanation:

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1 year ago

Read 2 more answers

Identify the oxidizing and reducing agents in the following: 2H+(aq) + H2O2(aq) + 2Fe2+(aq) → 2Fe3+(aq) + 2H2O(l)

schepotkina [342]

Answer : The oxidizing and reducing agents are, $H_2O_2$ and $Fe^{2+}$ .

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Reducing agent : It is defined as the agent which helps the other substance to reduce and itself gets oxidized. Thus, it will undergo oxidation reaction.

Oxidizing agent : It is defined as the agent which helps the other substance to oxidize and itself gets reduced. Thus, it will undergo reduction reaction.

The given redox reaction is:

$2H^+(aq)+H_2O_2(aq)+2Fe^{2+}(aq)\rightarrow 2Fe^{3+}(aq)+2H_2O(l)$

The half oxidation-reduction reactions are:

Oxidation reaction : $Fe^{2+}\rightarrow Fe^{3+}+1e^-$

Reduction reaction : $O^-+1e^-\rightarrow O^{2-}$

The oxidation state of oxygen in $H_2O_2$ and $H_2O$ is, (-1) and (-2) respectively.

In this reaction, 'Fe' is oxidized from oxidation (+2) to (+3) and 'O' is reduced from oxidation state (-1) to (-2). Hence, ' $Fe^{2+}$ ' act as a reducing agent and ' $H_2O_2$ ' act as a oxidizing agent.

Thus, the oxidizing and reducing agents are, $H_2O_2$ and $Fe^{2+}$ .

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1 year ago

How many liters of h2 gas, collected over water at an atmospheric pressure of 752 mm hg and a temperature of 21.0°c, can be made

natta225 [31]

Answer:
The balanced equation tells us that 1 mole of Zn will produce 1 mole of H2.
1.566 g Zn x (1 mole Zn / 65.38 g Zn) = 0.02395 moles Zn
0.02395 moles Zn x (1 mole H2 / 1 mole Zn) = 0.02395 moles H2 produced
Now use the ideal gas law to find the volume V.
P = 733 mmHg x (1 atm / 760 atm) = 0.964 atm
T = 21 C + 273 = 294 K
PV = nRT
V = nRT/ P = (0.02395 moles H2)(0.0821 L atm / K mole)(294 K) / (0.964 atm) = 0.600 L

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2 years ago

Calculate the percent activity of the radioactive isotope strontium-89 remaining after 5 half-lives.

maria [59]

The answer to this question would be: 3.125%

Half-life is the time needed for a radioactive molecule to decay half of its mass. In this case, the strontium-89 is already gone past 5 half lives. Then, the percentage of the mass left after 5 half-lives should be:
100%*(1/2^5)= 100%/32=3..125%

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1 year ago