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Naddik [55]
4 years ago
8

What is the molarity of a solution that has 2.50 moles of NaOH dissolved in 0.500 L of solution?

Chemistry
2 answers:
Yuliya22 [10]4 years ago
8 0

Answer:

3.0

Explanation:

const2013 [10]4 years ago
5 0

Answer: 5 is the molarity

Explanation:

The molarity formula is moles over liters and that in your case is 2.50 moles divided by .500 L which results in 5 which is your answear hope this helped god bless

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What is the name of this hydrocarbon? 2-dimethylyne 2-methylbutane 4-methylbutene 4,3-methylbutyne 4,4-methylbutane
Alisiya [41]
You have shown no structure, still we can work out for the answer :).

First Option: <span>2-dimethylyne: This name is incorrect, because it does not contain any parent chain name. 

Second Option: </span><span>2-methylbutane: I have drawn its structure below, compare it with your structure.

Third Option: </span>4-methylbutene:<span> I have drawn the structure for this name. But this name is also against IUPAC rules, and its correct name is 1-Pentene.

Fourth Option: </span><span>4,3-methylbutyne: Again incorrect name, 4,3 are two positions, but here only one substituent is given (methyl). I have drawn structure of 3,4-Dimethylbutene, this name is incorrect, and the correct name for this compound is 3-Methyl-1-pentene.

Fifth Option: </span><span>4,4-methylbutane: Again incorrect name, 4,4 means at 4 position two substituents, but in name only methyl is given. Anyhow lets make it two and draw a structure, for 4,4-Dimethylbutane. Ooops!! This name is also incorrect, and the correct name for this compound is 2-Methylpentane.

Result:
           Among all options only, option B (2-Methylbutane) is correctly named, and this will be the correct name for the structure given to you. </span>

8 0
3 years ago
Read 2 more answers
Select the statements below that are false and think about why they are not true.
Anna007 [38]

Answer:

a) and d) are false.

Explanation:

a) The second law of thermodynamics states that t<u>he entropy of the universe increases in a spon taneous  process and remains unchanged in an equilibrium process.</u>

For a spontaneous process:

ΔSuniv = ΔSsys + ΔSsurr > 0

For a spontaneous process, the second law says that ΔSuniv must be greater than  zero, but it does not place a restriction on either ΔSsys or ΔSsurr. Thus, it is possible  for either ΔSsys or ΔSsurr to be negative, as long as the sum of these two quantities  is greater than zero.

d) <u>A  reaction that does occur under the given set of conditions is called a</u> spontaneous  reaction. We observe spontaneous physical and chemical processes every day,  including many of the following examples:

• A waterfall runs downhill, but never up, spontaneously.

• A lump of sugar spontaneously dissolves in a cup of coffee, but dissolved sugar  does not spontaneously reappear in its original form.

• Heat flows from a hotter object to a colder one, but the reverse never happens  spontaneously.

5 0
3 years ago
Which term identifies a type of organic reaction?
lyudmila [28]
<span>The answer is (3) esterification. Note that deposition, distillation and sublimation are not chemical changes, they are physical changes, so you can solve this question by elimination if you know that. Regarding sterification it is a well known chemical reaction that consists on the formation of esters from carboxilic acids.</span>
3 0
4 years ago
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The oxidizing agent our bodies use to obtain energy from food is oxygen (from the air). If you breathe 17 times a minute (at res
nata0808 [166]

Answer:

<em>The volume of the air breath per day is =12240 L day ^ -1 and he volume of oxygen breathe each day is= 2570.4 L</em>

Explanation:

<em>Complete question: </em><em>The oxidizing agent our bodies use to obtain energy from food is oxygen (from the air). If you breathe 17 times a minute (at rest), taking in and exhaling 0.50 L of air with each breath, what volume of air do you breathe each day? Air is 21% oxygen by volume. what volume of oxygen do you breathe each day?</em>

<em>Solution</em>

<em>Given that,</em>

<em>The volume of the air with each breath = 0.50L</em>

<em>The frequency of breath is = 17 times per minute</em>

<em>the next step is to calculate the volume of the air breath per day</em>

<em>The volume of the breathed air = volume of air breathed/ breath * 17 breaths/min * 60 minutes / per hour * 24 hours/ per day</em>

<em>= 0.50L/ breath * 17 breath /min * 60 minutes / per hour * 24 hours/ per day</em>

<em>= 12240 L day ^ -1</em>

<em>The next step is to calculate the oxygen breathed per day:</em>

<em>The volume of oxygen breathed = volume of air breathed * percent oxygen in air/100</em>

<em>= 12240 L day ^ -1 * 21 /100</em>

<em>= 2570.4 L day ^ -1 </em>

<em>Therefore for each day, the volume of oxygen breathe is 2570.4 L</em>

5 0
4 years ago
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The molal freezing point constant for copper is 23 °C/m. If pure copper melts at 1083°C, what will be the melting point of a bra
nikitadnepr [17]

<u>Answer:</u> The freezing point of brass is 1028.57°C

<u>Explanation:</u>

We are given:

13.4 mass percent of zinc in brass

This means that 13.4 grams of zinc is present in 100 g of brass

Mass of copper = [100 - 13.4] g = 86.6 g

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Freezing point of pure copper = 1083°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point elevation constant = 23°C/m

m_{solute} = Given mass of solute (zinc) = 13.4 g

M_{solute} = Molar mass of solute (zinc) = 65.38  g/mol

W_{solvent} = Mass of solvent (copper) = 86.6 g

Putting values in above equation, we get:

1083-\text{Freezing point of solution}=1\times 23^oC/m\times \frac{13.4\times 1000}{65.38g/mol\times 86.6}\\\\\text{Freezing point of solution}=1028.57^oC

Hence, the freezing point of brass is 1028.57°C

8 0
3 years ago
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