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2 years ago

Determine the number of particles in 0.332 g of NaCl (molar mass NaCl=58.44 g/mol).

Chemistry

1 answer:

ira [324]2 years ago

8 0

Answer:

3.42×10²¹ of particles

Explanation:

To determine the number of particles we should know, that:

1 mol of anything contains 6.02×10²³ particles

Therefore we have to find out, how many moles corresponds to 0.332 g of NaCl.

The conversion is: 0.332 g / 58.44 g/mol = 0.00568 moles

The rule of three to determine the number of particles is:

1 mol of NaCl has 6.02×10²³ particles

0.00568 moles must have (0.00568 . 6.02×10²³ particles )/1 = 3.42×10²¹

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Convert 25 liters per hour to milliliters per second.

Dominik [7]

Answer:

6.94444

Explanation:

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2 years ago

Read 2 more answers

Modify the given fatty acid so that it represents the 18‑carbon fatty acid designated 18:2(Δ9,12). Draw any double bonds in the

Mekhanik [1.2K]

Answer:

See explanation

Explanation:

In this case, we have to remember the meaning of the nomenclature "18:2Δ9,12". Where 18 is the number of carbon atoms, 2 is the number of double bonds, and the numbers successive to Δ "delta" the position of the double bonds starting to count from the carboxylic -COOH end of the molecule.

In other words, the main functional group is a carboxylic acid. We have a total of 18 carbons. Additionally, we have 2 double bonds. On carbons 9 and 12.

Lets see figure 1

I hope it helps!

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2 years ago

Why are the average atomic masses of some of the elements in parentheses?

Pie

Because its only representing the most stable isotope of that element but its considered the atomic mass and not average atomic mass

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3 years ago

Protozoan infection is not likely to be transmitted by:

anygoal [31]

Hi!

The answer is breathing.

Protozoans are organisms that can multiply themselves in humans, and serious infections can develop from one organism. The most frequently transmission methods are contaminated food or fecal contact. Dog bites inoculate bacteria into deep tissue, and they also serve as a transmission method.

Breathing is not considered to be likely to transmit protozoan infection because protozoans rarely develop in the respiratory tract.

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2 years ago

Electrophilic addition of hbr to 3-methyl-2-hexene creates an asymmetric center at c-2. what is the product of this reaction?

AleksandrR [38]

This reaction would give rise to two products.

2-bromo-3-methylhexane, and
3-bromo-3-methylhexane.

However, 2-bromo-3-methylhexane would be more common than 3-bromo-3-methylhexane among the products.

The hydrogen atom in a hydrogen bromide molecule carries a partial positive charge. It is attracted to the double bond region with a high electron density. The hydrogen-bromine bond breaks when HBr gets too close to a double bond to produces a proton $\text{H}^{+}$ and a bromide ion $\text{Br}^{-}$ .

The proton would attack the double bond to produce a carbocation. It could attach itself to either the second or the third carbon atom.

$\phantom{\text{H}_3\text{C}-\text{CH}-\;}+\\\text{H}_3\text{C}-\text{CH}-\text{CH}-\text{CH}_2-\text{CH}_2-\text{CH}_3\\\phantom{\text{H}_3\text{C}-\;\;}|\phantom{\text{CH}-}\;}|\\\phantom{\text{H}_3\text{C}-\;} \text{H}\phantom{\text{CH}}\;\;}\text{CH}_3$
$\phantom{\text{H}_3\text{C}-\text{CH}-}\;\;\text{H}\\\phantom{\text{H}_3\text{C}-\;}+\phantom{\text{H}-\;}|\\\text{H}_3\text{C}-\text{CH}-\text{CH}-\text{CH}_2-\text{CH}_2-\text{CH}_3\\\phantom{\text{H}_3\text{C}-\;\;\;}\phantom{\text{CH}-}\;}|\\\phantom{\text{H}_3\text{C}-\;} \phantom{\text{H}}\phantom{\text{CH}}\;\;}\text{CH}_3$

Carbocations are unstable and might decompose over time. The first carbocation is more stable than the second for having three alkyl groups- i.e., straight carbon chains- attached to the center of the positive charge. Alkyl groups have stabilizing positive induction effect on positively-charged carbon. The second carbocation has only two, and is therefore not as stable. The first carbocation thus has the greatest chance to react with a bromide ion to produce a stable halocarbon.

Bromide ions are negatively charged. They attach themselves to carbocations at the center of positive charge. Adding a bromide ion to the first carbocation would produce 3-bromo-3-methylhexane whereas adding to the second produces 2-bromo-3-methylhexane.

The most likely product of this reaction is therefore 3-bromo-3-methylhexane.

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2 years ago