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Nataliya [291]
2 years ago
5

What’s the IUPAC name

Chemistry
1 answer:
hichkok12 [17]2 years ago
5 0

Answer:

Methyl pentanoate.

Explanation:

Hello there!

In this case, according to the given information, we can see the correct structure will be:

                             O

                              ||

CH3CH2CH2CH2COCH3

Which matches with the structure of an ester due to the -COO- functional group. In such a case, the first part of the name is in function of the right side of the ester, in this case, methyl, followed by the left side, pentanoate, as it has five carbon atoms and is an ester (similar to an inorganic salt, but organic) and therefore, the name will be methyl pentanoate.

Regards!

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How much energy is evolved during the formation of 197 g of Fe, according to the reaction below?
hoa [83]

Answer:

ΔH for formation of 197g Fe⁰ = 1.503 x 10³ Kj => Answer choice 'B'

Explanation:

Given Fe₂O₃(s) + 2Al⁰(s) => Al₂O₃(s) + 2Fe⁰(s) + 852Kj

197g Fe⁰ = (197g/55.85g/mol) = 3.527 mol Fe⁰(s)

From balanced standard equation 2 moles Fe⁰(s) => 852Kj, then ...

3.527 mole yield (a higher mole value) => (3.527/2) x 852Kj = 1,503Kj (a higher enthalpy value).

______

NOTE => If 2 moles Fe gives 852Kj (exo) as specified in equation, then a <u>higher energy value</u> would result if the moles of Fe⁰(s) is <u>higher than 2 moles</u>. The ratio of 3.638/2 will increase the listed equation heat value to a larger number because 197g Fe⁰(s) contains more than 2 moles of Fe⁰(s) => 3.527 mole Fe(s) in 197g.  Had the problem asked for the heat loss from <u>less than two moles Fe⁰(s)</u> - say 100g Fe⁰(s) (=1.79mole Fe⁰(s)) - then one would use the fractional ratio (1.79/2) to reduce the enthalpy value less than 852Kj.  

8 0
3 years ago
What is the limiting reactant when 19.9 g CuO react with 2.02 g H2?
Harlamova29_29 [7]

Answer:

Explanation:

use the equation

moles = mass/mr

=19.9/79.5

=0.250moles of CuO

then do the same for

H = 2.02/1

=2.02

so CuO is the limiting reagent because there is less amount of it.

Hope this helps  :)

4 0
2 years ago
A standard backpack is approximately 30cm x 30cm x 40cm. Suppose you find a hoard of pure gold while treasure hunting in the wil
Blizzard [7]

Explanation:

The dimensions of a standard backpack is 30cm x 30cm x 40cm

The mass of an average student is 70 kg

We know that, the density of gold is 19.3 g/cm³.

Let m be the mass of the backpack. So,

\text{density}=\dfrac{\text{mass}}{\text{volume}}\\\\m=d\times V\\\\m=19.3\ g/cm^3\times (30\times 30\times 40)\ cm^3\\\\m=694800\ g\\\\\text{or}\\\\m=694.8\ kg\approx 700\ kg

An average student has a mass of 70 kg. If we compare the mass of student and mass of backpack, we find that the backpack is 10 times of the mass of the student.

8 0
3 years ago
Complete each equation with a number that is true.<br>5 * _ =15
natali 33 [55]

The answer is 3. As 5 * 3 = 15.

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5 0
3 years ago
Read 2 more answers
The equilibrium constant for the dissolution of silver chloride (AgCl(s) Ag+(aq) + Cl–(aq)) has a value of 1.79 × 10–10. Which s
ryzh [129]

"Silver chloride is essentially insoluble in water" this statement is true for the equilibrium constant for the dissolution of silver chloride.

Option: b

<u>Explanation</u>:

As silver chloride is essentially insoluble in water but also show sparing solubility, its reason is explained through Fajan's rule. Therefore when AgCl added in water, equilibrium take place between undissolved and dissolved ions. While solubility product constant \left(\boldsymbol{K}_{s p}\right) for silver chloride is determined by equilibrium concentrations of dissolved ions. But solubility may vary also at different temperatures.  Complete solubility is possible in ammonia solution as it form stable complex as water is not good ligand for Ag+.  

To calculate \left(\boldsymbol{K}_{s p}\right) firstly molarity of ions are needed to be found with formula: \text { Molarity of ions }=\frac{\text { number of moles of solute }}{\text { Volume of solution in litres }}

Then at equilibrium cations and anions concentration is considered same hence:

\left[\mathbf{A} \mathbf{g}^{+}\right]=[\mathbf{C} \mathbf{I}]=\text { molarity of ions }

Hence from above data \left(\boldsymbol{K}_{s p}\right) can be calculated by: \left(\boldsymbol{K}_{s p}\right) = \left[\mathbf{A} \mathbf{g}^{+}\right] \cdot[\mathbf{C} \mathbf{I}]

6 0
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