Which is the best term to use when describing the energy of motion?

Match each term with its definition. 1) reaction coordinate

Andreyy89

1) reaction coordinate is : b.the plot of the reaction progress as a function of time
2) ΔG⁰ : c.the free energy of the reaction
3) ΔG⁺+ : a.the higher point on the energy curve

3 0

2 years ago

Problem Page Combustion of hydrocarbons such as pentane ( C5 H12 ) produces carbon dioxide, a "greenhouse gas." Greenhouse gases

agasfer [191]

Auuuuuhhhhhhh auhhh Frog

4 0

3 years ago

Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of sodium sulfa

notka56 [123]

Answer:

27%

Explanation:

Hello,

The following information is missing, but I found it: "1.92 g of sodium sulfate is produced from the reaction of 4.9 g of sulfuric acid and 7.8 g of sodium hydroxide" so the undergoing chemical reaction is:

$2NaOH+H_2SO_4-->Na_2SO_4+2H_2O$

Now, to compute the percent yield, we must first establish the limiting reagent to subsequently determine the theoretical yield of sodium sulfate because the real (1.92g) is already given, thus, we consider the following procedure:

$n_{NaOH}=7.8gNaOH*\frac{1molNaOH}{40gNaOH}=0.2molNaOH\\n_{H_2SO_4}=4.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.050molH_2SO_4\\$

- The moles of sodium hydroxide that completely react with 0.05 moles of sulfuric acid are:

$0.2molNaOH*\frac{1molH_2SO_4}{2molNaOH}=0.098molH_2SO_4$

As this number is higher than the previously computed 0.05 moles of available sulfuric acid, one states that the sulfuric acid is the limiting reagent. Now, the theoretical grams of sodium sulfate are found via:

$0.05molH_2SO_4*\frac{1molNa_2SO_4}{1mol H_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4} =7.1gNa_2SO_4$

Finally, the percent yield turns out into:

$Y=\frac{1.92g}{7.1g} *100$

$Y=27.0$ %

Best regards.

6 0

2 years ago

When comparing the two chair conformations for a monosubstituted cyclohexane ring, which type of substituent shows the greatest

Aleksandr [31]

Answer:

See the explanation

Explanation:

In this case, we have to keep in mind that in the monosubstituted product we only have to replace 1 hydrogen with another group. In this case, we are going to use the methyl group $CH_3$ .

In the axial position, we have a more steric hindrance because we have two hydrogens near to the $CH_3$ group. If we have more steric hindrance the molecule would be more unstable. In the equatorial positions, we don't any interactions because the $CH_3$ group is pointing out. If we don't have any steric hindrance the molecule will be more stable, that's why the molecule will the equatorial position.

See figure 1

I hope it helps!

6 0

2 years ago

What element’s valence electrons are located in 5s25p4?

IrinaK [193]

The element is TELLURIUM. Tellurium has atomic number 52 with the electronic configuration of [Kr] 4d10 5s2 5p4. The element belongs to group 16 elements on the periodic table. It exhibits various oxidation states: +2, +4, +6, and -2. +4 is its most common oxidation state.

5 0

2 years ago