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3 years ago

15

Which of the following changes has a decrease in entropy? 2LiClO3 (s) yields 3O2 (g) + 2LiCl (s) N2O4 (g) yields 2NO2 (g) C6H6 (

l) yields C6H6 (s) C2H5OH (l) + 3O2 (g) yields 2CO2 (g) + 3 H2O (g)

2 answers:

suter [353]3 years ago

8 0

I would have to go with n204

natali 33 [55]3 years ago

4 0

The answer is C6H6(I) yields C6H6(S).

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About _% of high school seniors report misusing prescription

mihalych1998 [28]

About 5% of high school seniors reportmisusing prescription

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4 years ago

The actual number of pennies in a jar is 218. You miscounted and came up with 215 pennies. What is your percent error?

Katyanochek1 [597]

Answer:

3%

Explanation:

Substract the actual error from the final and multiply by 100

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3 years ago

The four sets of lines in the hydrogen emission spectrum are known as Balmer, Brackett, Paschen, and Lyman. For each series, ass

arlik [135]

Answer:

Hydrogen spectrum

Explanation:

Balmer series - Observed in the visible region

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8 0

3 years ago

Explain how sodium and calcium react with water. explain with equations

34kurt

Reaction of sodium with water

Sodium metal reacts rapidly with water to form a colourless solution of sodium hydroxide (NaOH) and hydrogen gas (H2). The resulting solution is basic because of the dissolved hydroxide. The reaction is exothermic. During the reaction, the sodium metal may well become so hot that it catches fire and burns with a characteristic orange colour. The reaction is slower than that of potassium (immediately below sodium in the periodic table), but faster than that of lithium (immediately above sodium in the periodic table).

2Na(s) + 2H2O → 2NaOH(aq) + H2(g)

4 0

3 years ago

If you assume this reaction is driven to completion because of the large excess of one ion, what is the concentration of [Fe(SCN

viktelen [127]

Answer : The concentration of $[Fe(SCN)]^{2+}$ is, $4.32\times 10^{-4}M$

Explanation :

When we assume this reaction is driven to completion because of the large excess of one ion then we are assuming limiting reagent is $SCN^-$ and $Fe^{3+}$ is excess reagent.

First we have to calculate the moles of KSCN.

$\text{Moles of }KSCN=\text{Concentration of }KSCN\times \text{Volume of solution}$

$\text{Moles of }KSCN=0.00180M\times 0.006L=1.08\times 10^{-5}mol$

Moles of KSCN = Moles of $K^+$ = Moles of $SCN^-$ = $1.08\times 10^{-5}mol$

Now we have to calculate the concentration of $[Fe(SCN)]^{2+}$

$\text{Concentration of }[Fe(SCN)]^{2+}=\frac{\text{Moles of }[Fe(SCN)]^{2+}}{\text{Volume of solution}}$

Total volume of solution = (6.00 + 5.00 + 14.00) = 25.00 mL = 0.025 L

$\text{Concentration of }[Fe(SCN)]^{2+}=\frac{1.08\times 10^{-5}mol}{0.025L}=4.32\times 10^{-4}M$

Thus, the concentration of $[Fe(SCN)]^{2+}$ is, $4.32\times 10^{-4}M$

7 0

3 years ago