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blondinia [14]
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)
Chemistry
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 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

6 0
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

Brackett series - Observed in the infrared region

Paschen series - Observed in the infrared region

Lyman series - Observe in the Ultraviolet region.

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
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