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TEA [102]
3 years ago
15

There are two beakers that contain the same liquid substance at the same temperature. The larger beaker is 1,000ml and the small

er beaker is 100ml. How does the thermal energy of the liquid in the larger beaker compare to the thermal energy of the liquid in the smaller beaker?
A. The liquid in the larger beaker has the same amount of thermal energy as the liquid in thesmaller beaker.
B. The exact volume of liquid in each beaker must be known to compare the thermal energyof the liquids.
C. The liquid in the larger beaker has less thermal energy than the liquid in the smaller beaker.
D. The liquid in the larger beaker has more thermal energy than the liquid in the smaller beaker.
Chemistry
1 answer:
azamat3 years ago
6 0

Answer:

Answer choice B

Explanation:

Since you do not know the volume of the liquid in each beaker, the one in the smaller beaker could have more substance and therefore more thermal energy. If they had the same amount of substance, then the more voluminous one would radiate faster. However, since you do not know this, there is no way to tell. PM me if you have more questions. Hope this helps!

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When Wolverine’s 10-pound adamantium claws are dissolved in 100 mL of 10 M nitric acid, 10.7 grams of adamantium nitrate are rec
Mazyrski [523]

The percent yield is 71.3 %.

Explanation:

Percent yield is the measure to analyze the success percentage of any experiment .The percent yield of any experiment can be obtained by the ratio of actual or experimental value to expected or theoretical value multiplied with 100.

Percent Yield = \frac{Experimental Outcome}{Theoretical Outcome} *100

So, in the present problem, we have obtained 10.7 g of adamantium nitrate from Wolverine's 10 pound claws. So the actual value or the experimental value is the amount of adamantium nitrate obtained from Wolverine's claws.

Thus, the experimental outcome is 10.7 g. While we had expected to recover 15 g of adamantium nitrate. So the theoretical outcome is 15 g.

Percent yield = \frac{10.7}{15} * 100 = 71.3 %

Thus, the percent yield is 71.3 %.

7 0
3 years ago
The following reactions have the indicated equilibrium constants at a particular temperature: N2(g) + O2(g) ⇌ 2NO(g) Kc = 4.3 ×
Anuta_ua [19.1K]

Answer:

Kc=~1.49x10^3^4}

Explanation:

We have the reactions:

A: N_2_(_g_) + O_2_(_g_)  2NO_(_g_)~~~~~~Kc = 4.3x10^-^2^5

B: 2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~~~Kc = 6.4x10^9

Our <u>target reaction</u> is:

4NO_(_g_)  N_2_(_g_) + 2NO_2_(_g_)

We have NO_(_g_) as a reactive in the target reaction and  NO_(_g_) is present in A reaction but in the products side. So we have to<u> flip reaction A</u>.

A: 2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}

Then if we add reactions A and B we can obtain the target reaction, so:

A: 2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}

B: 2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~Kc=6.4x10^9

For the <u>final Kc value</u>, we have to keep in mind that when we have to <u>add chemical reactions</u> the total Kc value would be the <u>multiplication</u> of the Kc values in the previous reactions.

4NO_(_g_)  N_2_(_g_) + 2NO_2_(_g_)~~~Kc=\frac{6.4x10^9}{4.3x10^-^2^5}

Kc=~1.49x10^+^3^4}

3 0
3 years ago
Mintu needs to determine the missing ecosystem in the set.
alexandr1967 [171]
The answer is Wetland and Stream
4 0
3 years ago
Read 2 more answers
If the [H+] = 0.01 M, what is the pH of the solution, and is the solution a strong acid, weak acid, strong base, or weak base?
lana66690 [7]

Answer:

2, strong acid

Explanation:

Data obtained from the question. This includes:

[H+] = 0.01 M

pH =?

pH of a solution can be obtained by using the following formula:

pH = –Log [H+]

pH = –Log 0.01

pH = 2

The pH of a solution ranging between 0 and 6 is declared to be an acid solution. The smaller the pH value, the stronger the acid.

Since the pH of the above solution is 2, it means the solution is a strong acid.

5 0
3 years ago
Predict the splitting pattern for each of the labeled hydrogens in the following molecules. Assume that all coupling constants a
Ghella [55]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a) Splitting pattern for Ha= 2+1 , Triplet

For Proton Hb and Hc both are equivalent to each other. non equivalent protons n= 3

Splitting pattern for Hb and Hc= 3+1 , Quartet

For Proton Hd, number of non equivalent protons n= 0

Splitting pattern for Hd= 0+1 , Singlet

b) Splitting pattern for Ha= 2+1 , Triplet

For Proton Hb and Hc both are equivalent to each other. non equivalent protons n= 3

Splitting pattern for Hb and Hc= 3+1 , Quartet

For Proton Hd, number of non equivalent protons n= 0

Splitting pattern for Hd= 0+1 , Singlet

c) The IUPAC name is Butan-2-ol

Explanation:

Considering the first question the rule used for prediction of splitting pattern is n+1 (Pascal's Triangle rule), where n is number of H atom on the adjacent carbon which are non equivalent.

According to that for molecule 1 as shown on the second uploaded image

For Proton Ha, number of non equivalent protons n= 2

Splitting pattern for Ha= 2+1 , Triplet

For Proton Hb and Hc both are equivalent to each other. non equivalent protons n= 3

Splitting pattern for Hb and Hc= 3+1 , Quartet

For Proton Hd, number of non equivalent protons n= 0

Splitting pattern for Hd= 0+1 , Singlet

Considering the second question for Molecule 2 as shown on the third uploaded image  

For Proton Ha, number of non equivalent protons n= 1

Splitting pattern for Ha= 1+1=2 , Doublet

For Proton Hb, number of non equivalent protons n= 3

Splitting pattern for Hb= 3+1=4 , Quartet

For Proton Hc, number of non equivalent protons n= 3

Splitting pattern for Hc= 3+1=4 , Quartet

For Proton Hd, number of non equivalent protons n= 1

Splitting pattern for Ha= 1+1=2 , Doublet

Considering the third question

The name of the given molecule  is gotten according to longest carbon chain  = 4 (Prefix 'Butan')

Functional group = -OH (Suffix 'ol') at C-2

The IUPAC name is Butan-2-ol

3 0
3 years ago
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