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uysha [10]
3 years ago
8

Two glasses labeled A and B contain equal amounts of water at different temperatures. Kim put an antacid tablet into each of the

two glasses. The table shows the time taken by the tablet to dissolve completely in the two glasses.
Antacid Experiment
Glass --> Time
A --> 40 seconds
B --> 10 seconds

Which of the following statements is correct?

A. The water in Glass A is warmer than the water in Glass B; therefore, the water particles in Glass A are stationary.

B. The water in Glass A is cooler than the water in Glass B; therefore, the water particles in Glass A are stationary.

C. The water in Glass A is warmer than the water in Glass B; therefore, the particles in Glass A move faster.

D. The water in Glass A is cooler than the water in Glass B; therefore, the particles in Glass A move slower.
Chemistry
2 answers:
skelet666 [1.2K]3 years ago
6 0

Answer:

Option D. The water in Glass A is cooler than the water in Glass B; therefore, the particles in Glass A move slower.

Explanation:

Solubilities of solutes are enhanced when the temperature is increased.

From the experiment conducted,

It is evident that glass B temperature is higher than glass A temperature, because the solute dissolves faster in glass B than in glass A . This implies that glass A is cooler than glass B, hence the particles in A will move slower than that in B.

valentina_108 [34]3 years ago
5 0

Answer:

D. The water in Glass A is cooler than the water in Glass B; therefore, the particles in Glass A move slower

Explanation:

In science, temperature is defined as a measure of the average kinetic energy of the molecules of a substance. The more the velocity of the molecules of a substance, the greater their kinetic energy and the greater their temperature.

When molecules of solvent move faster, they collide more frequently with solute particles hence dissolution occurs faster at a higher temperature and molecular slowed compared to another system at lower temperature and lower molecular speed.

The higher temperature of particles in B will see to the faster dissolution of the antacid in B compared to A

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Suppose that in the citric acid cycle, oxalacetate is labeled with 14C in the carboxyl carbon farthest from the keto group. Wher
Nadusha1986 [10]

Answer:

All of alpha-ketoglutarate formed in the citric acid cycle would contain the radioactive ^{14}C while none of succinate would contain ^{14}C, and all of carbon dioxide released would contain ^{14}C.

Explanation:

When oxaloacetate in the citric acid cycle is labeled with ^{14}C in carboxyl carbon atom which is farthest from keto group, alpha ketoglutarate formed from this oxaloactetate has the full radioactive label. In the next step, the carboxylic group (that contains the ^{14}C) is eliminated in the form of the release of the carbon dioxide and succinate is formed. Succinate thus will not have radioactivity. CO_2 released had all the radioactivity.

8 0
4 years ago
An ideal gas is stored in a container at constant volume. If the temperature (T) were increased to 3T, what would be the change
Olin [163]
According to Gayle Lusac's law, pressure is proportional to absolute temperature of a gas. Thus:
P/T = constant

So if the temperature becomes 3T, the pressure would increase to 3P
6 0
3 years ago
Read 2 more answers
Please help with #2 and #3
Vaselesa [24]

Answer:

2. V_2=17L

3. V=82.9L

Explanation:

Hello there!

2. In this case, we can evidence the problem by which volume and temperature are involved, so the Charles' law is applied to:

\frac{V_2}{T_2}=\frac{V_1}{T_1}

Thus, considering the temperatures in kelvins and solving for the final volume, V2, we obtain:

V_2=\frac{V_1T_2}{T_1}

Therefore, we plug in the given data to obtain:

V_2=\frac{18.2L(22+273)K}{(45+273)K} \\\\V_2=17L

3. In this case, it is possible to realize that the 3.7 moles of neon gas are at 273 K and 1 atm according to the STP conditions; in such a way, considering the ideal gas law (PV=nRT), we can solve for the volume as shown below:

V=\frac{nRT}{P}

Therefore, we plug in the data to obtain:

V=\frac{3.7mol*0.08206\frac{atm*L}{mol*K}*273.15K}{1atm}\\\\V=82.9L

Best regards!

6 0
3 years ago
List the two main kinds of changes that you can observe when chemical reactions occur.
melisa1 [442]
Changes in energy and formation of new substances.
5 0
3 years ago
A volume of 40.0 mLmL of aqueous potassium hydroxide (KOHKOH) was titrated against a standard solution of sulfuric acid (H2SO4H2
Anni [7]

<u>Answer:</u> The concentration of KOH solution is 1.215 M

<u>Explanation:</u>

For the given chemical equation:

2KOH(aq.)+H_2SO_4(aq.)\rightarrow K_2SO_4(aq.)+2H_2O(l)

To calculate the concentration of base, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is KOH.

We are given:

n_1=2\\M_1=1.50M\\V_1=16.2mL\\n_2=1\\M_2=?M\\V_2=40.0mL

Putting values in above equation, we get:

2\times 1.50\times 16.2=1\times M_2\times 40.0\\\\M_2=\frac{2\times 1.50\times 16.2}{1\times 40.00}=1.215M

Hence, the concentration of KOH solution is 1.215 M

3 0
4 years ago
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