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Anestetic [448]

2 years ago

13

When an Alka-Seltzer® tablet is put in water, there is a chemical change because there are bubbles produced. The Alka-Seltzer® t

ablet slowly disappears over time. What would speed up the chemical change?
Adding more water would cause the chemical change to speed up.
Changing the color of the tablet would cause the chemical change to speed up.
Decreasing the temperature of the water would cause the chemical change to speed up.
Increasing the temperature of the water would cause the chemical change to speed up.

2 answers:

egoroff_w [7]2 years ago

4 0

It’s either increasing the temp or water. the temp increasing adds more energy bc heat = more energy. so i would go w increasing temp

Soloha48 [4]2 years ago

3 0

Answer:

Heyy

Explanation:

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A certain weak acid, HA, has a Ka value of 2.6×10−7. Calculate the percent ionization of HA in a 0.10 M solution.

Andrej [43]

Answer:

The percent ionization is 0,16%

Explanation:

The percent ionization is defined as the number of ions that exist in a substance.

$PI=\frac{[A-]}{[HA]} x100$

First, we find the [A-] using the ka equation

HA ⇄ $H^{+} + A^{-}$

[H+] = [A-]

$Ka=\frac{[H+][A-]}{[HA]}\\ \\$

since the ionization constant is very small we can assume that the final concentration of [HA] is the same

$Ka=\frac{[H+]^{2} }{[HA]} \\\\$

$[H+]=\sqrt[2]{Ka.[HA]} \\\\$

$[H+] =\sqrt{(2,610^{-7} )(0,1)} = 1,61210^{-4}$

Now we calculate the percent ionization using these values

$PI=\frac{1,61210^{-4} }{0,1} X100$

PI=0,16%

4 0

4 years ago

How are energy storage molecules affected if there is LESS carbon dioxide?

kipiarov [429]

Answer:

When there is less carbon (in the form of carbon dioxide) in abiotic matter, less carbon is available for producers making energy storage molecules. When there is more sunlight, producers can make more energy storage molecules from the carbon in carbon dioxide.

Hope this helps.

7 0

3 years ago

Consider the reaction: 2BrF3(g) --> Br2(g) + 3F2(g)

riadik2000 [5.3K]

Answer : The entropy change of reaction for 1.62 moles of $BrF_3$ reacts at standard condition is 217.68 J/K

Explanation :

The given balanced reaction is,

$2BrF_3(g)\rightarrow Br_2(g)+3F_2(g)$

The expression used for entropy change of reaction $(\Delta S^o)$ is:

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{Br_2}\times \Delta S_f^0_{(Br_2)}+n_{F_2}\times \Delta S_f^0_{(F_2)}]-[n_{BrF_3}\times \Delta S_f^0_{(BrF_3)}]$

where,

$\Delta S^o$ = entropy change of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

$\Delta S_f^0_{(Br_2)}$ = 245.463 J/mol.K

$\Delta S_f^0_{(F_2)}$ = 202.78 J/mol.K

$\Delta S_f^0_{(BrF_3)}$ = 292.53 J/mol.K

Now put all the given values in this expression, we get:

$\Delta S^o=[1mole\times (245.463J/K.mole)+3mole\times (202.78J/K.mole)}]-[2mole\times (292.53J/K.mole)]$

$\Delta S^o=268.74J/K$

Now we have to calculate the entropy change of reaction for 1.62 moles of $BrF_3$ reacts at standard condition.

From the reaction we conclude that,

As, 2 moles of $BrF_3$ has entropy change = 268.74 J/K

So, 1.62 moles of $BrF_3$ has entropy change = $\frac{1.62}{2}\times 268.74=217.68J/K$

Therefore, the entropy change of reaction for 1.62 moles of $BrF_3$ reacts at standard condition is 217.68 J/K

3 0

4 years ago

According to the theory of evolution, ______ traits will become common in a population over time.

Pavlova-9 [17]

I believe your answer is Favorable.
Hope this helps, and happy studying~!
~{Dunsforhands}

7 0

3 years ago

Can any person help me for this problem? Anyone's welcome

noname [10]

Answer:

The anode made of the impure copper

The cathode made of pure copper

The electrolyte of copper (II) sulfate CuSO₄ solution

The silver impurities at the anode due to their high tendency of accepting electrons and being a stronger reducing agent than either copper or zinc will remain relatively in place and relatively stable and will not actively take part in the oxidation reaction taking place at the anode

The zinc impurities will be the first element of the three metals to give up electrons and go into the solution as zinc ions due to their high tendency to loan out two electrons and become oxidized into Zn²⁺ ions

The drawing of the electrolytic cell created with Microsoft Visio is attached

Explanation:

5 0

3 years ago