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3 years ago

Sulfanomids warnings

Chemistry

1 answer:

cupoosta [38]3 years ago

4 0

Answer:

Sulfanomides Warnings

Sulfonamides may cause blood problems, especially if they're taken for a long period of time. These medicines can also cause a serious, even life-threatening, skin rash. Tell your doctor right away if you notice a rash or unusual skin changes.

Explanation:

<h3 /><h3>Hope this helps you !!</h3>

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Ede4ka [16]

Answer:

what is scientific model

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3 years ago

Calculate the number of moles in the following: 2.8 X 10^24 atoms of Cl2

vova2212 [387]

Answer:

<h3>The answer is 4.65 moles</h3>

Explanation:

To find the number of moles given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question

N = 2.8 × 10²⁴ atoms of Cl2

So we have

$n = \frac{2.8 \times {10}^{24} }{6.02 \times {10}^{23} } \\ = 4.65116279069...$

We have the final answer as

<h3>4.65 moles</h3>

Hope this helps you

7 0

3 years ago

In thermodynamics, we determine the spontaneity of a reaction by the sign of ΔG. In electrochemistry, spontaneity is determined

faltersainse [42]

Answer:

For A: The standard cell potential of the reaction is 4.4 V

For B: The standard Gibbs free energy of the reaction is $-8.50\times 10^5J$

For C: The reaction is spontaneous as written.

Explanation:

For A:

The given chemical reaction follows:

$2Li(s)+Cl_2(g)\rightarrow 2Li^+(aq.)+2Cl^-(aq.)$

The given half reaction follows:

Oxidation half reaction: $Li(s)\rightarrow Li^+(aq.)+e^-;E^o_{Li^+/Li}=-3.04V$ ( × 2)

Reduction half reaction: $Cl_2(g)+2e^-\rightarrow 2Cl^-(aq.);E^o_{Cl_2/2Cl^-}=+1.36V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction.

Here, chlorine will undergo reduction reaction will get reduced.

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=1.36-(-3.04)=4.4V$

Hence, the standard cell potential of the reaction is 4.4 V

For B:

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$

where,

n = number of electrons transferred = $2mol\text{ e}^-$

F = Faradays constant = $96500J/V.mol\text{ e}^-$

$E^o_{cell}$ = standard cell potential = 4.4 V

Putting values in above equation, we get:

$\Delta G^o=-2\times 96500\times 4.4=-849200J=-8.50\times 10^5J$

Hence, the standard Gibbs free energy of the reaction is $-8.50\times 10^5J$

For C:

For a reaction to be spontaneous, the standard Gibbs free energy change of the reaction must be negative.

From above, the standard Gibbs free energy change of the reaction is coming out to be negative.

Hence, the reaction is spontaneous as written.

7 0

3 years ago

Are the particles in a solid in constant motion, or are they "frozen in place?"

Tems11 [23]

The molecules are frozen in place but still vibrate

3 0

3 years ago

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A substance has half the number of particles as 12 grams of carbon 12, how many moles are in the substance?

gavmur [86]

the molar mass is how many grams it takes to make a full mole. The number in carbon 12 tells us that it takes 12 g to makes a mole, so one half of that would be option d, 0.5.

7 0

3 years ago

Read 2 more answers

Sulfanomids warnings ​

Sulfanomids warnings