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3 years ago

HARD QUESTION (really need help)

Chemistry

1 answer:

torisob [31]3 years ago

5 0

Answer:

answer is A

hope this helps

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Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e

notsponge [240]

<u>Answer:</u> The concentration of $Sn^{2+}$ in the cell is $9.0\times 10^{-3}M$

<u>Explanation:</u>

We are given:

<u>Oxidation half reaction:</u> $Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-$ $E^o_{Zn^{2+}/Zn}=-0.76V$

<u>Reduction half reaction:</u> $Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)$ $E^o_{Sn^{2+}/Sn}=-0.136V$

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

Here, tin will undergo reduction reaction and will get reduced.

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

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

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

Putting values in above equation, we get:

$E^o_{cell}=-0.136-(-0.76)=0.624V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]}{[Cu^{2+}]}$

where,

$E_{cell}$ = electrode potential of the cell = 0.660 V

$E^o_{cell}$ = standard electrode potential of the cell = +0.624 V

n = number of electrons exchanged = 2

$[Zn^{2+}]=2.5\times 10^{-3}M$

$[Sn^{2+}]$ = ?

Putting values in above equation, we get:

$0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})$

$[Sn^{2+}]=9.0\times 10^{-3}M$

Hence, the concentration of $Sn^{2+}$ ions is $9.0\times 10^{-3}M$

3 0

3 years ago

(a) What is the total volume (in L) of gaseous products, measured at 350°C and 735 torr, when an automobile engine burns 100. g

Anarel [89]

Answer:

Part A

The volume of the gaseous product is $V = 787L$

Part B

The volume of the the engine’s gaseous exhaust is $V_e = 2178 \ L$

Explanation:

Part A

From the question we are told that

The temperature is $T = 350^oC = 350 +273 =623K$

The pressure is $P = 735 \ torr = \frac{735}{760} = 0.967\ atm$

The of $C_8 H_{18} = 100.0g$

The chemical equation for this combustion is

$2 C_8 H_{18}_{(l)} + 25O_2_{(l)} ----> 16CO_2_{(g)} + 18 H_2 O_{(g)}$

The number of moles of $C_8 H_{18}$ that reacted is mathematically represented as

$n = \frac{mass \ of \ C_8H_{18} }{Molar \ mass \ of C_8H_{18} }$

The molar mass of $C_8 H_{18}$ is constant value which is

$M = 114.23 \ g/mole$

So $n = \frac{100 }{114.23} }$

$n = 0.8754 \ moles$

The gaseous product in the reaction is $CO_2_{(g)}$ and water vapour

Now from the reaction

2 moles of $C_8 H_{18}$ will react with 25 moles of $O_2$ to give (16 + 18) moles of $CO_2_{(g)}$ and $H_2 O_{(g)}$

1 mole of $C_8 H_{18}$ will react with 12.5 moles of $O_2$ to give 17 moles of $CO_2_{(g)}$ and $H_2 O_{(g)}$

This implies that

0.8754 moles of $C_8 H_{18}$ will react with (12.5 * 0.8754 ) moles of $O_2$ to give (17 * 0.8754) of $CO_2_{(g)}$ and $H_2 O_{(g)}$

So the no of moles of gaseous product is

$N_g = 17 * 0.8754$

$N_g = 14.88 \ moles$

From the ideal gas law

$PV = N_gRT$

making V the subject

$V = \frac{N_gRT}{P}$

Where R is the gas constant with a value $R = 0.08206 \ L\cdot atm /K \cdot mole$

Substituting values

$V = \frac{14.88* 0.08206 *623}{0.967}$

$V = 787L$

Part B

From the reaction the number of moles of oxygen that reacted is

$N_o = 0.8754 * 12.5$

$N_o = 10.94 \ moles$

The volume is

$V_o = \frac{10.94 * 0.08206 *623}{0.967}$

$V_o = 579 \ L$

No this volume is the 21% oxygen that reacted the 79% of air that did not react are the engine gaseous exhaust and this can be mathematically evaluated as

$V_e = V_o * \frac{0.79}{0.21}$

Substituting values

$V_e = 579 * \frac{0.79}{0.21}$

$V_e = 2178 \ L$

3 0

3 years ago

14 km is how many centimeters?

Yuki888 [10]

Answer:

1400000 cm

Explanation:

14 km equals to 1400000 cm (14 km = 1400000 cm)

PLEASE MARK ME AS THE BRAINLIEST... ;)

8 0

2 years ago

What is the density of an object that has a volume of 34.2 cm^3 and a mass of 19.6 g? (a) 0.573 g/cm^3 (b) 1.74 g/cm^3 (c) 670 g

KengaRu [80]

Answer: The density of the object will be $0.573g/cm^3$

Explanation:

Density is defined as the mass contained per unit volume.

$Density=\frac{mass}{Volume}$

Given : Mass of object = 19.6 grams

Volume of object= $34.2cm^3$

Putting in the values we get:

$Density=\frac{19.6g}{34.2cm^3}=0.573g/cm^3$

Thus density of the object will be $0.573g/cm^3$

8 0

3 years ago

Use a Punnett square. In rabbits, black fur is dominant over white fur. What percentage of offspring will be white if a heterozy

blsea [12.9K]

<h2>Answer 1:</h2>

Option D is correct: 0%

Percentage of offspring will be 0% white if a heterozygous male and homozygous dominant female have offspring.

<h2>Answer 2:</h2>

Option A is correct: 75%

Two parents who are heterozygous are expecting a child. the percentage that the child will have free ear lobes will be 75%.

<h2>Answer 3:</h2>

Option A is correct: 25%

Two parents who are heterozygous are expecting a child. 25 is the percentage that the child be homozygous dominant.

<h2>Answer 4:</h2>

Option D is correct: 25%

If a homozyous recessive and a heterozygous plant have offspring. 25 will percentage will have smooth seeds.

<h2>Answer 5:</h2>

Option D is correct: 0%

If a homozyous recessive and a heterozygous plant have offspring. There will 0% chances that offsprings will be homozygous dominant.

<h2>Answer 6:</h2>

Option D is correct: substitution

In this question one of base thymine at 7 number is substituted with another base Adenine.

<h2>Answer 7:</h2>

Option A is correct: Incomplete dominance

Incomplete dominance is a form of intermediate inheritance in which one allele for a specific trait is not completely expressed over its paired allele. This results in a third phenotype in which the expressed physical trait is a combination of the phenotypes of both alleles.

<h2>Answer 8:</h2>

Option A is correct: T T C G A A T C T G T A T G A C

After the insertion of a new base thymine the new sequence will be of option A.

<h2>Answer 9:</h2>

True

Yes this is true that "Environmental influences and genetics can determine a person's phenotype".

<h2>Answer 10:</h2>

Option B is correct: selective breeding

Humans have been choosing desired traits of livestock and pets for centuries by breeding organisms that they choose with the traits they desire. This process is called selective breeding.

<h2>Answer 11:</h2>

homologous chromosomes

karyotyping is a test that evaluates the number and structure of a person's chromosomes in order to detect abnormalities. Chromosomes are thread-like structures within each cell nucleus and contain the body's genetic blueprint.

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6 0

3 years ago