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2 years ago

5. How are the members of Kingdom Archaebacteria different from the

Chemistry

1 answer:

yan [13]2 years ago

5 0

Answer:

5. C

6. B

Explanation:

sana nakatulong

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The product gas is then passed through a concentrated solution of KOH to remove the CO2. After passage through the KOH solution,

Kamila [148]

Answer: The mass percent of nitrogen gas in the compound is 13.3 %

Explanation:

Assuming the chemical equation of the compound forming product gases is:

$\text{Compound}\xrightarrow[CuO(s)]{Hot}N_2(g)+CO_2(g)+H_2O(g)$

Now, the product gases are treated with KOH to remove carbon dioxide.

We are given:

$p_{Total}=726torr\\P_{water}=23.8torr\\$

So, pressure of nitrogen gas will be = $p_{Total}-p_{water}=726-23.8=702.2torr$

To calculate the number of moles of nitrogen, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of nitrogen gas = 702.2 torr = 0.924 atm (Conversion factor: 1 atm = 760 torr)

V = Volume of nitrogen gas = 31.8 mL = 0.0318 L (Conversion factor: 1 L = 1000 mL)

T = Temperature of nitrogen gas = $25^oC=[25+273]K=298K$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles of nitrogen gas = ?

Putting values in above equation, we get:

$0.924atm\times 0.0318L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 298K\\n_{mix}=\frac{0.924\times 0.0318}{0.0821\times 298}=0.0012mol$

To calculate the mass of nitrogen gas, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of nitrogen gas = 28 g/mol

Moles of nitrogen gas = 0.0012 moles

Putting values in above equation, we get:

$0.0012mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=(0.0012mol\times 28g/mol)=0.0336g$

To calculate the mass percent of nitrogen gas in compound, we use the equation:

$\text{Mass percent of nitrogen gas}=\frac{\text{Mass of nitrogen gas}}{\text{Mass of compound}}\times 100$

Mass of compound = 0.253 g

Mass of nitrogen gas = 0.0336 g

Putting values in above equation, we get:

$\text{Mass percent of nitrogen gas}=\frac{0.0336g}{0.253g}\times 100=13.3\%$

Hence, the mass percent of nitrogen gas in the compound is 13.3 %

8 0

3 years ago

5) Chlorine has 7 valence electrons. Neon has 8 valence electrons. Which on is more likely to undergo a

kari74 [83]

Answer:

well, chlorine will undergo in a chemical reaction

Explanation:

neon has completed valence shells, so it neither need electron and or it also don't need to donate the electrons. hence it will not participate in any chemical reaction and it is stable in nature.

whereas chlorine need one more electron to complete it's valence shell and in order to be stable , so it will try to gain one electron , hence

it will take place in a chemical reaction

6 0

2 years ago

Why do acids and bases neutralize eachother?

timofeeve [1]

Answer:

Explanation:

When an acid and a base are placed together, they react to neutralize the acid and base properties, producing a salt. The H(+) cation of the acid combines with the OH(-) anion of the base to form water. The compound formed by the cation of the base and the anion of the acid is called a salt.

5 0

2 years ago

The standard reduction potentials of the following half-reactions are given in Appendix E in the textbook:

german

Answer:

For 1: The largest positive cell potential is of cell having 1st and 4th half reactions.

For 2: The standard electrode potential of the cell is 1.539 V

For 3: The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

Explanation:

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

We are given:

$Ag^++(aq.)+e^-\rightarrow Ag(s);E^o_{Ag^+/Ag}=0.799V\\\\Cu^{2+}+(aq.)+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.337V\\\\Ni^{2+}(aq.)+2e^-\rightarrow Ni(s);E^o_{Ni^{2+}/Ni}=-0.28V\\\\Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o_{Cr^{3+}/Cr}=-0.74V$

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:

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

Cell having 1st and 2nd half reactions:

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Copper will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-0.337=0.462V$

Cell having 1st and 3rd half reactions:

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-(-0.28)=1.079V$

Cell having 1st and 4th half reactions:

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-(-0.74)=1.539V$

Cell having 2nd and 3rd half reactions:

Copper has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.337-(-0.28)=0.617V$

Cell having 3rd and 4th half reactions:

Nickel has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

$E^o_{cell}=-0.28-(-0.74)=0.46V$

Hence,

For 1: The largest positive cell potential is of cell having 1st and 4th half reactions.

For 2: The standard electrode potential of the cell is 1.539 V

For 3: The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

8 0

3 years ago

How to write a balanced chemical equation from empirical formula?

netineya [11]

Let's use the example: H2O ---> H2 + O2
We find how many elements of a product are on one side and how many elements on the other side.
Reactant: H=2 O=1
Product: H=2 O=2
We need to make the same amount of hydrogen and oxegyn atoms on each side, regardless of how high the numbers are, and we do this by adding coefficients to the compounds.

Reactant: H=4 O=2
Product : H=4 O=2
2 H2O---> 2 H2 + O2

8 0

2 years ago