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3 years ago

Complete and balance the equation for this reaction in acidic solution. zns+no3-

1 answer:

8 0

The equation given in question is as follow,

   ZnS + NO₃⁻   → Zn²⁺ + S + NO

Oxidation state of elements in reactant side are as follow,
Zn = +2
S = -2
N = +5
O = -2
Oxidation state of elements in product side are as follow,
Zn = +2
S = 0
N = +2
O = -2
So, In this reaction Nitrogen is Reduced while Sulfur is Oxidized.

Now Split the reaction into two half cell reactions,

Reduction Reaction,
ZnS → S + 2e⁻
Oxidation Reaction,
NO₃⁻ + 3e⁻   → NO

As the oxygen atoms are not balance, So, in acidic medium add H⁺ on the side having greater number of Oxygen atom and H₂O on the side having less number of Oxygen atoms,
Hence,
H⁺ + NO₃⁻ + 3e⁻    → NO + H₂O
Now Balance the reaction,

4H⁺ + NO₃⁻ + 3e⁻    → NO + 2H₂O

So, write both half cell equations as,

ZnS → S + 2e⁻ --------- (1)
4H⁺ + NO₃⁻ + 3e⁻    → NO + 2H₂O -------(2)

Multiply eq. 1 with 3 and eq. 2 with 2 to equalize the electron.
So,
3ZnS → 3S + 6e⁻

8H⁺ + 2NO₃⁻ + 6e⁻    → 2NO + 4H₂O
_____________________________________(e⁻ cancelled)

3ZnS + 2NO₃⁻ + 8H⁺ → 3Zn + 3S + 2NO + 4H₂O

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Compare your circulatory system to a busy street in a city or town.

White raven [17]

City individuals -different cells. Group of individuals - Organ systems. City Road System- Circulatory system.

Explanation:

Any city has population. The individuals of the population can be considered as the cells that are present in a human body. As a city cannot be formed with individuals, the human body cannot survive without cells. These individuals will belong to different profession and they go to an organisation for doing various work.

These can be compared with the organ system of human body. These organ systems will be performing different functions. The network of any rod in a city can be compared with the circulatory system of human body. As it helps individuals to reach their home of any destination, the presence of arteries and veins can be compared with the highways and capillaries can be associated with side ways of the road system.

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3 years ago

Who knows this I need it

arsen [322]

1. In the first image, you can see two cacti and a javelina in the background.

2. In picture one, neither of the plants are being disturbed. In picture two, however, the smaller plant is indeed being disturbed by the javelina.

3. The javelina is not eating the cactus on the right because it is stronger in color (?).

4. The cactus the javelina was eating has disappeared.

5. In picture 3, the idle cactus blossomed.

5 0

3 years ago

Calculate the mass of methane that must be burned to provide enough heat to convert 242.0 g of water at 26.0°C into steam at 101

Anarel [89]

Answer: The mass of methane burned is 12.4 grams.

Explanation:

The chemical equation for the combustion of methane follows:

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CO_2(g))})+(2\times \Delta H^o_f_{(H_2O(g))})]-[(1\times \Delta H^o_f_{(CH_4(g))})+(2\times \Delta H^o_f_{(O_2(g))})]$

We are given:

$\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.51kJ/mol\\\Delta H^o_f_{(CH_4(g))}=-74.81kJ/mol\\\Delta H^o_f_{O_2}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-393.51))+(2\times (-241.82))]-[(1\times (-74.81))+(2\times (0))]\\\\\Delta H^o_{rxn}=-802.34kJ$

The heat calculated above is the heat released for 1 mole of methane.

The process involved in this problem are:

$(1):H_2O(l)(26^oC)\rightarrow H_2O(l)(100^oC)\\\\(2):H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)\\\\(3):H_2O(g)(100^oC)\rightarrow H_2O(g)(101^oC)$

Now, we calculate the amount of heat released or absorbed in all the processes.

For process 1:

$q_1=mC_p,l\times (T_2-T_1)$

where,

$q_1$ = amount of heat absorbed = ?

m = mass of water = 242.0 g

$C_{p,l}$ = specific heat of water = 4.18 J/g°C

$T_2$ = final temperature = $100^oC$

$T_1$ = initial temperature = $26^oC$

Putting all the values in above equation, we get:

$q_1=242.0g\times 4.18J/g^oC\times (100-(26))^oC=74855.44J$

For process 2:

$q_2=m\times L_v$

where,

$q_2$ = amount of heat absorbed = ?

m = mass of water or steam = 242 g

$L_v$ = latent heat of vaporization = 2257 J/g

Putting all the values in above equation, we get:

$q_2=242g\times 2257J/g=546194J$

For process 3:

$q_3=mC_p,g\times (T_2-T_1)$

where,

$q_3$ = amount of heat absorbed = ?

m = mass of steam = 242.0 g

$C_{p,g}$ = specific heat of steam = 2.08 J/g°C

$T_2$ = final temperature = $101^oC$

$T_1$ = initial temperature = $100^oC$

Putting all the values in above equation, we get:

$q_3=242.0g\times 2.08J/g^oC\times (101-(100))^oC=503.36J$

Total heat required = $q_1+q_2+q_3=(74855.44+546194+503.36)=621552.8J=621.552kJ$

To calculate the number of moles of methane, we apply unitary method:

When 802.34 kJ of heat is needed, the amount of methane combusted is 1 mole

So, when 621.552 kJ of heat is needed, the amount of methane combusted will be = $\frac{1}{802.34}\times 621.552=0.775mol$

To calculate the number of moles, we use the equation:

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

Molar mass of methane = 16 g/mol

Moles of methane = 0.775 moles

Putting values in above equation, we get:

$0.775mol=\frac{\text{Mass of methane}}{16g/mol}\\\\\text{Mass of methane}=(0.775mol\times 16g/mol)=12.4g$

Hence, the mass of methane burned is 12.4 grams.

8 0

3 years ago

Determine the volume of a substance that has a density of 1.84 g/ml and a mass of 12 grams

vazorg [7]

Hey there!:

density = mass / volume

1.84 g/mL = 12 g / V

V = 12 / 1.84

V = 6.521 mL

hope this helps!

5 0

3 years ago

Photosynthesis takes carbon dioxide, water, and sunshine and turns them into____?

RUDIKE [14]

Photosynthesis takes carbon dioxide, water and sunshine and turn them into glucose and oxygen

Explanation
Glucose is the principal carbohydrate formed through photosynthesis, Through the process of photosynthesis sunshine, carbon dioxide and water to form glucose and oxygen

Equation for reaction is as follows

CO2+ H2O + light energy → C6H12O6 +O2

C6H12O6 is the glucose

3 0

3 years ago