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Tems11 [23]
2 years ago
14

HELP ME OUT PLEASE!!!!

Chemistry
2 answers:
Sunny_sXe [5.5K]2 years ago
7 0

Answer:

D Im 90% Sure

Explanation:

If its not right i owe you one I did this one before

PolarNik [594]2 years ago
5 0

Answer:

Answer D. Picture II shows a chemical change, because the same substance changes form

Explanation:

This is the temperature that water molecules slow down enough to stick to each other and form a solid crystal

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Calculate for the following electrochemical cell at 25°C, Pt H2(g) (1.0 atm) H (0.010 M || Ag (0.020 M) Ag if E (H) - +0.000 V a
viva [34]

Answer : The correct option is, (b) +0.799 V

Solution :

The values of standard reduction electrode potential of the cell are:

E^0_{[H^{+}/H_2]}=+0.00V

E^0_{[Ag^{+}/Ag]}=+0.799V

From the cell representation we conclude that, the hydrogen (H) undergoes oxidation by loss of electrons and thus act as anode. Silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : H_2\rightarrow 2H^{+}+2e^-    

Reaction at cathode (reduction) : Ag^{+}+1e^-\rightarrow Ag    

The balanced cell reaction will be,  

H_2+2Ag^{+}\rightarrow 2H^{+}+2Ag

Now we have to calculate the standard electrode potential of the cell.

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

E^o_{cell}=E^o_{[Ag^{+}/Ag]}-E^o_{[H^{+}/H_2]}

E^o_{cell}=(+0.799V)-(+0.00V)=+0.799V

Therefore, the standard cell potential will be +0.799 V

4 0
3 years ago
Consider the reaction: CaCO3(s) à CaO(s) + CO2(g)
Vsevolod [243]

Answer:

131.5 kJ

Explanation:

Let's consider the following reaction.

CaCO₃(s) → CaO(s) + CO₂(g)

First, we will calculate the standard enthalpy of the reaction (ΔH°).

ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g) ) - 1 mol × ΔH°f(CaCO₃(s) )

ΔH° = 1 mol × (-634.9 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1207.6 kJ/mol)

ΔH° = 179.2 kJ

Then, we calculate the standard entropy of the reaction (ΔS°).

ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g) ) - 1 mol × S°(CaCO₃(s) )

ΔS° = 1 mol × (38.1 J/mol.K) + 1 mol × (213.8 J/mol.K) - 1 mol × (91.7 J/mol.K)

ΔS° = 160.2 J/K = 0.1602 kJ/K

Finally, we calculate the standard Gibbs free energy of the reaction at T = 25°C = 298 K.

ΔG° = ΔH° - T × ΔS°

ΔG° = 179.2 kJ - 298 K × 0.1602 kJ/K

ΔG° = 131.5 kJ

6 0
3 years ago
Is fireworks exploding a chemical change?
zalisa [80]

Answer:

Yes

Explanation:

The fireworks explode

7 0
2 years ago
Read 2 more answers
Silver is a white metal that is an excellent conductor. Silver tarnishes when exposed to air and light. The density of silver is
Stells [14]

Answer:

c tarnishes in air

Explanation:

After silver has been exposed to air that contains sulphur gases, discoloration would occur. there would be darkening that is caused by the reaction with gases.When any silver object tarnishes, it brings about a disfiguring of that object. Hydrogen sulphide would be needed for this to happen. silver sulphide is black and a if a thin layer should form on any surface, it ill darken it. This  is what we refer to as tarnishing.

3 0
3 years ago
Read 2 more answers
A solution is made by dissolving 0.565 g of potassium nitrate in enough water to make up 250. mL of solution. What is the molari
aalyn [17]
<h3>Molar mass of Potassium Nitrate:-</h3>

\\ \large\sf\longmapsto KNO_3

\\ \large\sf\longmapsto 39u+14u+3(16u)

\\ \large\sf\longmapsto 53u+48u

\\ \large\sf\longmapsto 101u

\\ \large\sf\longmapsto 101g/mol

Now

\boxed{\sf No\:of\:moles=\dfrac{Given\:mass}{Molar\:mass}}

\\ \large\sf\longmapsto No\:of\:moles=\dfrac{0.565}{101}

\\ \large\sf\longmapsto No\:of\:moles=0.005mol

We know

\boxed{\sf Molarity=\dfrac{Moles\:of\:solute}{Vol\:of\:Solution\:in\:L}}

\\ \large\sf\longmapsto Molarity=\dfrac{0.005}{\dfrac{250}{1000}L}

\\ \large\sf\longmapsto Molarity=\dfrac{0.005}{0.250}

\\ \large\sf\longmapsto Molarity=0.02M

8 0
2 years ago
Read 2 more answers
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