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3 years ago

Why doesn't water change into ice at 10 degree Celsius?

2 answers:

3 0

The fusion on water is an enxothermic change but at this temp, the temp-entropy product is outweighed by the change in enthalpy.

scoray [572]3 years ago

3 0

the answer is b for plato

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Arthur is performing a lab which requires a 0.2 M solution of calcium hydroxide. However, the only calcium hydroxide solution av

aev [14]

Answer:

He needs to add 4 mL of the 0.5 M solution to 6 mL of water.

7 0

3 years ago

a flask was filled with SO2 at a partial pressure of 0.409 atm and O2 at a partial pressure of 0.601 atm. The following gas-phas

jarptica [38.1K]

Answer:

The equilibrium partial pressure of O2 is 0.545 atm

Explanation:

Step 1: Data given

Partial pressure of SO2 = 0.409 atm

Partial pressure of O2 = 0.601 atm

At equilibrium, the partial pressure of SO2 was 0.297 atm.

Step 2: The balanced equation

2SO2 + O2 ⇆ 2SO3

Step 3: The initial pressure

pSO2 = 0.409 atm

pO2 = 0.601 atm

pSO3 = 0 atm

Step 4: Calculate the pressure at the equilibrium

pSO2 = 0.409 - 2X atm

pO2 = 0.601 - X atm

pSO3 = 2X

pSO2 = 0.409 - 2X atm = 0.297

X = 0.056 atm

pO2 = 0.601 - 0.056 = 0.545 atm

pSO3 = 2*0.056 = 0.112 atm

Step 5: Calculate Kp

Kp = (pSO3)²/((pO2)*(pSO2)²)

Kp = (0.112²) / (0.545 * 0.297²)

Kp = 0.261

The equilibrium partial pressure of O2 is 0.545 atm

3 0

3 years ago

Molecule of dissimilar elements with a net charge remaining is called

MariettaO [177]

It is called a polyatomic ion.

Hope this helps!!!

3 0

3 years ago

As we move UP a energy pyramid, the amount of energy available to each level of consumers

cricket20 [7]

Answer:

well we do need up to explain the neergyu field on the world you probaly think im spamming but i thing you need up

Explanation:

7 0

2 years ago

For the Zn - Cu^2+ voltaic cell Zn(s) + Cu^2+(aq, 1M) + Cu(s) E degree _cell = 1.10 V Given that the standard reduction potentia

Fittoniya [83]

Answer : The value of $E^o_{(Cu^{2+}/Cu)}$ is, 0.34 V

Explanation :

Here, copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Zn\rightarrow Zn^{2+}+2e^-$

Reduction half reaction: $Cu^{2+}+2e^-\rightarrow Cu$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.

The overall balanced equation of the cell is,

$Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu$

To calculate the $E^o_{(Cu^{2+}/Cu)}$ of the reaction, we use the equation:

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

$E^o_{cell}=E^o_{(Cu^{2+}/Cu)}-E^o_{(Zn^{2+}/Zn)}$

Putting values in above equation, we get:

$1.10V=E^o_{(Cu^{2+}/Cu)}-(-0.76V)$

$E^o_{(Cu^{2+}/Cu)}=0.34V$

Hence, the value of $E^o_{(Cu^{2+}/Cu)}$ is, 0.34 V

8 0

2 years ago