1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
artcher [175]
3 years ago
9

True or false; when something heats up, New energy is created, and when something cool down, Energy is destroyed.

Chemistry
1 answer:
Ann [662]3 years ago
4 0

Answer:

This is false.

Explanation:

This statement directly opposes the first law of thermodynamics which states that energy can neither be created nor destroyed, but can transfer from one source to another.

You might be interested in
Consider the reaction N2(g) + 2O2(g)2NO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr
zysi [14]

<u>Answer:</u> The value of \Delta S^o for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]

For the given chemical reaction:

N_2+2O_2\rightarrow 2NO_2

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

\Delta S^o_{rxn}=[(2\times \Delta S^o_{(NO_2(g))})]-[(1\times \Delta S^o_{(N_2(g))})+(2\times \Delta S^o_{(O_2(g))})]

We are given:

\Delta S^o_{(NO_2(g))}=240.06J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(N_2)}=191.61J/K.mol

Putting values in above equation, we get:

\Delta S^o_{rxn}=[(2\times (240.06))]-[(1\times (191.61))+(2\times (205.14))]\\\\\Delta S^o_{rxn}=-121.77J/K

Entropy change of the surrounding = - (Entropy change of the system) = -(-121.77) J/K = 121.77 J/K

We are given:

Moles of nitrogen gas reacted = 1.90 moles

By Stoichiometry of the reaction:

When 1 mole of nitrogen gas is reacted, the entropy change of the surrounding will be 121.77 J/K

So, when 1.90 moles of nitrogen gas is reacted, the entropy change of the surrounding will be = \frac{121.77}{1}\times 1.90=231.36 J/K

Hence, the value of \Delta S^o for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

7 0
3 years ago
What is the formula for an alkene containing six carbon atoms?
Charra [1.4K]

Answer:

C

Explanation:

Cyclohexane is a cycloalkane with the molecular formula C₆H₁₂. Cyclohexane is non-polar.

3 0
3 years ago
a block of wood has a density of 0.95 g/cm3. will it float or sink when placed on a liquid with a density of 0.88 g/mL? explain
vesna_86 [32]
Since 1mL=1cm^3 the wood would sink due to it being more dense. I.e. 0.95>0.88
3 0
3 years ago
Read 2 more answers
If the initial temperature of a movable cylinder was 50 degrees Celsius
slega [8]

Answer:

8.45 L

Explanation:

From the question given above, the following data were obtained:

Initial temperature (T₁) = 50 °C

Initial pressure (P₁) = 2 atm

Initial volume (V₁) = 5 L

Final temperature (T₂) = 0 °C

Final pressure (P₂) = 1 atm

Final volume (V₂) =?

Next, we shall convert celsius temperature to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 50 °C

Initial temperature (T₁) = 50 °C + 273

Initial temperature (T₁) = 323 K

Final temperature (T₂) = 0 °C

Final temperature (T₂) = 0 °C + 273

Final temperature (T₂) = 273 K

Finally, we shall determine the new volume. This can be obtained as follow:

Initial temperature (T₁) = 323 K

Initial pressure (P₁) = 2 atm

Initial volume (V₁) = 5 L

Final temperature (T₂) = 273 k

Final pressure (P₂) = 1 atm

Final volume (V₂) =?

P₁V₁ / T₁ = P₂V₂ / T₂

2 × 5 / 323 = 1 × V₂ / 273

10 / 323 = V₂ / 273

Cross multiply

323 × V₂ = 10 × 273

323 × V₂ = 2730

Divide both side by 323

V₂ = 2730 / 323

V₂ = 8.45 L

Thus, the new volume is 8.45 L

5 0
3 years ago
Which is the most likely to be reduced?
NISA [10]

Answer : The correct option is, Zn^{2+}

Explanation :

  • Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. That means, the loss of electrons takes place.

Or we can say that, oxidation reaction occurs when a reactant losses electrons in the reaction.

  • Reduction reaction : It is defined as the reaction in which a substance gains electrons. That means, the gain of electrons takes place.

Or we can say that, reduction reaction occurs when a reactant gains electrons in the reaction.

According to the electrochemical series, Zn^{2+} most likely to be reduced because

Hence, the ion most likely to be reduced is Zn^{2+}.

3 0
2 years ago
Other questions:
  • How is the law of conservation of mass applied to<br> ecology?
    8·2 answers
  • Hypothesis for isotopes and atomic mass
    6·1 answer
  • 1. The density of solid Cu is 8.96g/cm3 . How many atoms are present per cubic centimeter(cm3) of Cu ?
    12·1 answer
  • 2NaHCO3--&gt;Na2CO3+H2O+CO2
    7·1 answer
  • Why do you use 6.02*10^23 here?? #21???
    8·2 answers
  • Bob and Ada are camping and have sweet potatoes wrapped in foil to eat. Ada put her sweet potato in the bonfire for a while, but
    8·1 answer
  • 4.
    9·1 answer
  • Is lemon juice a asid base or neutral?<br>plzz help mee​
    5·2 answers
  • 5. What is common among the following phenomena?
    8·1 answer
  • you are carrying out the following reaction: o2 2h2 −−&gt; 2h2o you start with 6.0 moles of oxygen and 4.0 moles of hydrogen. ho
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!