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monitta
3 years ago
11

How would you expect energy to change during a chemical reaction?

Chemistry
1 answer:
Basile [38]3 years ago
6 0
B. The energy will be converted into another form. In chemical reactions energy is usually changed to heat or light energy. This is an example of the First Law of Thermodynamics which states that energy can not be created or destroyed.
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garri49 [273]

Answer:

Mg would blow off. AI would be affective to copper but not to MG

Explanation:

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3 years ago
*blank* carry blood away from the heart, while *blank* carry blood back to the heart.
worty [1.4K]

Answer:

i think its C im not so sure

Explanation:

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A piece of solid carbon dioxide with a mass of 5.50 grams is placed in a 10 Liter vessel that already contains 705 torr at 24 Ce
alex41 [277]

total pressure = ( 705 torr + 231.525torr)  =   936.525torr

7 0
3 years ago
Read 2 more answers
If 11.0 g of ccl3f is enclosed in a 1.1 −l container, will any liquid be present?if so, what mass of liquid?
Anastasy [175]
Considering that CCL3F gas behave like an ideal gas then we can use the Ideal Gas Law 
<span>PV = nRT, however is an approximation and not the only way to resolve this problem with the given data..So,at the end of the solution I am posting some sources for further understanding and a expanded point of view. </span>

<span>Data: P= 856torr, T = 300K, V= 1.1L, R = 62.36 L Torr / KMol </span>

<span>Solving and substituting in the Gas equation for n = PV / RT = (856)(1.1L) /( 62.36)(300) = 0.05 Mol. This RESULT is of any gas. To tie it up to our gas we need to look for its molecular weight:MW of CCL3F = 137.7 gm/mol.  </span>

<span>Then : 0.05x 137.5 = 6.88gm of vapor </span>

<span>If we sustract the vapor weight from the TOTAL weight of liquid we have: 11.5gm - 6.88gm = 4.62 gm of liquid.d</span>
8 0
3 years ago
For the reaction C2H4(g) + H2O(g) --&gt; CH3CH2OH(g)
Dominik [7]

Answer : The value of equilibrium constant for this reaction at 262.0 K is 3.35\times 10^{2}

Explanation :

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

where,

\Delta G^o = standard Gibbs free energy  = ?

\Delta H^o = standard enthalpy = -45.6 kJ = -45600 J

\Delta S^o = standard entropy = -125.7 J/K

T = temperature of reaction = 262.0 K

Now put all the given values in the above formula, we get:

\Delta G^o=(-45600J)-(262.0K\times -125.7J/K)

\Delta G^o=-12666.6J=-12.7kJ

The relation between the equilibrium constant and standard Gibbs free energy is:

\Delta G^o=-RT\times \ln k

where,

\Delta G^o = standard Gibbs free energy  = -12666.6 J

R = gas constant  = 8.314 J/K.mol

T = temperature  = 262.0 K

K = equilibrium constant = ?

Now put all the given values in the above formula, we get:

-12666.6J=-(8.314J/K.mol)\times (262.0K)\times \ln k

k=3.35\times 10^{2}

Therefore, the value of equilibrium constant for this reaction at 262.0 K is 3.35\times 10^{2}

3 0
4 years ago
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