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Diano4ka-milaya [45]
3 years ago
14

Which of these is an example of investigating an intensive property?

Chemistry
1 answer:
Nonamiya [84]3 years ago
4 0
I think the answer is B
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When an acid reacts with a base what compounds are formed?
Alecsey [184]
(C)When acid reacts with base, a salt and a water formed.
8 0
3 years ago
CH4 (g) yields C(g) + 4H (g) (reaction for expansion)
anyanavicka [17]

Answer:

Here's what I find.

Explanation:

\rm CH$_{4}$(g) $\, \rightleftharpoons \,$ C(g) + 4H(g)

\text{Same} =\begin{cases}1. & \text{The number of atoms of each element}\\2. & \text{The state of the methane molecules}\\4. &\text{The state of the carbon atoms}\\\end{cases}

\text{Different} =\begin{cases}3. & \text{The enthalpy change of the reaction}\\\end{cases}

The sign of ΔH changes when you reverse the reaction.  

5 0
4 years ago
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The equation represents the decomposition of a generic diatomic element in its standard state. 12X2(g)⟶X(g) Assume that the stan
aliina [53]

Answer:

K^{2000K}=0.774\\\\K^{3000K}=12.56

Explanation:

Hello,

In this case, considering the reaction, we can compute the Gibbs free energy of reaction at each temperature, taking into account that the Gibbs free energy for the diatomic element is 0 kJ/mol:

\Delta _rG=\Delta _fG_{X}-\frac{1}{2} \Delta _fG_{X_2}=\Delta _fG_{X}

Thus, at 2000 K:

\Delta _rG=\Delta _fG_{X}^{2000K}=4.25kJ/mol

And at 3000 K:

\Delta _rG=\Delta _fG_{X}^{3000K}=-63.12kJ/mol

Next, since the relationship between the equilibrium constant and the Gibbs free energy of reaction is:

K=exp(-\frac{\Delta _rG}{RT} )

Thus, at each temperature we obtain:

K^{2000K}=exp(-\frac{4250J/mol}{8.314\frac{J}{mol\times K}*2000K} )=0.774\\\\K^{3000K}=exp(-\frac{-63120J/mol}{8.314\frac{J}{mol\times K}*3000K} )=12.56

In such a way, we can also conclude that at 2000 K reaction is unfavorable (K<1) and at 3000 K reaction is favorable (K>1).

Best regards.

4 0
3 years ago
A 150.0 mL sample of a 1.50 M solution of CuSO4 is mixed with a 150.0 mL sample of 3.00 M KOH in a coffee cup calorimeter. The t
svp [43]
Mols CuSO4 = M x L = 1.50 x 0.150 = 0.225 
<span>mols KOH = 3.00 x 0.150 = 0.450 </span>
<span>specific heat solns = specific heat H2O = 4.18 J/K*C </span>

<span>CuSO4 + 2KOH = Cu(OH)2 + 2H2O </span>
<span>q = mass solutions x specific heat solns x (Tfinal-Tinitial) + Ccal*deltat T </span>
<span>q = 300g x 4.18 x (31.3-25.2) + 24.2*(31.3-25.2) </span>
<span>dHrxn in J/mol= q/0.225 mol CuSO4 </span>
<span>Then convert to kJ/mol


</span>
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4 years ago
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Read the information in the table below:
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Exothermic Reactions

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4 years ago
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