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Nadya [2.5K]
3 years ago
13

When the concentration of a system is changed:

Chemistry
1 answer:
Alex Ar [27]3 years ago
8 0

Answer:

The concentrations change to maintain the original value of K.

Explanation:

The concentration changes to maintain the original value of K when the concentration of the system changes.

This is the effect concentration has on a reaction at equilibrium.

Le Chatelier's principle proposes that "if any conditions of a system in equilibrium is changed, the system will adjust itself in order to annul the effect of the change".

  • The equilibrium constant K is temperature dependent.
  • An increase in concentration of a specie favors the direction that uses it up and vice versa.
  • This does not change the value of the equilibrium constant.
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12. What is the frequency of a photon with an energy of 3.03 x 10-19 J?
bazaltina [42]

Answer:

u=4.57x10^5GHz

Explanation:

Hello.

In this case, given the formula:

E=h*u

Whereas E is the energy, h the Planck's constant and u the frequency of the photon. Thus, solving for it, we obtain:

u=\frac{E}{h}=\frac{3.03x10^{-19}J}{6.63x10^{-34}J*s}\\  \\u=4.57x10^{14}s^{-1}

Or also:

u=4.57x10^{14}Hz*\frac{1GHz}{1x10^9Hz}\\ \\u=4.57x10^5GHz

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5 0
3 years ago
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noname [10]

Answer: answer what

Explanation:

7 0
3 years ago
Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reaction:P4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°
miskamm [114]

<u>Answer:</u> The \Delta H^o_{rxn} for the reaction is -1835 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

P_4(g)+10Cl_2(g)\rightarrow 4PCl_5(s)      \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) PCl_5(s)\rightarrow PCl_3(g)+Cl_2(g)    \Delta H_1=157kJ   ( × 4)

(2) P_4(g)+6Cl_2(g)\rightarrow 4PCl_3(g)     \Delta H_2=-1207kJ

The expression for enthalpy of the reaction follows:

\Delta H^o_{rxn}=[4\times (-\Delta H_1)]+[1\times \Delta H_2]

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(4\times (-157))+(1\times (-1207))=-1835kJ

Hence, the \Delta H^o_{rxn} for the reaction is -1835 kJ.

6 0
3 years ago
List three examples of high energy waves that are harmful to living things:
lord [1]
UV rays, gamma rays, and x rays
Hope this help
5 0
3 years ago
Given the unbalanced equation below, answer the following: Calculate the number of liters of 3.00 M lead (II) iodide solution pr
mr_godi [17]

The number of liters of 3.00 M lead (II) iodide : 0.277 L

<h3>Further explanation</h3>

Reaction(balanced)

Pb(NO₃)₂(aq) + 2KI(aq) → 2KNO₃(aq) + PbI₂(s)

moles of KI = 1.66

From the equation, mol ratio of KI : PbI₂ = 2 : 1, so mol PbI₂ :

\tt \dfrac{1}{2}\times 1.66=0.83

Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution

\large \boxed {\bold {M ~ = ~ \dfrac {n} {V}}}

Where

M = Molarity

n = Number of moles of solute

V = Volume of solution

So the number of liters(V) of 3.00 M lead (II) iodide-PbI₂ (n=0.83, M=3):

\tt V=\dfrac{n}{M}\\\\V=\dfrac{0.83}{3}\\\\V=0.277~L

5 0
3 years ago
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