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Dmitriy789 [7]
3 years ago
5

+

Chemistry
2 answers:
Alexxx [7]3 years ago
7 0
The correct awnser is A- 0.174
Alex3 years ago
6 0

Answer: 0.714 moles of PbI_2 will be produced from 237.1 g of potassium iodide

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of potassium iodide}=\frac{237.1g}{166g/mol}=1.428moles

The balanced chemical reaction is:

Pb(NO_3)_2+2KI\rightarrow PbI_2+2KNO_3

According to stoichiometry :

2 moles of KI produce =  1 mole of PbI_2

Thus moles of KI will require=\frac{1}{2}\times 1.428=0.714moles  of PbI_2

Thus 0.714 moles of PbI_2 will be produced from 237.1 g of potassium iodide

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Question 4 of 10
Luba_88 [7]

Answer:

C. P = nRT

Explanation:

PV = nRT, where n is a number of moles and R is the universal gas constant, R = 8.31 J/mol ⋅ K.

Hope this helps :)

7 0
3 years ago
Which pH value represents a solution with the lowest OHM<br> ion concentration?
Arlecino [84]

pH value 1 represents a solution with the lowest OH⁻ion concentration.

<u>Explanation:</u>

pH is given by the expression as the negative logarithm to the base 10 of the concentration of hydrogen ions.

pH = -log₁₀[H⁺]

If the pH is lower than 7, pH < 7 then it is acidic

If the pH = 7, then it is neutral

If the pH > 7, then it is basic

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7 0
3 years ago
Calculate Ho298 for the process
Inga [223]

Explanation:

As per the Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

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

Sb + \frac{3}{2}Cl_2 \rightarrow SbCl_{3}    \Delta H^0_1 =  -314 kJ  ..........(1)

SbCl_{3} + Cl_2 \rightarrow SbCl_{5}    \Delta H^0_2 = -80kJ   ..............(2)

The final reaction is as follows:  

Sb + \frac{5}{2}Cl_{2} \rightarrow SbCl_{5}  \Delta H^0_3 = ?  .............(3)

Therefore, adding (1) and (2) we get the final equation (3) and value of \Delta H^{0}_{3} at 298 K will be as follows.

             \Delta H^{0}_{3} = \Delta H^{0}_{1} + \Delta H^{0}_{2}    

                       = -314 kJ + (-80) kJ

                       = -394 kJ

Thus, we can conclude that H^{o} at 298 K for the given process is -394 kJ.

4 0
3 years ago
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Ilia_Sergeevich [38]

Answer:

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Explanation:

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