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Ray Of Light [21]
4 years ago
8

HELPPP The sum of percentage compositions must equal to 100 ±0.1. True False

Chemistry
2 answers:
Iteru [2.4K]4 years ago
4 0
False it cannot be above 100%
Diano4ka-milaya [45]4 years ago
3 0

Answer:

True

Explanation:

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Glittering in the light
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Manganese metal can be obtained by the reaction of manganese dioxide with aluminium. 4Al(s) + 3MnOz(s) → 2 Al2O3(s) + 3 Mn(s) Ca
Snezhnost [94]

Answer : The enthalpy change of reaction is -1800 kJ

Explanation :

According to 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.

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. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given final reaction is,

4Al(s)+3MnO_2(s)\rightarrow 2Al_2O_3(s)+3Mn(s)    \Delta H=?

The intermediate balanced chemical reaction will be,

(1) 2Al(s)+\frac{3}{2}O_2(g)\rightarrow Al_2O_3(s)    \Delta H_1=-1680kJ/mole

(2) Mn(s)+O_2(g)\rightarrow MnO_2(s)    \Delta H_2=-520kJ/mole

First we will multiply reaction 1  by 2 and reverse reaction of reaction 2 by 3 then adding both the equation, we get :

The expression for final enthalpy is,

\Delta H=[n\times \Delta H_1]+[n\times (-\Delta H_2)]

where,

n = number of moles

\Delta H=[2mole\times (-1680kJ/mole)]+[3\times -(-520kJ/mole)]

\Delta H=-1800kJ

Therefore, the enthalpy change of reaction is -1800 kJ

8 0
3 years ago
Find an expression for the change in entropy when two blocks of the same substance of equal mass, one at the temperature Th and
Gre4nikov [31]

Explanation:

Relation between entropy change and specific heat is as follows.

            \Delta S = C_{p} log (\frac{T_{2}}{T_{1}})

The given data is as follows.

     mass = 500 g,         C_{p} = 24.4 J/mol K

     T_{h} = 500 K,          T_{c} = 250 K               

   Mass number of copper = 63.54 g /mol

Number of moles = \frac{mass}{/text{\molar mass}}

                                 = \frac{500}{63.54}

                                 = 7.86 moles

Now, equating the entropy change for both the substances as follows.

     7.86 \times 24.4 \times [T_{f} - 250] = 7.86 \times 24.4 \times [500 -T_{f}]

       T_{f} - 250 = 500 - T_{f}

          2T_{f} = 750

So,       T_{f} = 375^{o}C

  • For the metal block A,  change in entropy is as follows.

         \Delta S = C_{p} log (\frac{T_{2}}{T_{1}})

              = 24.4 log [\frac{375}{500}]

              = -3.04 J/ K mol

  • For the block B,  change in entropy is as follows.

         \Delta S = C_{p} log (\frac{T_{2}}{T_{1}})

                  = 24.4 log [\frac{375}{250}]

                  = 4.296  J/Kmol

And, total entropy change will be as follows.

                       = 4.296 + (-3.04)

                      = 1.256 J/Kmol

Thus, we can conclude that change in entropy of block A is -3.04 J/ K mol  and change in entropy of block B is 4.296  J/Kmol.

8 0
3 years ago
Determine the final concentration when 250mL of 3M solution is dilluted to a final volume of 850mL. (Enter Number)​
LenKa [72]

Answer:

M₂ = 0.9 M

Explanation:

Given data:

Initial Volume = V₁ = 250 mL

Initial molarity = M₁= 3 M

Final volume = 850 mL

Final molarity = ?

Solution:

M₁V₁ = M₂V₂

M₂ = M₁V₁ /V₂

M₂ = 3 M × 250 mL / 850 mL

M₂ = 750 M / 850

M₂ = 0.9 M

7 0
4 years ago
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