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Diano4ka-milaya [45]
3 years ago
7

What is the difference between soluble/insoluble and miscible/immiscible?

Chemistry
1 answer:
pashok25 [27]3 years ago
3 0
Soluble means it’s able to be dissolved while insoluble means it’s not. Miscible means it is able to form a homogenous mixture with another miscible object, immiscible means the opposite.

hope that helps!
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konstantin123 [22]
When a change in PH = 10^-ΔPH
so the change = 10^-3.2
change depends on two factor 0.00063 (10^-3.2)and factor 1585 (10^3.2)depending on the way which the change goes.if PH change from PH=0 to PH= 3.2 so the change is decreasing from concentration from 1 mol to 0.00063 and if PH change from PH = 3.2 to PH=0 so the change is increasing by a factor of 1585.
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A person runs 500 ft in 50 seconds how fast are they moving
Nezavi [6.7K]

10 miles per hour because 500 divided by 50 is 10

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Giving brainliest please help with this question
siniylev [52]

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Explanation:

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3 years ago
Given these reactions, X ( s ) + 1 2 O 2 ( g ) ⟶ XO ( s ) Δ H = − 668.5 k J / m o l XCO 3 ( s ) ⟶ XO ( s ) + CO 2 ( g ) Δ H = +
qwelly [4]

<u>Answer:</u> The \Delta H^o_{rxn} for the reaction is -1052.8 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:

X(s)+\frac{1}{2}O_2(g)+CO_2(g)\rightarrow XCO_3(s)      \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) X(s)+\frac{1}{2}O_2(g)\rightarrow XO(s)    \Delta H_1=-668.5kJ

(2) XCO_3(s)\rightarrow XO(s)+CO_2     \Delta H_2=+384.3kJ

The expression for enthalpy of the reaction follows:

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

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-668.5))+(1\times (-384.3))=-1052.8kJ

Hence, the \Delta H^o_{rxn} for the reaction is -1052.8 kJ.

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3 years ago
How many moles are in 155 grams of aluminum
Aliun [14]

Answer:

wads

Explanation:

wadsdw

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