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2 years ago

Find the mass in grams of 1.40x10^23 molecules of n2

1 answer:

7 0

<span>Avogadro's number represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23 units / mole. This number can be used to convert the number of atoms or molecules into number of moles. We calculate as follows:

</span>1.40x10^23 molecules of N2 ( 1 mol / 6.022 x 10^23 molecules ) ( 28.02 g / mol ) = 6.51 g N2

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For each pair of gases, select one that most likely has the highest rate of effusion. Use the periodic table if necessary

brilliants [131]

You will do the highest one i think

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2 years ago

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2CH4(g)⟶C2H4(g)+2H2(g)

Rasek [7]

Answer : The enthalpy change for the reaction is, 201.9 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 balanced reaction of $CH_4$ will be,

$2CH_4(g)\rightarrow C_2H_4(g)+2H_2(g)$ $\Delta H^o=?$

The intermediate balanced chemical reaction will be,

(1) $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)$ $\Delta H_1=-890.3kJ$

(2) $C_2H_4(g)+H_2(g)\rightarrow C_2H_6(g)$ $\Delta H_2=-136.3kJ$

(3) $2H_2(g)+O_2(g)\rightarrow 2H_2O(l)$ $\Delta H_3=-571.6kJ$

(4) $2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(l)$ $\Delta H_4=-3120.8kJ$

Now we will multiply the reaction 1 by 2, revere the reaction 2, reverse and half the reaction 3 and 4 then adding all the equations, we get :

(1) $2CH_4(g)+4O_2(g)\rightarrow 2CO_2(g)+4H_2O(l)$ $\Delta H_1=2\times (-890.3kJ)=-1780.6kJ$

(2) $C_2H_6(g)\rightarrow C_2H_4(g)+H_2(g)$ $\Delta H_2=-(-136.3kJ)=136.3kJ$

(3) $H_2O(l)\rightarrow H_2(g)+\frac{1}{2}O_2(g)$ $\Delta H_3=-\frac{1}{2}\times (-571.6kJ)=285.8kJ$

(4) $2CO_2(g)+3H_2O(l)\rightarrow C_2H_6(g)+\frac{7}{2}O_2(g)$ $\Delta H_4=-\frac{1}{2}\times (-3120.8kJ)=1560.4kJ$

The expression for enthalpy of the reaction will be,

$\Delta H^o=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4$

$\Delta H=(-1780.6kJ)+(136.3kJ)+(285.8kJ)+(1560.4kJ)$

$\Delta H=201.9kJ$

Therefore, the enthalpy change for the reaction is, 201.9 kJ

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2 years ago

What is the mass in grams of 7.5 x 10^15 atoms of nickel?

satela [25.4K]

Atomic mass Ni = 58.69 a.m.u

58.69 g ----------------- 6.02x10²³ atoms
?? g --------------------- 7.5x10¹⁵ atoms

58.69x (7.5x10¹⁵) / 6.02x10²³

=> 7.31x10⁻⁷ g

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The volume of blood plasma in adults is 3.1 L. it's density is 1.03 g/cm3. Approximately how many pounds of blood plasma are the

e-lub [12.9K]

The important thing in this question is the unit. The mass equals density * volume. 3.1 L = 3.1 * 10^3 cm3. So the mass is 3.193*10^3 g. 1 pound = 453.95 g. So the answer is 7.04 pounds.

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2 years ago

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10. A solution contains 130 grams of KNO3 dissolved in 100 grams of water When 3 more grams of KNO3 is added, none of it dissolv

Murljashka [212]

Answer:

Option B

Explanation:

We will check the solubility graph for potassium nitrate, KNO 3. Based on the graph it can be said that the temperature of solution when 130 grams of KNO3 dissolves in 100 grams of water is near to 65 degree Celsius. Now if three grams of solute is increased then the temperature of the solution will increase by a degree or so and hence the most probable temperature would be 68 degree Celsius.

Hence, option B is correct

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2 years ago

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