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Anit [1.1K]
2 years ago
9

determine the maximum amount of NaN03 that was produced during the experiment. Explain how you determined the amount

Chemistry
1 answer:
Angelina_Jolie [31]2 years ago
7 0

Answer:

9 moles of NaNO3 is obtained

Explanation:

The balanced chemical reaction equation for the reaction is;

Al(NO3)3 + 3NaCl-------> 3NaNO3 + AlCl3

Now, we have to determine the limiting reactant. The limiting reactant yields the least amount of NaNO3.

1 mole of Al(NO3)3 yields 3 moles of NaNO3

4 moles of Al(NO3)3 yields 4 * 3/1 = 12 moles of NaNO3

Also,

3 moles of NaCl yields 3 moles of NaNO3

9 moles of NaCl yields 9 * 3/3 = 9 moles of NaNO3

Hence, NaCl  is the limiting reactant and 9 moles of NaNO3 is obtained.

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The Bohr atomic model, relying on quantum mechanics, built upon the Rutherford model to explain the orbits of electrons.
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2 years ago
Un anillo, de masa 90 gramos, contiene 59,1% de oro. ¿Cuál es el valor del anillo si cada mol de oro vale S/.1800?. Considere el
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Answer:

S/.486 es el valor del anillo

Explanation:

Para hallar el precio del anillo se deben encontrar las moles de oro que contiene este.

Si el anillo es de 90g y solo el 59.1% contiene oro, la cantidad de oro en gramos es:

90g × 59.1% = 53.19g Oro en el anillo

Ahora, para convertir los gramos de oro a moles se debe usar la masa atómica del oro (197g/mol), así:

53.19g × (1mol / 197g) = <em><u>0.27 moles de oro contiene el anillo</u></em>.

Ya que cada mol de oro cuesta S/.1800, 0.27 moles de oro (Y por lo tanto, el anillo) costarán:

0.27mol × (S/.1800 / 1mol oro) =

<h3>S/.486 es el valor del anillo</h3>
8 0
2 years ago
A gas cylinder of volume 5.00 l contains 1.00 g of ar and 0.500 g of ne. the temperature is 275 k. find the partial pressure of
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<span>11.3 kPa The ideal gas law is PV = nRT where P = Pressure V = Volume n = number of moles R = Ideal gas constant (8.3144598 L*kPa/(K*mol) ) T = Absolute temperature We have everything except moles and volume. But we can calculate moles by starting with the atomic weight of argon and neon. Atomic weight argon = 39.948 Atomic weight neon = 20.1797 Moles Ar = 1.00 g / 39.948 g/mol = 0.025032542 mol Moles Ne = 0.500 g / 20.1797 g/mol = 0.024777375 mol Total moles gas particles = 0.025032542 mol + 0.024777375 mol = 0.049809918 mol Now take the ideal gas equation and solve for P, then substitute known values and solve. PV = nRT P = nRT/V P = 0.049809918 mol * 8.3144598 L*kPa/(K*mol) * 275 K/5.00 L P = 113.8892033 L*kPa / 5.00 L P = 22.77784066 kPa Now let's determine the percent of pressure provided by neon by calculating the percentage of neon atoms. Divide the number of moles of neon by the total number of moles. 0.024777375 mol / 0.049809918 mol = 0.497438592 Now multiply by the pressure 0.497438592 * 22.77784066 kPa = 11.33057699 kPa Round the result to 3 significant figures, giving 11.3 kPa</span>
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How do you prove law of conservation of mass??
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3 0
3 years ago
Acetylene burns in air according to the following equation: C2H2(g) + 5 2 O2(g) → 2 CO2(g) + H2O(g) ΔH o rxn = −1255.8 kJ Given
professor190 [17]

Answer:  -227 kJ

Explanation:

The balanced chemical reaction is,

C_2H_2(g)+\frac{5}{2}O_2(g)\rightarrow 2CO_2(g)+H_2O(g)

The expression for enthalpy change is,

\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+ n_{H_2O}\times \Delta H_{H_2O})]-[(n_{C_2H_2}\times \Delta H_{C_2H_2})+(n_{O_2}\times \Delta H_{O_2})]

where,

n = number of moles

\Delta H_{O_2}=0 (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

-1255.8=[(2\times -393.5)+(1\times -241.8)]-[(1\times \Delta H_{C_2H_2})+(\frac{5}{2}\times 0)]

-1255.8=[(-787)+(-241.8)]-[(1\times \Delta H_{C_2H_2})+(0)]

\Delta H_{C_2H_2}=-227kJ

Therefore, the enthalpy change for C_2H_2 is -227 kJ.

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