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Tutor profile: Keller J.

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Keller J.
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Questions

Subject: Spanish

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Question:

Completa el párrafo con la forma correcta de los verbos en la lista: Ser, Depender, tener, Generar. El calentamiento global actual (1) un tipo de cambio climático sin precedentes, y está (2) una cascada de efectos secundarios en nuestro sistema climático. Son estos efectos secundarios, como los cambios en el nivel del mar a lo largo de costas muy pobladas y la retirada mundial de los glaciares de montaña de los que (3) millones de personas para el agua potable y la agricultura, que probablemente (4) un impacto mucho mayor en la sociedad que el cambio de temperatura.

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Keller J.
Answer:

El calentamiento global actual $$ \underline{\bf{es}} $$ (1) un tipo de cambio climático sin precedentes, y está $$ \underline{\bf{generando}} $$ (2) una cascada de efectos secundarios en nuestro sistema climático. Son estos efectos secundarios, como los cambios en el nivel del mar a lo largo de costas muy pobladas y la retirada mundial de los glaciares de montaña de los que $$ \underline{\bf{dependen}} $$ (3) millones de personas para el agua potable y la agricultura, que probablemente $$ \underline{\bf{tengan}} $$ (4) un impacto mucho mayor en la sociedad que el cambio de temperatura.

Subject: Chemistry

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Question:

A sample of $$ O_2 $$ has a volume of 3.52 L at 27°C and 800 torr. How many $$ O_2 $$ molecules does it contain?

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Keller J.
Answer:

Given the formula of the ideal gases: $$ PV = nRT $$ Solve "n" $$ n = \frac{PV}{RT} $$ Values: $$ P = 800\, torr \times ​\frac{1\, atm}{760\, torr} = 1.0526\, atm $$ $$ V = 3.52\, L $$ $$ T = 27°C + 273 = 300°K $$ $$ R = 0.08206 \frac{atm \times L}{mol \times °K} $$ Replacing:  $$ n = \frac{1.0526\, atm \times 3.52\, L}{0.08206 \frac{atm \times L}{mol \times °K} \times 300°K} = 0.1505\, mol\, of\, O_2 $$ Convert $$ 0.1505\, mol\, of\, O_2 \, to\, molecules\, of\, O_2 $$ $$ 0.1505\, mol\, of\, O_2 \times \frac{6.022 \times 10^{23}\, molecules\, of\, O_2}{1\, mol} = 9.063 \times 10^{22}\, molecules\, of\, O_2$$ The sample contain $$ 9.063 \times 10^{22}\, molecules\, of\, O_2$$

Subject: Basic Chemistry

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Question:

How many liters of a 0.300 M of sucrose $$ (C_{12} H_{22} O_{11} ) $$ solution contain 1.2 kg of sucrose?

Inactive
Keller J.
Answer:

First calculate the molar mass of sucrose: $$ (C_{12} H_{22} O_{11} ) $$ Where: $$ C = 12\, atoms \times 12\, g/mole = 144\, \frac{g}{mole} $$ $$ H = 22\, atoms \times 1\, g/mole = 22\, \frac{g}{mole} $$ $$ O = 11\, atoms \times 16\, g/mole = 176\, \frac{g}{mole} $$ Then: $$ M.M_{sucrose} = 144\, \frac{g}{mole} + 22\, \frac{g}{mole} + 176\. \frac{g}{mole} = 342\, \frac{g}{mole} $$ Next convert 1.2 kg of sucrose to moles of sucrose: $$ 1.2\, Kg\, of\, sucrose \times \frac{1000\, g\, of\, sucrose}{1\, Kg\, of\, sucrose} \times \frac{1\, mole\, of\, sucrose}{342\, g\, of\, sucrose} = 3.51\, moles\, of\, sucrose $$ Finally, given the formula to calculate the molarity: $$ M = \frac{n}{V (L)} $$ Solve "V" for this: $$ V = \frac{n}{M}$$ Replace the values: $$ V = \frac{3.51\, moles\, of\, sucrose}{0.300 \frac{moles\, of\, sucrose}{L}} = 11.7\, L $$ 11.7 L of sucrose solution contain 1.2 Kg of sucrose.

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