Michelle W.

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Basic Chemistry

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Question:

For the reaction: $$2CO + O_2 \to 2CO_2$$ What is the theoretical yield of $$CO_2$$ in grams from $$5g \, O_2$$ and excess $$CO$$?

Michelle W.

Answer:

We start with the $$5g$$ of $$O_2$$. Next, we convert to moles using the molecular weight, $$32g/mol$$, from the periodic table. Then, we convert from moles of $$O_2$$ to moles of $$CO_2$$ using the ratios in the chemical equation given in the problem. Lastly, we convert from moles of $$CO_2$$ to grams of $$CO_2$$ using its molecular weight, $$44g/mol$$. This process is shown in the equation below, resulting in an answer of $$13.75 g \, CO_2$$: $$\frac{5g \, O_2}{}\times\frac{mol \, O_2}{32g \, O_2}\times\frac{2mol \, CO2}{1 mol \,O_2}\times\frac{44g\, CO_2}{mol\, CO_2} = 13.75 g \, CO_2$$

Calculus

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Question:

Evaluate the integral: $$\int\limits_\pi^{2\pi} cos(x) + 2x\, dx$$

Michelle W.

Answer:

Integrate both terms: $$sin(x) + x^2 |_\pi^{2\pi}$$ Plug in the limits: $$sin(2\pi) + (2\pi)^2 - (sin(\pi) + (\pi)^2)$$ Simplify terms: $$0 + 4\pi^2 - (0 + \pi^2)$$ Subtract: $$3\pi^2$$

Algebra

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Question:

Solve the following equation for $$x$$: $$ 3x^2 + 2(x+2) - 2 = 3 $$

Michelle W.

Answer:

Distribute over parentheses: $$3x^2 + 2x + 4 - 2 = 3 $$ Combine terms on the left: $$3x^2 + 2x + 2 = 3$$ Subtract 3 from both sides: $$3x^2 + 2x - 1 = 0 $$ Factorize: $$(3x - 1)(x + 1) = 0$$ Solve left parentheses for x: $$3x - 1 = 0$$ $$3x = 1$$ $$x = 1/3$$ Solve right parenthesis for x: $$x + 1 = 0$$ $$x = -1$$ Therefore, $$x = -1, 1/3$$

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