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# Tutor profile: Hunter B.

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Hunter B.
Student at the University of Oklahoma
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## Questions

### Subject:Inorganic Chemistry

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Question:

Suppose an algaecide for cleaning swimming pools contains 50 grams (g) of anhydrous copper(II) sulfate, CuSO$$_4$$. If the molecular weight of copper(II) sulfate is 160 grams per mol, how many molecules of CuSO$$_4$$ are in the algaecide?

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Hunter B.

In order to answer this question, we first need to know what information we are given. From the question, we know we have 50 grams of copper(II) sulfate and we know its molar mass is 160 grams per mol. Using this information, we need to find the number of molecules in 50 grams of copper(II) sulfate. To solve this, we first need to get the grams of copper(II) sulfate into moles of copper(II) sulfate. To do this, we need to divide the known mass of copper(II) sulfate, 50 g, by the known molar mass of copper(II) sulfate, 160 g per mol, which gives us 50 g CuSO$$_4$$ $$\bullet \frac{1}{160 \frac{g}{mol}}=0.313$$ moles of copper(II) sulfate. We then need to multiply the moles of copper(II) sulfate by Avogadro's number, 6.022 X $$10^{23}$$, which gives us: 0.313 moles of copper(II) sulfate $$\bullet \frac{6.022 X 10^{23}moles \hspace{.08cm} of \hspace{.05cm} copper(II) sulfate}{1 \hspace{.08cm} mol \hspace{.05cm} of \hspace{.08cm}copper(II) \hspace{.05cm}sulfate} =$$ 1.88 X 10$$^{23}$$ molecules of copper(II) sulfate. So, there are 1.88 X 10 $$^{23}$$ molecules of copper(II) sulfate in the algaecide.

### Subject:Chemistry

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Question:

Consider the equilibrium reaction $$2A+B\rightleftharpoons 4C$$. After multiplying the reaction by a multiple of 2, what is the new equilibrium equation and what is the new equilibrium-constant expression, $$K_c$$?

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Hunter B.

To find the new equilibrium equation, we simply multiply the coefficient of each species by 2, yielding $$4A+2B \rightleftharpoons 8C$$. To find the new equilibrium-constant expression, we need to recall that the parent formula for the equilibrium-constant expression, the reaction quotient $$Q_c$$, is as follows: $$mA+nB\rightleftharpoons xC+yD$$ = $$\frac{[C]^x[D]^y}{[A]^m[B]^n}$$. Which gives us $$K_c$$ = $$\frac{[C]^8}{[A]^4[B]^2}$$.

### Subject:Algebra

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Question:

Solve the following equation for $$x$$. $$2x+5=-3x-10$$.

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Hunter B.

In order to solve this problem for $$x$$, we first need to collect all of the like terms. To collect the $$x$$ terms, we can either subtract 2$$x$$ from both sides or we can add 3$$x$$ to both sides. By adding 3$$x$$ to each side, we are left with $$5x+5=-10$$. We now need to collect the like terms without the variable $$x$$. To do this, we can either subtract 5 from each side or we can add 10 to each side. By adding 10 to each side, we are left with $$5x+15=0$$. Now, we need to separate these two terms so that we can solve for $$x$$. To do this, we subtract 15 from each side leaving us with $$5x=-15$$. Then, the last step is to solve for $$x$$. By dividing each side by 5, we are left with the answer of $$x=-3$$.

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