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# Tutor profile: Desiree O.

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Desiree O.
Student of Clemson University Studying Environmental Engineering
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## Questions

### Subject:MATLAB

TutorMe
Question:

Write a program where a user will get 3 chances to enter a 1x5 vector. If the user inputs a 5x1 vector, the vector will be transposed and a warning will be issued that says "This vector is 5x1 not a 1x5! It will be fixed". If the user enters a vector with any other incorrect dimensions, the program will give the user a total of 3 tries ask the user to input the correct vector. The user does not, the program will be terminated with a message that says "You have run out of tries".

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Desiree O.

clear clc vector=input('Enter a 1x5 vector of numbers: '); [R,C]=size(vector); if R==5 && C==1 warning('This vector is 5x1 not a 1x5! It will be fixed'); vector=vector'; [R,C]=size(vector); end c=1; while R~=1 || C~=5 if c==3 && (R~=1 || C~=5) error('You have run out of tries'); end if R==5 && C==1 warning('This vector is 5x1 not a 1x5! It will be fixed'); vector=vector'; [R,C]=size(vector); end if R~=1 || C~=5 warning('This vector does not have the right dimensions, try again!'); vector=input('Enter a 1x5 vector of numbers: '); [R,C]=size(vector); end end

### Subject:Chemistry

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Question:

You have $$25 mL$$ of $$0.05M$$ $$Na_2CO_3$$ and $$10 mL$$ of $$0.002M$$ $$Ca(NO_3)_2$$ mixed together. Will a precipitate form? The likely precipitate, if any, is $$CaCO_3$$. $$K_{sp}=8.7\times 10^{-9}$$

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Desiree O.

$$CaCO_{3(s)} \longleftrightarrow Ca^{2+}_{(aq)} + CO^{2-}_{3(aq)}$$ $$K_{sp}= [Ca^{+2}][CO^{2-}_{3}] =8.7\times 10^{-9}$$ *the concentration of a pure solid or liquid is not included in the equilibrium expression $$25mL$$ of $$0.05M$$ of $$NaCO_3$$ $$0.025L\times\frac{0.05 mol} {L}$$ = $$0.00125 mols$$ $$NaCO_3$$ = $$0.00125 mols$$ $$CO^{2-}_3$$ $$10mL$$ of $$0.002M$$ of $$Ca(NO_3)_2$$ $$0.01L\times\frac{0.002 mol} {L}$$ = $$2\times 10^{-5} mols$$ $$Ca(NO_3)_2$$ = $$2\times 10^{-5} mols$$ $$Ca^{+2}$$ The total volume is $$25mL + 10mL = 35mL$$ $$[CO^{-2}_3] = \frac {0.00125mol} {0.035L} = 0.0357M\quad$$ $$[Ca^{+2}] = \frac {2\times 10^{-5} mols} {0.035L} = 5.714\times 10^{-4}M$$ Find the reaction quotient, $$Q$$, which has the same algebraic form as $$K_{sp}$$, but it is evaluated with current concentrations and not equilibrium concentrations, so $$Q = [Ca^{+2}][CO^{2-}_{3}] = (5.714\times 10^{-4}) (0.0357) = 2.04\times 10^{-5}$$ Since $$Q>K_{sp}$$, the reaction moves from right to left, so precipitate IS being formed.

### Subject:Calculus

TutorMe
Question:

Find the instantaneous rate of change of $$f(x)=x^2$$ at $$x=3$$ using limits.

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Desiree O.

The instantaneous rate of change of $$f$$ at $$x=a$$ is $$\lim\limits_{x\to a} \displaystyle\frac{f(x)-f(a)} {x-a}$$. $$f(x)=x^2,\quad x=3$$ $$\lim\limits_ {x\to 3} \displaystyle\frac{f(x)-f(3)} {x-3}$$ $$\lim\limits_ {x\to 3} \displaystyle\frac {x^2-9} {x-3}$$ $$\lim\limits_ {x\to 3} \displaystyle\frac {(x+3)(x-3)} {x-3}$$ $$\lim\limits_ {x\to 3} \displaystyle (x+3)$$ $$\lim\limits_ {x\to 3} \displaystyle (x+3) = 6$$

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