The proportion of the sodium carbonate, $p$, which has dissolved by time $t$ seconds is given by the formula $ p=\\frac{bt-at^2}{c}$. The aim is to calculate the proportion of sodium carbonate in a solution at a given time and vice versa.
", "licence": "None specified"}, "statement": "A small volume of sodium carbonate is added to a beaker of water. The proportion of the sodium carbonate, $p$, which has dissolved by time $t$ seconds is given by:
\n\\[p=\\frac{\\var{b}t-\\var{a}t^2}{\\var{c}}\\]
", "advice": "a) We need to substitute $t=\\var{t_1}$
\\[ p=\\frac{\\var{b}\\times\\var{t_1}-\\var{a}\\times \\var{t_1}^2}{\\var{c}}\\\\ \\implies p=\\frac{\\var{b}\\times\\var{t_1}-\\var{a}\\times \\var{t_1^2}}{\\var{c}} \\\\ \\implies p=\\frac{\\simplify{{b}*{t_1}}-\\simplify{{a}*{t_1^2}}}{\\var{c}} \\\\ \\implies p=\\frac{\\simplify{{b}*{t_1}-{a}*{t_1^2}}}{\\var{c}} \\\\ \\implies p=\\simplify{{b*t_1-a*t_1^2}/{c}} ~~~~ \\text{or} ~~~~ \\var{ans_a_rounded}~\\text{(2d.p.)} \\]
b) We first need to transform the percentage value which we were given into a proportion written as a decimal.
\n\\[ \\var{p_1*100}\\%=\\var{p_1}\\]
\nThen, substituting $p=\\var{p_1}$ we get the following equation to solve:
\n\\[ \\var{p_1}=\\frac{\\var{b}t-\\var{a}t^2}{\\var{c}} \\\\ \\implies \\var{c}\\times
\\var{p_1}=\\var{b}t-\\var{a}t^2 \\\\ \\implies \\simplify[unitFactor]{{a}t^2-{b}t+{c*p_1}}=0\\]
Using the quadratic formula
\n\\[ t=\\frac{-b\\pm \\sqrt{b^2-4a c}}{2a}\\]
\nwe get the solutions:
\n\\[ t_1=\\var{anst_1_rounded} \\\\ t_2=\\var{anst_2_rounded}\\]
\nHowever, we are solving the equation to find a time which is $0\\leq p\\leq \\var{tmax}$. Therefore, the final answer is only $t=\\var{ans_t}$.
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\n$p=$ [[0]]
\nGive your answer rounded to 2 decimal places.
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\n$t=$ [[0]] sec.
\nGive your answer rounded to 2 decimal places.
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