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The rate of a certain chemical reaction is directly proportional to th
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27 Nov 2010, 18:06
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The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present. If the concentration of chemical B is increased by 100 percent, which of the following is closest to the percent change in the concentration of chemical A required to keep the reaction rate unchanged? (A) 100% decrease (B) 50% decrease (C) 40% decrease (D) 40% increase (E) 50% increase
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Re: The rate of a certain chemical reaction is directly proportional to th
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28 Nov 2010, 02:12
The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present. If the concentration of chemical B is increased by 100 percent, which of the following is closest to the percent change in the concentration of chemical A required to keep the reaction rate unchanged?(A) 100% decrease (B) 50% decrease (C) 40% decrease (D) 40% increase (E) 50% increase NOTE: Put directly proportional in nominator and inversely proportional in denominator. \(RATE=\frac{A^2}{B}\), (well as it's not the exact fraction it should be multiplied by some constant but we can ignore this in our case). We are told that B increased by 100%, hence in denominator we have 2B. We want the rate to be the same. As rate is directly proportional to the SQUARE of A, A should also increase (nominator) by x percent and increase of A in square should be 2. Which means x^2=2, x=~1.41, which is approximately 40% increase. \(R=\frac{A^2}{B}=\frac{(1.4A)^2}{2B}\) Answer: D. To elaborate more: \(a\) is directly proportional to \(b\) means that as the absolute value of \(b\) gets bigger, the absolute value of \(b\) gets bigger too, so there is some nonzero constant \(x\) such that \(a=xb\); \(a\) is inversely proportional to \(b\) means that as the absolute value of \(b\) gets bigger, the absolute value of \(a\) gets smaller, so there is some nonzero constant constant \(y\) such that \(a=\frac{y}{b}\). So, when we are told that the rate (R) is directly proportional to the square of A and inversely proportional to B we can write \(R=\frac{(A^2x)y}{B}\). Hope it's clear.
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Re: The rate of a certain chemical reaction is directly proportional to th
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26 Dec 2016, 14:26
This question can also be solved by plugging in values for a and b and comparing the results:  The rate of this reaction at first is \(r=\frac{A^2}{B}\)  If \(A=1\) and \(B=2\) the result of the equation is \(r=\frac{1}{2}\)  When \(B\) is increased by 100% the equation becomes \(r=\frac{A^2}{2B}\)  Since the question asks by how much \(A\) would need to increase in order for the reaction rate to remain the same, the second equation can be rewritten as \(\frac{1}{2}=\frac{A^2}{4}\) where \(r=\frac{1}{2}\), the same as in the first equation.  The second equation can then be rewritten as \(2=A^2\) and then as \(\sqrt{2}=A\)  The percent increase from \(A=1\) in the first equation to \(A=\sqrt{2}\) in the second equation is approximately 40% since\(\sqrt{2}=1.4\)  The answer is




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The rate of a certain chemical reaction is directly proportional to th
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08 Oct 2019, 10:38
afyl128 wrote: The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present. If the concentration of chemical B is increased by 100 percent, which of the following is closest to the percent change in the concentration of chemical A required to keep the reaction rate unchanged?
(A) 100% decrease (B) 50% decrease (C) 40% decrease (D) 40% increase (E) 50% increase x is directly proportional to y, translated into math: \(x = ky\), where k is a constant. x is inversely proportional to z, translated into math: \(x = \frac{k}{z}\), where is k is a constant. x is directly proportional to y and inversely proportional to z, translated into math: \(x = \frac{ky}{z}\), where k is a constant. In the problem above, the rate is directly proportional to the square of A and inversely proportional to B. Thus: \(R = \frac{{kA²}}{B}\) Original values:Let: A = 10 B = 1 k = 1 Then: \(R = \frac{1*10²}{1}\) R = 100 New values:Since the rate doesn't change, R = 100. B increased by 100% = 2. k = 1. (Since k is a constant.) Solving for A, we get: \(100 = \frac{1*A²}{2}\) \(200 = A²\) \(A = √200 = √100 * √2 = 10√2 ≈ 10(1.4) = 14\) Percent increase in A:(increase in A)/(original A) \(= \frac{1410}{10} = \frac{4}{10} =\) 40%
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Re: The rate of a certain chemical reaction is directly proportional to th
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07 Sep 2017, 16:32
Accountant wrote: The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present. If the concentration of chemical B is increased by 100%, which of the following is closest to the percent change in the concentration of chemical A required to keep the reaction rate unchanged?
A. 100% decrease B. 50% decrease C. 40% decrease D. 40% increase E. 50% increase We can let n = the rate of a certain chemical reaction, a = the concentration of chemical A, and b = the concentration of chemical B. We are given that the rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present, so, for some positive constant k, we have: n = ka^2/b When b is increased by 100 percent, b becomes 2b. To keep the reaction rate unchanged, we can let a become c, so we have: ka^2/b = kc^2/(2b) 2bka^2 = bkc^2 2a^2 = c^2 c = √(2a^2) c = a√2 Since √2 ≈ 1.4, c ≈ 1.4a or approximately 140% of a, i.e., a 40% increase in the concentration of chemical A. Answer: D
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Re: The rate of a certain chemical reaction is directly proportional to th
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28 Nov 2010, 09:56
Bunuel wrote: afyl128 wrote: Bunuel wrote: NOTE: Put directly proportional in nominator and inversely proportional in denominator. \(RATE=\frac{A^2}{B}\), (well as it's not the exact fraction it should be multiplied by some constant but we can ignore this in our case).
We are told that B increased by 100%, hence in denominator we have 2B. We want the rate to be the same. As rate is directly proportional to the SQUARE of A, A should also increase (nominator) by x percent and increase of A in square should be 2. Which means x^2=2, x=~1.41, which is approximately 40% increase. \(R=\frac{A^2}{B}=\frac{(1.4A)^2}{2B}\)
Answer: D. Brunel, why do you put the directly proportional in the nominator and inversely proportional in the denoimnator? \(a\) is directly proportional to \(b\) means that as the absolute value of \(a\) gets bigger, the absolute value of \(b\) gets bigger too, so there is some nonzero constant \(x\) such that \(a=xb\); \(a\) is inversely proportional to \(b\) means that as the absolute value of \(a\) gets bigger, the absolute value of \(b\) gets smaller, so there is some nonzero constant constant \(y\) such that \(a=\frac{y}{b}\). So, when we are told that the rate (R) is directly proportional to the square of A and inversely proportional to B we can write \(R=\frac{(A^2x)y}{B}\). Hope it's clear. Thanks =) do you have any links to similar questions? i'm very shaky on these



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Re: The rate of a certain chemical reaction is directly proportional to th
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28 Nov 2010, 10:27
afyl128 wrote: Thanks =) do you have any links to similar questions? i'm very shaky on these
DS problems about this concept: dsquestion93667.htmlogproportionalindex63570.htmlPS problems about this concept: easyproportionquestion88971.htmlvic80941.htmlHope it helps.
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Re: The rate of a certain chemical reaction is directly proportional to th
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30 Sep 2018, 09:21
Accountant wrote: The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present. If the concentration of chemical B is increased by 100%, which of the following is closest to the percent change in the concentration of chemical A required to keep the reaction rate unchanged?
A. 100% decrease B. 50% decrease C. 40% decrease D. 40% increase E. 50% increase
\(rate = {\text{cte}} \cdot \frac{{{{\left[ A \right]}^2}}}{{\left[ B \right]}}\,\,\,\,\,\left( {{\text{cte}} \ne 0} \right)\,\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{rate}}\,\,{\text{unchanged}}} \,\,\,\,\,{\text{cte}} \cdot \frac{{{{\left[ {{A_{\,\,2}}} \right]}^2}}}{{\left[ {{B_{\,2}}} \right]}} = {\text{cte}} \cdot \frac{{{{\left[ {{A_{\,\,1}}} \right]}^2}}}{{\left[ {{B_{\,1}}} \right]}}\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{cte}}\,\, \ne \,\,{\text{0}}} \,\,\,\,\,\,{\left( {\frac{{\left[ {{A_{\,2}}} \right]}}{{\left[ {{A_{\,1}}} \right]}}} \right)^{\,2}} = \frac{{\left[ {{B_{\,2}}} \right]}}{{\left[ {{B_{\,1}}} \right]}}\) \(\left[ {{B_{\,2}}} \right] = \left[ {2{B_{\,1}}} \right]\,\,\,\,\,\,;\,\,\,\,\,\,\left[ {{A_{\,2}}} \right] = \left[ {k{A_{\,1}}} \right]\,\,\,\,\,\,\,\left( {k > 0} \right)\) \(?\,\, \cong \,\,k  1\) \({\left( {\frac{{\left[ {{A_{\,2}}} \right]}}{{\left[ {{A_{\,1}}} \right]}}} \right)^{\,2}} = \frac{{\left[ {{B_{\,2}}} \right]}}{{\left[ {{B_{\,1}}} \right]}}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\left\{ \begin{gathered} {k^2} = 2 \hfill \\ \sqrt 2 \cong 1.41 \hfill \\ \end{gathered} \right.\,\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{k\,\, > \,\,0} \,\,\,\,\,\,?\,\, = \,\,\sqrt 2  1\,\,\, \cong \,\,\,0.41\,\,\, = \,\,\,41\% \,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\sqrt 2  1\,\, > 0\,\,\,\, \Rightarrow \,\,\,{\text{increase}}} \right)\) This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio.
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Re: The rate of a certain chemical reaction is directly proportional to th
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Re: The rate of a certain chemical reaction is directly proportional to th
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08 Oct 2019, 10:16
afyl128 wrote: The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present. If the concentration of chemical B is increased by 100 percent, which of the following is closest to the percent change in the concentration of chemical A required to keep the reaction rate unchanged?
(A) 100% decrease (B) 50% decrease (C) 40% decrease (D) 40% increase (E) 50% increase Given: The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present. Asked: If the concentration of chemical B is increased by 100 percent, which of the following is closest to the percent change in the concentration of chemical A required to keep the reaction rate unchanged? Rate of chemical reaction = constant * (concentration of chemical A)^2 / (concentration of chemical B) r = k * Ca^2 / Cb Cb' = 2Cb r = k Ca'^2/2Cb Ca'^2 = 2Ca^2 Ca' = \sqrt 2 * Ca Ca' = 1.4 * Ca 40% increase in concentration of chemical A IMO D
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Re: The rate of a certain chemical reaction is directly proportional to th
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08 Oct 2019, 12:43
Given: r = (k.(Ca)^2)/Cb (where Ca is conc. of chemical A and Cb is conc. of chemical B) Cb increases by 100% => Cb' = 2Cb (Cb'is the new conc. of chemical B) r is same => (Ca)^2 / Cb = (Ca')^2/Cb' (Ca'is the new conc. of chemical A) Cb' = 2Cb (Ca')^2 = 2(Ca)^2 => Ca' = 1.41Ca (Implying 40% increase in existing conc. of chemical A) Ans: D
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Re: The rate of a certain chemical reaction is directly proportional to th
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05 Nov 2019, 03:03
Hi Bunuel ScottTargetTestPrep Could anyone clear my small doubt? Why did you guys ignore the constant? Nonetheless, isn't there a possibility for two different constants for both A and B? Thank you, Dablu



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Re: The rate of a certain chemical reaction is directly proportional to th
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07 Nov 2019, 11:06
gurudabl wrote: Hi Bunuel ScottTargetTestPrep Could anyone clear my small doubt? Why did you guys ignore the constant? Nonetheless, isn't there a possibility for two different constants for both A and B? Thank you, Dablu My solution did not ignore the constants; I used "k" to denote the constant of proportionality, but it canceled out later. The constant can, on the other hand, be ignored for this question because you are looking for a value which will keep a ratio unchanged. If the value (A^2)/B is unchanged without the constant, it will still be unchanged with the constant (which is (k*A^2)/B). In regards to why there are not different constants for A and B, there actually is. If you kept B fixed, then R = p*A^2 for some p and if you kept A fixed, then R = s/B for some s. The values of p and s are not necessarily equal and you will obtain different values of p and s depending on which value of B and A you are fixing. When A and B are both varying, then the equation becomes R = k*(A^2)/B where k is yet another constant. Whatever value you choose for A and B, this k will not change.
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Re: The rate of a certain chemical reaction is directly proportional to th
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07 Nov 2019, 22:58
Thank you ScottTargetTestPrep! Quote: If you kept B fixed, then R = p*A^2 for some p and if you kept A fixed, then R = s/B for some s. The values of p and s are not necessarily equal and you will obtain different values of p and s depending on which value of B and A you are fixing. When A and B are both varying, then the equation becomes R = k*(A^2)/B where k is yet another constant. Whatever value you choose for A and B, this k will not change. Or we can simply say, p and s can be written as p*s = k in the equation and this constant k will stay the same even if the value of B were to double. p*A^2 * s/ B = p*A^2 * s/2B p and s can multiply can be another constant "k", which will stay the same even when B doubles. I think I got it! Thank you for giving a great explanation and keeping up with my queries.



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Re: The rate of a certain chemical reaction is directly proportional to th
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08 Nov 2019, 13:22
gurudabl wrote: Thank you ScottTargetTestPrep! Quote: If you kept B fixed, then R = p*A^2 for some p and if you kept A fixed, then R = s/B for some s. The values of p and s are not necessarily equal and you will obtain different values of p and s depending on which value of B and A you are fixing. When A and B are both varying, then the equation becomes R = k*(A^2)/B where k is yet another constant. Whatever value you choose for A and B, this k will not change. Or we can simply say, p and s can be written as p*s = k in the equation and this constant k will stay the same even if the value of B were to double. p*A^2 * s/ B = p*A^2 * s/2B p and s can multiply can be another constant "k", which will stay the same even when B doubles. I think I got it! Thank you for giving a great explanation and keeping up with my queries. Happy to help!
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