Q.How many mL of 0.1 M HCl are required to react completely with 1 g mixture of $Na_2CO_3$ and $NaHCO_3$ containing equimolar amounts of both?

Showing the 12 most recent of 31 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.Given below are two statements Statement-I: Rate of a first order reaction (A $\rightarrow$ p) decreases with time Statement-II: Rate of a zero-order reaction (B $\rightarrow$ p) decreases with time The correct answer is (A) Both statements I and II are correct (B) Both statements I and II are not correct (C) Statement I is correct, but statement II is not correct (D) Statement I is not correct, but statement II is correct

   ›Reveal solutionSolution

   For a first‑order reaction, the rate is proportional to the concentration of reactant, which falls exponentially, so the rate decreases with time. For a zero‑order reaction, the rate is constant (independent of concentration) until the reactant is exhausted, so the rate does not decrease with time. Therefore Statement I is correct, Statement II is not correct.

   Concept & Intuition

   The rate of a reaction tells us how fast the concentration of reactant (or product) changes per unit time. The key is how that rate depends on the remaining concentration of the reactant.

   • In a first‑order reaction, rate = $k[A]$. As the reaction proceeds, $[A]$ drops, so the rate drops proportionally.
   • In a zero‑order reaction, rate = $k$ (a constant). The rate does not depend on how much reactant is left; it stays the same until the reactant is suddenly gone.

   Thus the behaviour of the rate over time is fundamentally different for these two orders.

   Step‑by‑step reasoning

   1. Statement I: First‑order reaction For $A \rightarrow p$, the rate law is

   $\text{rate} = -\frac{d[A]}{dt} = k[A]$

   The integrated form gives $[A] = [A]_0 e^{-kt}$. Substituting back:

   $\text{rate} = k[A]_0 e^{-kt}$

   Since $e^{-kt}$ decreases as $t$ increases, the rate clearly decreases with time. So Statement I is correct.

   2. Statement II: Zero‑order reaction For $B \rightarrow p$, the rate law is $\text{rate} = -\frac{d[B]}{dt} = k$ …

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2. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.In leaching alumina, concentrated solution of 'X' and in leaching gold, dilute solution of 'Y' are used as leaching agents. X and Y respectively are (A) KCN, NaOH (B) NaOH, KCN (C) Ca(OH)$_2$, NaCN (D) NaCl, KCN

   ›Reveal solutionSolution

   The key idea is that alumina is leached with a concentrated base (NaOH) via the Bayer process, while gold is leached with a dilute cyanide solution (KCN or NaCN) via the MacArthur-Forrest process. The correct pair is NaOH for X and KCN for Y, which corresponds to option (B).

   The question tests your knowledge of two classic industrial leaching processes: the Bayer process for purifying bauxite (alumina ore) and the cyanidation process for extracting gold from its ores. The trick is remembering that the concentration of the leaching agent is opposite in each case — concentrated for alumina, dilute for gold — and that the specific chemicals are a strong base and a cyanide salt.

   1. Identify the process for alumina (X). Alumina (Al₂O₃) is extracted from bauxite ore using the Bayer process. The leaching agent is a concentrated solution of sodium hydroxide (NaOH). Hot, concentrated NaOH reacts with amphoteric alumina to form soluble sodium aluminate:

   $\text{Al}_2\text{O}_3 + 2\text{NaOH} + 3\text{H}_2\text{O} \rightarrow 2\text{NaAl(OH)}_4$

   Impurities like iron oxides remain insoluble and are filtered off. So X must be NaOH.

   2. Identify the process for gold (Y). Gold is extracted from its ores using the cyanidation process (MacArthur-Forrest process). The leaching agent is a dilute solution of sodium cyanide (NaCN) or potassium cyanide (KCN). Gold is oxidized and forms a soluble complex:

   $4\text{Au} + 8\text{NaCN} + \text{O}_2 + 2\text{H}_2\text{O} \rightarrow 4\text{Na[Au(CN)}_2] + 4\text{NaOH}$

   The solution is kept dilute (typically 0.1–0.2% NaCN) to avoid excessive consumption of cyanide and to maintain safety. So Y is a cyanide salt — either NaCN or KCN.

   3. Match with the options.
      • (A) KCN, NaOH → Wrong order (X would be KCN, Y would be NaOH). …

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3. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.At 27 $^\circ$C, the ratio of RMS velocity and most probable velocity of SO$_2$ is (A) $\sqrt{3} : \sqrt{2}$ (B) $\sqrt{2} : \sqrt{3}$ (C) $\sqrt{3} : \sqrt{5}$ (D) $\sqrt{5} : \sqrt{3}$

   ›Reveal solutionSolution

   The ratio of RMS velocity to most probable velocity for any ideal gas is a fixed constant, independent of temperature and molar mass, equal to $\sqrt{3} : \sqrt{2}$.

   The key idea is that both RMS velocity and most probable velocity are derived from the Maxwell–Boltzmann distribution of molecular speeds. Their ratio depends only on the mathematical form of these averages, not on the gas identity or temperature. So for SO₂ at 27 °C, the ratio is the same as for any ideal gas.

   1. Recall the formulas

      For an ideal gas at temperature $T$ and molar mass $M$:

      • RMS velocity: $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$
      • Most probable velocity: $v_{\text{mp}} = \sqrt{\frac{2RT}{M}}$

   2. Form the ratio

   $\frac{v_{\text{rms}}}{v_{\text{mp}}} = \frac{\sqrt{\frac{3RT}{M}}}{\sqrt{\frac{2RT}{M}}} = \sqrt{\frac{3}{2}}$

   The $RT/M$ cancels completely.

   3. Interpret the ratio This gives $v_{\text{rms}} : v_{\text{mp}} = \sqrt{3} : \sqrt{2}$. Notice that temperature (27 °C) and the gas (SO₂) are irrelevant — the ratio is universal. …

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4. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.The sulphide ore of zinc is X and carbonate ore of zinc is Y. X and Y are respectively (A) Kaolinite, siderite (B) Siderite, calamine (C) Calamine, sphalerite (D) Sphalerite, calamine

   ›Reveal solutionSolution

   The sulphide ore of zinc is Sphalerite ($ZnS$) and the carbonate ore of zinc is Calamine ($ZnCO_3$). Therefore, option (D) is the correct choice.

   Concept and Intuition

   Ores are naturally occurring rocks or minerals from which a metal can be extracted profitably. Different metals are found in various chemical forms within these ores. For zinc, two common types of ores are sulphide ores and carbonate ores.

   • Sulphide ores are minerals where the metal is chemically bonded with sulphur. These ores are typically processed by roasting (heating in air) to convert the sulphide to an oxide, which can then be reduced to the metal.
   • Carbonate ores are minerals where the metal is chemically bonded with the carbonate ion ($CO_3^{2-}$). These ores are typically processed by calcination (heating in the absence of air) to convert the carbonate to an oxide, which can then be reduced.

   To solve this problem, we need to recall the common names and chemical compositions of zinc's sulphide and carbonate ores.

   Step-by-Step Solution

   1. Identify common zinc ores:

      Zinc is primarily extracted from a few key minerals. The most important ones are:

      • Sphalerite (also known as Zinc blende)
      • Calamine (also known as Smithsonite)
      • Zincite

   2. Determine the sulphide ore of zinc (X):

      The most common sulphide ore of zinc is Sphalerite.

      The chemical formula for Sphalerite is $ZnS$.

      Therefore, X is Sphalerite.

   3. Determine the carbonate ore of zinc (Y):

      The most common carbonate ore of zinc is Calamine.

      The chemical formula for Calamine is $ZnCO_3$.

      Therefore, Y is Calamine.

   4. Match X and Y with the given options:

      We have identified X as Sphalerite and Y as Calamine. Let's check the options: …

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5. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.At T(K), the following data is obtained for the decomposition of A(g), which follows first order kinetics A(g) $\rightarrow$ B(g) + C(g)

   [!FORMULA] $\begin{array}{|c|c|} \hline \text{Time (sec)} & \text{Total pressure (atm)} \\ \hline 0 & 0.5 \\ 100 & 0.6 \\ x & 0.65 \\ \hline \end{array}$

   What is $x$ (in sec)? $(\log 1.25 = 0.097 ; \log 1.4285 = 0.1549)$ (A) 180 (B) 160 (C) 140 (D) 200

   ›Reveal solutionSolution

   The key idea is to relate total pressure to the concentration of reactant A using stoichiometry, then apply the first‑order integrated rate law. The missing time $x$ is found to be 160 seconds, corresponding to option (B).

   We are told the decomposition of A(g) follows first‑order kinetics. For a first‑order reaction, the rate depends only on the concentration (or partial pressure) of A. The integrated law is:

   $\ln \frac{[A]_0}{[A]_t} = k t$

   But the data gives total pressure, not the partial pressure of A. So we must first convert total pressure into the partial pressure of A at each time.

   ---

   1. Relate total pressure to partial pressure of A

   The reaction is:

   $A(g) \rightarrow B(g) + C(g)$

   Initially, only A is present. Let the initial pressure of A be $P_0 = 0.5$ atm.

   At any time $t$, let the decrease in pressure of A be $p$ atm. Then:

   • Pressure of A remaining: $P_A = P_0 - p$
   • Pressure of B formed: $p$
   • Pressure of C formed: $p$

   Total pressure at time $t$:

   $P_{\text{total}} = (P_0 - p) + p + p = P_0 + p$

   So:

   $p = P_{\text{total}} - P_0$

   And therefore:

   $P_A = P_0 - p = P_0 - (P_{\text{total}} - P_0) = 2P_0 - P_{\text{total}}$

   This is the key relation: the partial pressure of A at any time is twice the initial pressure minus the total pressure.

   ---

   2. Find the rate constant $k$ using the data at $t = 100$ sec

   At $t = 0$: $P_A(0) = 0.5$ atm

   At $t = 100$ sec: $P_{\text{total}} = 0.6$ atm

   $P_A(100) = 2(0.5) - 0.6 = 1.0 - 0.6 = 0.4 \text{ atm}$

   For a first‑order reaction (using pressures instead of concentrations, since $P \propto n$ at constant $T$ and $V$):

   $k = \frac{1}{t} \ln \frac{P_A(0)}{P_A(t)}$

   $k = \frac{1}{100} \ln \frac{0.5}{0.4} = \frac{1}{100} \ln(1.25)$

   Given $\log 1.25 = 0.097$, we convert to natural log: $\ln(1.25) = 2.303 \times 0.097$

   $k = \frac{2.303 \times 0.097}{100} = \frac{0.223391}{100} = 0.00223391 \text{ s}^{-1}$

   We can keep it symbolic for now: $k = \frac{\ln(1.25)}{100}$.

   ---

   3. Use the same relation to find $x$ when total pressure is 0.65 atm …

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6. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.Compared to cane sugar, artificial sweetener (X) has maximum sweetness value and (Y) has minimum sweetness value. X and Y respectively are (A) Alitame, Saccharin (B) Alitame, Aspartame (C) Aspartame, Sucralose (D) Saccharin, Sucralose

   ›Reveal solutionSolution

   The question asks for the artificial sweetener with the maximum sweetness value and the one with the minimum sweetness value among common options. The correct pair is Alitame (maximum) and Saccharin (minimum), which corresponds to option (A).

   The key here is understanding the relative sweetness of artificial sweeteners compared to cane sugar (sucrose). Sweetness is measured on a scale where sucrose is assigned a value of 1. Artificial sweeteners are many times sweeter, so a higher number means you need less of it to achieve the same sweetness.

   Let’s recall the approximate sweetness values of the common artificial sweeteners mentioned:

   1. Saccharin – about 300–500 times sweeter than sucrose. It’s one of the oldest, but not the most potent.
   2. Aspartame – about 200 times sweeter than sucrose. It’s less sweet than saccharin.
   3. Sucralose – about 600 times sweeter than sucrose. This is quite high.
   4. Alitame – about 2000 times sweeter than sucrose. This is among the most potent artificial sweeteners known.

   Now, the question asks for the one with maximum sweetness value and the one with minimum sweetness value among the given options. From the list above, Alitame has the highest (2000×), and Saccharin (300–500×) is lower than Sucralose (600×) and Aspartame (200×). Wait — check carefully: Aspartame is 200×, which is actually lower than Saccharin’s 300–500×. So the minimum among these four is Aspartame, not Saccharin.

   But the options pair X (max) and Y (min). Let’s examine each: …

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7. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.At 300 K, the diffusion rate of one mole of an ideal gas is $0.082~\mathrm{L~s^{-1}}$. What is the pressure (in atm) of this gas which can diffuse in 100 s? (R = $0.082~\mathrm{L~atm~mol^{-1}~K^{-1}}$) (A) 1 (B) 3 (C) 5 (D) 2

   ›Reveal solutionSolution

   Volume diffused in $100\ \mathrm{s}$ is $8.2\ \mathrm{L}$; then $P=\dfrac{nRT}{V}=\dfrac{1\times0.082\times300}{8.2}=3\ \mathrm{atm}$.

   The diffusion rate is $0.082\ \mathrm{L\,s^{-1}}$, so the volume of gas that diffuses in $100\ \mathrm{s}$ is

   $V = 0.082\ \mathrm{L\,s^{-1}}\times100\ \mathrm{s}=8.2\ \mathrm{L}.$ …

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8. TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
   Q.What is the correct equation that relates kinetic energy ($E_{ke}$) and pressure (P) of one mole of an ideal gas? (V = Volume) (A) $P = \dfrac{2E_{ke}}{3V}$ (B) $P = \dfrac{3V}{2E_{ke}}$ (C) $P = \dfrac{3V^2}{2E_{ke}}$ (D) $P = \dfrac{2(E_{ke})^2}{3V}$

   ›Reveal solutionSolution

   The pressure of an ideal gas arises from the collisions of its molecules with the container walls, and this pressure is directly related to the total translational kinetic energy of the gas molecules. For one mole of an ideal gas, the correct relationship is $P = \dfrac{2E_{ke}}{3V}$.

   The relationship between the macroscopic properties of a gas, like pressure and volume, and the microscopic properties of its constituent molecules, like their kinetic energy, is established by the kinetic theory of gases. This theory provides a model for how gas molecules behave and how their motion translates into observable phenomena.

   The core idea is that gas molecules are in constant, random motion. When these molecules collide with the walls of their container, they exert a force. The sum of these forces over the area of the walls gives rise to the pressure of the gas. The faster and more massive the molecules, and the more frequently they collide, the greater the pressure.

   Simultaneously, the kinetic energy of these molecules is directly related to the temperature of the gas. For an ideal gas, the total internal energy is purely translational kinetic energy. By linking the microscopic definition of pressure to the total kinetic energy, we can derive the required relationship.

   Here's the step-by-step derivation:

   1. Pressure from Kinetic Theory: According to the kinetic theory of gases, the pressure ($P$) exerted by an ideal gas in a container of volume ($V$) is given by the formula:

   $P = \frac{1}{3} \frac{N}{V} m \overline{v^2}$

   where:
   *   $N$ is the total number of gas molecules.
   *   $m$ is the mass of a single gas molecule.
   *   $\overline{v^2}$ is the mean square speed of the gas molecules.
   

   2. Total Kinetic Energy of the Gas:

   The translational kinetic energy of a single molecule is $\frac{1}{2} m v^2$. Therefore, the average translational kinetic energy of a single molecule is $\frac{1}{2} m \overline{v^2}$.

   The total translational kinetic energy ($E_{ke}$) of all $N$ molecules in the gas is the sum of the kinetic energies of individual molecules.

   $E_{ke} = N \times \left( \frac{1}{2} m \overline{v^2} \right)$

   $E_{ke} = \frac{1}{2} N m \overline{v^2}$

   3. Relating Pressure and Total Kinetic Energy: From the expression for total kinetic energy, we can isolate the term $N m \overline{v^2}$: …

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9. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.For a first order reaction, a plot of $\ln k$ (y-axis) and $\dfrac{1}{T}$ (x-axis) gave the straight line with slope equal to $-10^3 k$ and intercept equal to $2.303$ (on y-axis). What is the activation energy ($E_a$ in kJ mol$^{-1}$) of the reaction? (Given R = $8.314$ J mol$^{-1}$K$^{-1}$) (A) 8.314 (B) 2.303 (C) 2303 (D) 83.14

   ›Reveal solutionSolution

   The Arrhenius equation gives $\ln k = \ln A - \frac{E_a}{R}\cdot\frac{1}{T}$, so the slope of the $\ln k$ vs $1/T$ plot is $-\frac{E_a}{R}$. Equating this to the given slope $-10^3$ yields $E_a = 8.314$ kJ mol$^{-1}$, which corresponds to option (A).

   The key idea is the Arrhenius equation in its linearized form. For a first-order reaction (or any reaction), the temperature dependence of the rate constant $k$ is given by:

   $k = A e^{-E_a/(RT)}$

   Taking natural logs:

   $\ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T}$

   This is a straight line: $y = c + m x$, where $y = \ln k$, $x = 1/T$, intercept $c = \ln A$, and slope $m = -\frac{E_a}{R}$.

   So the slope directly gives the activation energy. No need to involve the intercept except to check consistency.

   ---

   1. Identify the slope from the problem The problem states: the slope equals $-10^3 k$. Wait — careful: the notation says “slope equal to $-10^3 k$”. Here $k$ is not the rate constant; it’s likely a typographical shorthand for “$10^3$” meaning $1000$. In many textbooks, they write $10^3$ as $10^3$ (i.e., $1000$). So the slope is $-1000$ (units: K, because $1/T$ has units K$^{-1}$, and $\ln k$ is dimensionless). So:

   $\text{slope} = -1000\ \text{K}$

   2. Relate slope to activation energy From the Arrhenius plot:

   $\text{slope} = -\frac{E_a}{R}$

   Therefore:

   $-\frac{E_a}{R} = -1000$

   Cancel the negatives:

   $\frac{E_a}{R} = 1000$

   3. Solve for $E_a$ Given $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$:

   $E_a = 1000 \times 8.314 = 8314\ \text{J mol}^{-1}$

   Convert to kJ mol$^{-1}$:

   $$$$

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10. TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQ
    Q.At 400 K, the following graph is obtained for x moles of an ideal gas. x is equal to (R = gas constant, P = pressure, V = volume) (A) $\dfrac{m}{400R}$ (B) $\dfrac{400R}{m}$ (C) $400mR$ (D) $\dfrac{1}{400mR}$

    ›Reveal solutionSolution

    A plot of $P$ against $1/V$ at constant $T$ is a straight line through the origin whose slope is $xRT$. With slope $=m$ and $T = 400$ K, $x = m/(400R)$ — option (A).

    The concept first

    Boyle's law says $P \propto 1/V$ at constant temperature — but plotted as $P$ vs $V$ that is a hyperbola, which is useless for extracting a number by eye. The standard trick in physical chemistry is to linearise: choose the axes so that the law becomes a straight line, because a straight line has just two readable parameters, a slope and an intercept, and each of them is a physical quantity.

    Here the axes chosen are $P$ (vertical) against $V^{-1}$ (horizontal). Start from the ideal gas equation for $x$ moles:

    $PV = xRT$

    and solve for $P$:

    $P = xRT \left(\frac{1}{V}\right)$

    Compare with $y = m'x' + c$:

    • $y \equiv P$,
    • $x' \equiv 1/V$,
    • slope $m' \equiv xRT$,
    • intercept $c = 0$ — which is why the line in the figure passes through the origin. (That zero intercept is itself a check that the gas is ideal.)

    The key realisation: at a fixed temperature, $R$ and $T$ are constants, so the slope is proportional to the amount of gas. Double the moles and the line gets twice as steep.

    Step-by-step

    1. Write the ideal gas equation for $x$ moles:

    $PV = xRT$

    2. Cast it in the form of the plotted variables ($P$ on the $y$-axis, $V^{-1}$ on the $x$-axis):

    $P = (xRT)\cdot V^{-1}$

    3. Identify the slope. Comparing with $y = (\text{slope})\,x'$: $\text{slope} = xRT$ …

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11. TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQ
    Q.A $\rightarrow$ P is a first order reaction. The change in concentration of A with time is shown below. The instantaneous rate at points X, Y and Z is $R_X$, $R_Y$, and $R_Z$ respectively. What is the correct order of $R_X$, $R_Y$ and $R_Z$? [FIGURE] (A) $R_Y > R_X > R_Z$ (B) $R_Y > R_Z > R_X$ (C) $R_X > R_Y > R_Z$ (D) $R_Z > R_X > R_Y$

    ›Reveal solutionSolution

    A first-order rate is $k[A]$, so the rate simply tracks the concentration. Y has the highest $[A]$ (steepest part), then X, then Z on the flat tail: $R_Y > R_X > R_Z$ — option (A).

    The concept first

    Two ideas meet in this question, and seeing that they are the same idea is the whole point.

    Idea 1 — the rate law. For $A \to P$ obeying first-order kinetics,

    $R = -\frac{d[A]}{dt} = k[A]$

    The rate is proportional to how much A is still there. Early on, plenty of A ⇒ fast; late on, little A ⇒ slow. This self-limiting behaviour is what generates the characteristic exponential decay

    $[A]_t = [A]_0 e^{-kt}$

    which is exactly the curve drawn: steep at first, then flattening asymptotically towards the time axis.

    Idea 2 — the graph. The instantaneous rate at a point is the magnitude of the slope of the tangent to the $[A]$ vs $t$ curve at that point. On a decaying curve, the steepest tangent is at the top-left and the tangent becomes almost horizontal on the tail.

    Put the two together: $-\dfrac{d[A]}{dt} = k[A]$ says the steepness of the curve is proportional to the height of the curve. So you can answer either by reading heights or by reading slopes — they must agree, and that agreement is first-order kinetics.

    Step-by-step

    1. Locate the three points on the curve. Moving along the curve from the top-left downwards: Y is on the steep upper part (earliest time), X is on the middle, less-steep part, and Z is far out on the flat tail (latest time).
    2. Read off the concentrations. Because the curve falls monotonically, $[A]_Y > [A]_X > [A]_Z$ …

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12. TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQ
    Q.The number average molecular weight ($M_n$) of a polymer is 1500. In this polymer, 800 molecules of molar mass 1000, 100 molecules of molar mass 2000 and $x$ molecules of molar mass 5000 are present. What is the value of $x$? (A) 200 (B) 400 (C) 100 (D) 50

    ›Reveal solutionSolution

    The number‑average molecular weight is the total mass of all molecules divided by the total number of molecules. Setting up that ratio equal to 1500 and solving for the unknown count $x$ gives $x = 100$, which corresponds to option (C).

    The key idea is that number‑average molecular weight $M_n$ is defined as

    $M_n = \frac{\sum N_i M_i}{\sum N_i}$

    where $N_i$ is the number of molecules of molar mass $M_i$.

    We are given $M_n = 1500$, three groups of molecules with known counts and masses, and one unknown count $x$. We simply plug into the definition and solve.

    ---

    1. Write the total number of molecules There are 800 molecules of mass 1000, 100 molecules of mass 2000, and $x$ molecules of mass 5000.

    $\text{Total number} = 800 + 100 + x = 900 + x$

    2. Write the total mass of the polymer sample

    $\text{Total mass} = (800 \times 1000) + (100 \times 2000) + (x \times 5000)$

    Compute the known parts:

    $800 \times 1000 = 800\,000,\qquad 100 \times 2000 = 200\,000$

    So total mass = $800\,000 + 200\,000 + 5000x = 1\,000\,000 + 5000x$.

    3. Set up the number‑average equation

    $M_n = \frac{\text{Total mass}}{\text{Total number}} = \frac{1\,000\,000 + 5000x}{900 + x} = 1500$

    4. Solve for $x$ Multiply both sides by $(900 + x)$:

    $1\,000\,000 + 5000x = 1500(900 + x)$

    Expand the right side:

    $1\,000\,000 + 5000x = 1\,350\,000 + 1500x$

    Subtract $1500x$ from both sides:

    $1\,000\,000 + 3500x = 1\,350\,000$

    Subtract $1\,000\,000$:

    $$$$

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