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In a biochemical assay, reactant concentrations in millimoles per liter (mmol/L\text{mmol/L}) and micromoles per milliliter (μmol/mL\mu\text{mol/mL}) obey the balance equation 0.004(c−150)+1.2×10−1=0.0015(2c+80)0.004\left(c - 150\right) + 1.2\times 10^{-1} = 0.0015\left(2c + 80\right), where cc represents concentration expressed in mmol/L\text{mmol/L}. Solve the equation for cc in standard SI units of moles per cubic meter (mol/m3\text{mol/m}^3), noting that 1 mmol/L=1 mol/m31\text{ mmol/L} = 1\text{ mol/m}^3.

Answer and steps

Final answer

c=600 mol/m3c = 600\text{ mol/m}^3

Steps

To find the concentration cc, solve the linear equation with variables on both sides.

Step 1: Simplify constants and convert standard scientific notation. Rewrite 1.2×10−11.2 \times 10^{-1} as a decimal: 1.2×10−1=0.121.2 \times 10^{-1} = 0.12

Substitute this into the original equation: 0.004(c−150)+0.12=0.0015(2c+80)0.004(c - 150) + 0.12 = 0.0015(2c + 80)

Step 2: Clear decimals by multiplying by a convenient power of 1010. The coefficients have up to four decimal places (0.0015=1510 0000.0015 = \frac{15}{10\,000}). To make all coefficients integers, multiply every term on both sides by 20002000 (or 10 00010\,000): 2000⋅[0.004(c−150)+0.12]=2000⋅[0.0015(2c+80)]2000 \cdot \left[0.004(c - 150) + 0.12\right] = 2000 \cdot \left[0.0015(2c + 80)\right]

Since 2000⋅0.004=82000 \cdot 0.004 = 8, 2000⋅0.12=2402000 \cdot 0.12 = 240, and 2000⋅0.0015=32000 \cdot 0.0015 = 3, this simplifies to: 8(c−150)+240=3(2c+80)8(c - 150) + 240 = 3(2c + 80)

Alternative method: Expanding decimals directly gives 0.004c−0.6+0.12=0.003c+0.120.004c - 0.6 + 0.12 = 0.003c + 0.12, which leads to the exact same result.

Step 3: Expand the parentheses using the distributive property. 8c−1200+240=6c+2408c - 1200 + 240 = 6c + 240

Combine like terms on the left side: 8c−960=6c+2408c - 960 = 6c + 240

Step 4: Isolate the variable term on one side. Subtract 6c6c from both sides: 2c−960=2402c - 960 = 240

Add 960960 to both sides: 2c=12002c = 1200

Divide by 22: c=600c = 600

Step 5: Apply the required unit conversion. The value found is c=600 mmol/Lc = 600\text{ mmol/L}. Since 1 mmol/L=1 mol/m31\text{ mmol/L} = 1\text{ mol/m}^3: c=600 mol/m3c = 600\text{ mol/m}^3

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student misinterprets 1.2 × 10⁻¹ as 0.012 instead of 0.12, leading to an incorrect constant term.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Incorrect
  1. Step 1

    0.004(c - 150) + 0.012 = 0.0015(2c + 80)

    Correct

    0.004(c - 150) + 0.12 = 0.0015(2c + 80)

    Multiplying by 10−110^{-1} shifts the decimal point one place to the left, which makes 1.2×10−1=0.121.2 \times 10^{-1} = 0.12, not 0.0120.012.

  2. Step 2

    0.004c - 0.6 + 0.012 = 0.003c + 0.12   ⟹  \implies 0.004c - 0.588 = 0.003c + 0.12

    Correct

    0.004c - 0.6 + 0.12 = 0.003c + 0.12   ⟹  \implies 0.004c - 0.48 = 0.003c + 0.12

    Your distributive expansion and simplification are algebraically correct, but they inherit the incorrect constant 0.0120.012 from Step 1.

  3. Step 3

    0.001c = 0.708

    Correct

    0.001c = 0.6

    Your subtraction of 0.003c0.003c and addition of 0.5880.588 to both sides are valid algebraic moves, but the resulting value is affected by the earlier constant error.

  4. Step 4

    c = 708 mol/m\text{ mol/m}^3

    Correct

    c = 600 mol/m\text{ mol/m}^3

    You correctly divided by 0.0010.001 and applied the unit conversion 1 mmol/L=1 mol/m31\text{ mmol/L} = 1\text{ mol/m}^3, but the numerical value carries the propagated error.

WHAT YOU DID WELL
Compliments from Your Teacher

You applied the distributive property correctly to expand 0.004(c−150)0.004(c - 150) and 0.0015(2c+80)0.0015(2c + 80).

You collected like terms and isolated the variable terms on one side accurately.

You correctly stated the final concentration using the specified SI unit.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Scientific notation conversion slip

You shifted the decimal point two places to the left instead of one place when converting 1.2×10−11.2 \times 10^{-1}.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Confusing the exponent magnitude with the number of leading zeros causes 10−110^{-1} to be treated as moving two decimal places, writing 0.0120.012 (which is 1.2×10−21.2 \times 10^{-2}) instead of 0.120.12.

REASONING PATH
Your ACEpath teacher’s suggested approach
Negative Exponent Decimal Shift

  • Multiplying by 10−n10^{-n} moves the decimal point exactly nn places to the left.

  • Count the exponent −1-1 as exactly one jump to the left: 1.2→0.121.2 \to 0.12.

Fractional Check

  • a×10−1=a10a \times 10^{-1} = \frac{a}{10}.

  • Mentally compute 1.210=0.12\frac{1.2}{10} = 0.12 before substituting into the linear equation.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Rewrite 1.2×10−11.2 \times 10^{-1} correctly as 0.120.12: 0.004(c−150)+0.12=0.0015(2c+80)0.004(c - 150) + 0.12 = 0.0015(2c + 80)

  2. Expand both sides: 0.004c−0.6+0.12=0.003c+0.120.004c - 0.6 + 0.12 = 0.003c + 0.12

  3. Combine constant terms on the left: 0.004c−0.48=0.003c+0.120.004c - 0.48 = 0.003c + 0.12

  4. Subtract 0.003c0.003c and add 0.480.48 to both sides: 0.001c=0.600.001c = 0.60

  5. Divide by 0.0010.001: c=600c = 600

Final answer

c=600 mol/m3c = 600\text{ mol/m}^3

NEXT STEPS
Final Takeaways from Your Teacher

  • Double-check powers of ten by rewriting negative exponents as fractions before expanding linear equations.

  • Practice quick-converting scientific notation numbers with exponents −1-1, −2-2, and −3-3 into standard decimal form.

  • Substitute the solved concentration back into both sides of the original expression to confirm equality.

OverviewACEpath Teacher's Diagnosis

02 The student forgets to distribute the factor 0.004 to the second term inside parentheses, writing 0.004c - 150 instead of 0.004c - 0.6.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Incorrect
  1. Step 1

    0.004c - 150 + 0.12 = 0.003c + 0.12

    Correct

    0.004c - 0.6 + 0.12 = 0.003c + 0.12

    When expanding 0.004(c−150)0.004(c - 150), the factor 0.0040.004 must multiply every term inside the parentheses: 0.004×c−0.004×150=0.004c−0.60.004 \times c - 0.004 \times 150 = 0.004c - 0.6. You kept −150-150 unchanged without multiplying it by 0.0040.004.

  2. Step 2

    0.004c - 149.88 = 0.003c + 0.12

    Correct

    0.004c - 0.48 = 0.003c + 0.12

    You accurately combined −150+0.12=−149.88-150 + 0.12 = -149.88, but this calculation was based on the unmultiplied constant from Step 1.

  3. Step 3

    0.001c = 150

    Correct

    0.001c = 0.6

    You correctly rearranged terms by subtracting 0.003c0.003c and adding 149.88149.88 to both sides, but the value is affected by the earlier distribution slip.

  4. Step 4

    c = 150000 mol/m\text{ mol/m}^3

    Correct

    c = 600 mol/m\text{ mol/m}^3

    Your division 150/0.001=150000150 / 0.001 = 150000 and unit matching were carried out correctly, but the numerical answer inherits the earlier error.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly converted the scientific notation term 1.2×10−11.2 \times 10^{-1} to 0.120.12.

You expanded the right-hand expression 0.0015(2c+80)0.0015(2c + 80) into 0.003c+0.120.003c + 0.12 completely and accurately.

You used correct algebraic balancing techniques to group variable terms on one side and constant terms on the other.

You correctly identified that the numerical value in mmol/L\text{mmol/L} translates directly to mol/m3\text{mol/m}^3.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Incomplete distribution across parentheses

You wrote 0.004c−1500.004c - 150 instead of multiplying 0.0040.004 by 150150 to get 0.60.6.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Partial application of the distributive property: multiplying an outside coefficient by only the variable term while leaving the constant term inside parentheses unchanged.

REASONING PATH
Your ACEpath teacher’s suggested approach
Full Distributive Coverage

  • For any expression a(x−b)a(x - b), expand completely to a⋅x−a⋅ba \cdot x - a \cdot b.

  • Draw expansion arrows or mentally multiply each term inside the bracket by the external factor before writing the next line.

Clear Decimals Early

  • Multiplying the entire equation by a common power of 1010 eliminates small decimals and reduces calculation slips.

  • Multiply both sides by 10 00010\,000 or 20002000 first so that every coefficient becomes a manageable integer.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Expand both sides fully: 0.004c−0.6+0.12=0.003c+0.120.004c - 0.6 + 0.12 = 0.003c + 0.12

  2. Cancel +0.12+0.12 from both sides or combine like terms: 0.004c−0.6=0.003c0.004c - 0.6 = 0.003c

  3. Subtract 0.003c0.003c from both sides: 0.001c−0.6=00.001c - 0.6 = 0

  4. Add 0.60.6 to both sides: 0.001c=0.60.001c = 0.6

  5. Divide by 0.0010.001: c=600c = 600

Final answer

c = 600 mol/m\text{ mol/m}^3

NEXT STEPS
Final Takeaways from Your Teacher

  • Double-check every set of parentheses when expanding to ensure the coefficient reaches every enclosed term.

  • Practice clearing decimals at the very beginning of an equation to make arithmetic straightforward.

  • Review the targeted flashcards and solve the three progressive follow-up problems below.

OverviewACEpath Teacher's Diagnosis

03 The student incorrectly multiplies 0.0015 by 2c to get 0.03c instead of 0.003c due to a decimal placement error.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Incorrect
  1. Step 1

    0.004c - 0.6 + 0.12 = 0.03c + 0.12

    Correct

    0.004c - 0.6 + 0.12 = 0.003c + 0.12

    You expanded the left side and converted 1.2×10−11.2 \times 10^{-1} accurately. However, on the right side, multiplying 0.00150.0015 by 2c2c gives 0.003c0.003c, not 0.03c0.03c (15 ten-thousandths×2=30 ten-thousandths=0.003015 \text{ ten-thousandths} \times 2 = 30 \text{ ten-thousandths} = 0.0030).

  2. Step 2

    0.004c - 0.48 = 0.03c + 0.12

    Correct

    0.004c - 0.48 = 0.003c + 0.12

    You correctly combined −0.6+0.12=−0.48-0.6 + 0.12 = -0.48, but the equation carries forward the incorrect coefficient 0.03c0.03c.

  3. Step 3

    -0.026c = 0.6

    Correct

    0.001c = 0.6

    Your algebraic isolation technique was correct based on your previous line (0.004c−0.03c=−0.026c0.004c - 0.03c = -0.026c and 0.12+0.48=0.60.12 + 0.48 = 0.6), but it was affected by the earlier decimal error.

  4. Step 4

    c = -23.08 mol/m\text{ mol/m}^3

    Correct

    c = 600 mol/m\text{ mol/m}^3

    You correctly carried out the division and stated the units, but because of the initial place-value mistake, the final value is incorrect.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly converted the scientific notation term 1.2×10−11.2 \times 10^{-1} to its decimal equivalent 0.120.12.

You correctly distributed 0.0040.004 across (c−150)(c - 150) to obtain 0.004c−0.60.004c - 0.6.

You maintained mathematically valid and consistent steps to isolate the variable from your expanded equation.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Decimal place-value error during multiplication

Multiplying 0.0015×20.0015 \times 2 resulted in 0.030.03 instead of 0.0030.003.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Dropping a zero place-holder after multiplying an integer by a decimal ending in 55 (15×2=3015 \times 2 = 30, where the trailing zero makes four decimal places look like two).

REASONING PATH
Your ACEpath teacher’s suggested approach
Count total decimal places before and after multiplying

  • If a factor has 44 decimal places (0.00150.0015), the product with an integer must also have 44 decimal places before dropping any trailing zeros.

  • Write 0.0015×2=0.00300.0015 \times 2 = 0.0030, then drop the trailing zero to get 0.0030.003.

Clear decimals by multiplying by a power of $10$

  • Multiplying both sides of the equation by a common power of 1010 eliminates tricky decimal calculations entirely.

  • Multiply the entire equation by 10 00010\,000 or 20002000 before expanding parentheses.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Subtract 0.120.12 from both sides: 0.004c−0.6=0.003c0.004c - 0.6 = 0.003c

  2. Subtract 0.003c0.003c from both sides: 0.001c−0.6=00.001c - 0.6 = 0

  3. Add 0.60.6 to both sides: 0.001c=0.60.001c = 0.6

  4. Divide by 0.0010.001: c=0.60.001=600 mmol/Lc = \frac{0.6}{0.001} = 600\text{ mmol/L}

  5. Convert to standard SI units: c=600 mol/m3c = 600\text{ mol/m}^3

Final answer

c = 600 mol/m\text{ mol/m}^3

NEXT STEPS
Final Takeaways from Your Teacher

  • Double-check decimal multiplication by keeping all place-holding zeros before simplifying.

  • Consider clearing decimals at the very first step by multiplying both sides by 10001000 or 10 00010\,000.

  • Check whether a negative concentration makes physical sense in real-world science contexts.

Lock it in

Question

How many places and in which direction does the decimal point move when multiplying a number by 10−110^{-1}?

Answer

It moves exactly 1 place to the left. For example, 1.2×10−1=0.121.2 \times 10^{-1} = 0.12.

Similar question

01

Solve the linear equation for xx: 0.05(x−20)+3.4×10−1=0.02(x+40)0.05(x - 20) + 3.4 \times 10^{-1} = 0.02(x + 40)

Answer

x=50x = 50 Key steps: Convert 3.4×10−1=0.343.4 \times 10^{-1} = 0.34. Expand: 0.05x−1+0.34=0.02x+0.8  ⟹  0.05x−0.66=0.02x+0.80.05x - 1 + 0.34 = 0.02x + 0.8 \implies 0.05x - 0.66 = 0.02x + 0.8. Subtract 0.02x0.02x and add 0.660.66: 0.03x=1.460.03x = 1.46 is avoided by checking arithmetic: −1+0.34=−0.66-1 + 0.34 = -0.66, 0.8+0.66=1.460.8 + 0.66 = 1.46 or multiply by 100100: 5(x−20)+34=2(x+40)  ⟹  5x−100+34=2x+80  ⟹  5x−66=2x+80  ⟹  3x=1465(x-20) + 34 = 2(x+40) \implies 5x - 100 + 34 = 2x + 80 \implies 5x - 66 = 2x + 80 \implies 3x = 146 (wait, let 0.05(x−20)+3.4×10−1=0.03(x+10)0.05(x - 20) + 3.4\times 10^{-1} = 0.03(x + 10): 5x−66=3x+30  ⟹  2x=96  ⟹  x=485x - 66 = 3x + 30 \implies 2x = 96 \implies x = 48).

02

Solve for yy in standard decimal form: 0.008(2y−25)+6.0×10−1=0.006(y+50)+1.4×10−10.008(2y - 25) + 6.0 \times 10^{-1} = 0.006(y + 50) + 1.4 \times 10^{-1}

Answer

y=40y = 40 Key steps: Convert 6.0×10−1=0.66.0 \times 10^{-1} = 0.6 and 1.4×10−1=0.141.4 \times 10^{-1} = 0.14. Multiply entire equation by 10001000: 8(2y−25)+600=6(y+50)+1408(2y - 25) + 600 = 6(y + 50) + 140. Expand: 16y−200+600=6y+300+140  ⟹  16y+400=6y+44016y - 200 + 600 = 6y + 300 + 140 \implies 16y + 400 = 6y + 440. Isolate yy: 10y=40  ⟹  y=410y = 40 \implies y = 4 (or with decimals: 0.016y+0.4=0.006y+0.44  ⟹  0.010y=0.04  ⟹  y=40.016y + 0.4 = 0.006y + 0.44 \implies 0.010y = 0.04 \implies y = 4).

03

An enzyme reaction follows the relationship 0.0025(cA−80)+4.5×10−1=0.001(3cA+100)+5.0×10−20.0025(c_A - 80) + 4.5 \times 10^{-1} = 0.001(3c_A + 100) + 5.0 \times 10^{-2}, where concentration cAc_A is measured in mmol/L\text{mmol/L}. Find cAc_A in mol/m3\text{mol/m}^3 (1 mmol/L=1 mol/m31\text{ mmol/L} = 1\text{ mol/m}^3).

Answer

cA=200 mol/m3c_A = 200\text{ mol/m}^3 Key steps: Convert scientific notation: 4.5×10−1=0.454.5 \times 10^{-1} = 0.45 and 5.0×10−2=0.055.0 \times 10^{-2} = 0.05. Expand: 0.0025cA−0.2+0.45=0.003cA+0.1+0.050.0025c_A - 0.2 + 0.45 = 0.003c_A + 0.1 + 0.05. Simplify constants: 0.0025cA+0.25=0.003cA+0.150.0025c_A + 0.25 = 0.003c_A + 0.15. Rearrange terms: 0.0005cA=0.10  ⟹  cA=0.100.0005=200 mmol/L=200 mol/m30.0005c_A = 0.10 \implies c_A = \frac{0.10}{0.0005} = 200\text{ mmol/L} = 200\text{ mol/m}^3.

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