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Calculate the density of a mixture given the densities and volumes of its components.

Chemistry

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Question 1:
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Question 2: Calculate the density of the mixture.

  1. Calculate the mass of acid.

    Mass of acid=Density of acid×Volume of acid\text{Mass of acid} = \text{Density of acid} \times \text{Volume of acid} Mass of acid=1.08 g/cm3×200 cm3=216 g\text{Mass of acid} = 1.08 \text{ g/cm}^3 \times 200 \text{ cm}^3 = 216 \text{ g}
  2. Calculate the mass of alcohol.

    Mass of alcohol=Density of alcohol×Volume of alcohol\text{Mass of alcohol} = \text{Density of alcohol} \times \text{Volume of alcohol} Mass of alcohol=0.8 g/cm3×300 cm3=240 g\text{Mass of alcohol} = 0.8 \text{ g/cm}^3 \times 300 \text{ cm}^3 = 240 \text{ g}
  3. Calculate the total mass of the mixture.

    Total mass=Mass of acid+Mass of alcohol\text{Total mass} = \text{Mass of acid} + \text{Mass of alcohol} Total mass=216 g+240 g=456 g\text{Total mass} = 216 \text{ g} + 240 \text{ g} = 456 \text{ g}
  4. Calculate the total volume of the mixture.

    Total volume=Volume of acid+Volume of alcohol\text{Total volume} = \text{Volume of acid} + \text{Volume of alcohol} Total volume=200 cm3+300 cm3=500 cm3\text{Total volume} = 200 \text{ cm}^3 + 300 \text{ cm}^3 = 500 \text{ cm}^3
  5. Calculate the density of the mixture.

    Density of mixture=Total massTotal volume\text{Density of mixture} = \frac{\text{Total mass}}{\text{Total volume}} Density of mixture=456 g500 cm3=0.912 g/cm3\text{Density of mixture} = \frac{456 \text{ g}}{500 \text{ cm}^3} = 0.912 \text{ g/cm}^3

    The density of the mixture is 0.912 g/cm3\boxed{0.912 \text{ g/cm}^3}.

    Question 3: Find the magnitude of vector OM.

  6. Find the coordinates of point M.

    Point M divides line PQ in the ratio 2:1. Given P(5, 1) and Q(11, 4).
    Using the section formula M=(nx1+mx2m+n,ny1+my2m+n)M = \left(\frac{nx_1 + mx_2}{m+n}, \frac{ny_1 + my_2}{m+n}\right), where (x1,y1)=(5,1)(x_1, y_1) = (5, 1), (x2,y2)=(11,4)(x_2, y_2) = (11, 4), m=2m=2, n=1n=1.

    Mx=1×5+2×112+1=5+223=273=9M_x = \frac{1 \times 5 + 2 \times 11}{2+1} = \frac{5 + 22}{3} = \frac{27}{3} = 9 My=1×1+2×42+1=1+83=93=3M_y = \frac{1 \times 1 + 2 \times 4}{2+1} = \frac{1 + 8}{3} = \frac{9}{3} = 3

    So, the coordinates of M are (9,3)(9, 3).

  7. Determine the vector OM.

    The vector OM is the position vector of M, which is (93)\begin{pmatrix} 9 \\ 3 \end{pmatrix}.

  8. Calculate the magnitude of vector OM.

    ∣OM⃗∣=92+32|\vec{OM}| = \sqrt{9^2 + 3^2} ∣OM⃗∣=81+9|\vec{OM}| = \sqrt{81 + 9} ∣OM⃗∣=90|\vec{OM}| = \sqrt{90} ∣OM⃗∣=9×10=310|\vec{OM}| = \sqrt{9 \times 10} = 3\sqrt{10}

    The magnitude of vector OM is 310\boxed{3\sqrt{10}}.

    Question 4:
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    Question 5:
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    Question 6: A line passes through points (2, 5) and has a gradient of 2.

    a) Determine its equation in the form y=mx+cy=mx+c.

  9. Use the given gradient m=2m=2 and the point (x,y)=(2,5)(x,y) = (2,5) in the equation y=mx+cy=mx+c.

    5=2(2)+c5 = 2(2) + c 5=4+c5 = 4 + c c=5−4c = 5 - 4 c=1c = 1
  10. Write the equation of the line.

    y=2x+1y = 2x + 1

    The equation of the line is y=2x+1\boxed{y = 2x + 1}.

    b) Find the angle it makes with x-axis.

  11. Use the relationship between gradient and angle.

    The gradient m=tan⁡θm = \tan \theta, where θ\theta is the angle the line makes with the positive x-axis.

    tan⁡θ=2\tan \theta = 2 θ=arctan⁡(2)\theta = \arctan(2) θ≈63.43∘\theta \approx 63.43^\circ

    The angle the line makes with the x-axis is 63.43∘\boxed{63.43^\circ} (to two decimal places).

    Question 7: Find the value of P when Q=5.

  12. Write the general equation for P.

    P is partly constant and partly varies as the cube of Q.

    P=k+cQ3P = k + cQ^3

    where kk and cc are constants.

  13. Use the given conditions to form a system of equations.

    When Q=1,P=23Q=1, P=23:

    23=k+c(1)3  ⟹  23=k+c(Equation 1)23 = k + c(1)^3 \implies 23 = k + c \quad \text{(Equation 1)}

    When Q=2,P=44Q=2, P=44:

    44=k+c(2)3  ⟹  44=k+8c(Equation 2)44 = k + c(2)^3 \implies 44 = k + 8c \quad \text{(Equation 2)}
  14. Solve the system of equations for kk and cc.

    Subtract Equation 1 from Equation 2:

    (44−23)=(k+8c)−(k+c)(44 - 23) = (k + 8c) - (k + c) 21=7c21 = 7c c=217=3c = \frac{21}{7} = 3

    Substitute c=3c=3 into Equation 1:

    23=k+323 = k + 3 k=23−3=20k = 23 - 3 = 20
  15. Write the specific equation for P.

    P=20+3Q3P = 20 + 3Q^3
  16. Find P when Q=5.

    P=20+3(5)3P = 20 + 3(5)^3 P=20+3(125)P = 20 + 3(125) P=20+375P = 20 + 375 P=395P = 395

    The value of P when Q=5 is 395\boxed{395}.

    Question 8: Determine the translation vector T and find the coordinates of A' and B'.
    Given vertices A(1, 2), B(3, 5), C(4, 1). Image of C is C'(6, -2).

    a) Determine the translation vector T.

  17. Use the formula Image = Original + Translation Vector.

    C′=C+TC' = C + T (6−2)=(41)+T\begin{pmatrix} 6 \\ -2 \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \end{pmatrix} + T T=(6−2)−(41)T = \begin{pmatrix} 6 \\ -2 \end{pmatrix} - \begin{pmatrix} 4 \\ 1 \end{pmatrix} T=(6−4−2−1)=(2−3)T = \begin{pmatrix} 6-4 \\ -2-1 \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \end{pmatrix}

    The translation vector is T=(2−3)\boxed{T = \begin{pmatrix} 2 \\ -3 \end{pmatrix}}.

    b) Find the co-ordinates of A' and B' under the translation vector T.

  18. Find A'.

    A′=A+TA' = A + T A′=(12)+(2−3)=(1+22−3)=(3−1)A' = \begin{pmatrix} 1 \\ 2 \end{pmatrix} + \begin{pmatrix} 2 \\ -3 \end{pmatrix} = \begin{pmatrix} 1+2 \\ 2-3 \end{pmatrix} = \begin{pmatrix} 3 \\ -1 \end{pmatrix}

    So, A' is (3,−1)(3, -1).

  19. Find B'.

    B′=B+TB' = B + T B′=(35)+(2−3)=(3+25−3)=(52)B' = \begin{pmatrix} 3 \\ 5 \end{pmatrix} + \begin{pmatrix} 2 \\ -3 \end{pmatrix} = \begin{pmatrix} 3+2 \\ 5-3 \end{pmatrix} = \begin{pmatrix} 5 \\ 2 \end{pmatrix}

    So, B' is (5,2)(5, 2).
    The coordinates are A′(3,−1)\boxed{A'(3, -1)}

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