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2 years ago

In lab, you calculate the density of an iron rod to be 7.30 g/cm3. The accepted value

Chemistry

1 answer:

OLga [1]2 years ago

6 0

Answer:

Explanation:

The percentage error of a certain measurement can be found by using the formula

$P(\%) = \frac{error}{actual \: \: number} \times 100\% \\$

From the question

actual density = 7.80 g/cm³

error = 7.30 - 7.80 = 0.5

We have

$p(\%) = \frac{0.5}{7.8} \times 100 \\ = 6.410256...$

We have the final answer as

Hope this helps you

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If a dose of antacid containing 28 g Al(OH)3 reacts to produce 44 g AlCl3, what is the percentage yield of AlCl3

MArishka [77]

Answer:

96%

Explanation:

Step 1: Write the balanced neutralization reaction

Al(OH)₃ + 3 HCl ⇒ AlCl₃ + 3 H₂O

Step 2: Calculate the theoretical yield of AlCl₃

According to the balanced equation, the mass ratio of Al(OH)₃ to AlCl₃ is 81.03:133.34.

28 g Al(OH)₃ × 133.34 g AlCl₃/81.03 g Al(OH)₃ = 46 g AlCl₃

Step 3: Calculate the percent yield of AlCl₃

The real yield of AlCl₃ is 44 g. We can calculate the percent yield using the following expression.

%yield = real yield / theoretical yield × 100%

%yield = 44 g / 46 g × 100% = 96%

3 0

2 years ago

For doping silicon with boron, silicon specimen was kept in gaseous atmosphere containing B2O3 that maintained the B concentrati

Kay [80]

Solution :

From Fick's law:

$\frac{D_{AB}}{\Delta z } \times (C_{A1}-C_{A2})=N_a$

Mass balance: Exits = Accumulation

$-N_A A = \frac{dm}{dt}$

$-N_A A = \frac{dVp}{dt}$

$-N_A A = \frac{dV}{dt}p$

$-N_A A = \frac{dhA}{dt}$

$-N_A A = \frac{dh}{dt} \times Ap$

From the last step, area cancels out and thus leaves :

$-N_A = \frac{dh}{dt} \times p$

So now we can substitute the $N_A$ by the Fick's law

$\frac{D_{AB}}{\Delta z } \times (C_{A1}-C_{A2})=\frac{dh}{dt} p$

Substituting the values we get

$=\frac{-4 \times 10^{-13}}{0.0001} \times (3 \times 10^{26} - C_{A2}) = \frac{dh}{dt} \times 2.46$

$=-4 \times 10^{-9} \times( 3 \times 10^{26} - C_{A2}) = 0.0001 \times 2.46$

$= -7800 \times 4 \times 10^{-9} \times (3 \times 10^{26}-C_{A2})=0.000246$ $=-(3 \times 10^{26}-C_{A2}) = 7.8846 \times 100 \times \frac{1}{69.62} \times 6.022 \times 10^{23}$

$C_{A2} = 6.85 \times 10^{28} \ \text{ boron atoms} /m^3$

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2 years ago