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2 years ago

Calculate these volumes.

Chemistry

2 answers:

MrMuchimi2 years ago

7 0

Check photo for answers!

Westkost [7]2 years ago

3 0

Answer:

For a: The volume of graphite is $5.02cm^3$

For b: The volume of bromine is $13.54cm^3$

Explanation:

To calculate volume of a substance, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$ ......(1)

For a:

Density of graphite = $2.25g/cm^3$

Mass of graphite = 11.3 g

Putting values in equation 1, we get:

$2.25g/cm^3=\frac{11.3g}{\text{Volume of graphite}}\\\\\text{Volume of graphite}=\frac{11.3g}{2.25g/cm^3}=5.02cm^3$

Hence, the volume of graphite is $5.02cm^3$

For b:

Density of bromine = $2.928g/cm^3$

Mass of bromine = 39.657 g

Putting values in equation 1, we get:

$2.928g/cm^3=\frac{39.657g}{\text{Volume of bromine}}\\\\\text{Volume of bromine}=\frac{39.657g}{2.928g/cm^3}=13.54cm^3$

Hence, the volume of bromine is $13.54cm^3$

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A thermometer having first-order dynamics with a time constant of 1 min is placed in a temperature bath at 100oF. After the ther

sveticcg [70]

Answer:

(a) See below

(b) 103.935 °F; 102.235 °F

Explanation:

The equation relating the temperature to time is

$T = T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )$

1. Calculate the thermometer readings after 0.5 min and 1 min

(a) After 0.5 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-0.5/1} \right )\\ & = & 100 + 10\left (1 - e^{-0.5} \right )\\ & = & 100 + 10 (1 - 0.6065)\\ & = & 100 + 10(0.3935)\\ & = & 100 + 3.935\\ & = & 103.935\,^{\circ}F\\\end{array}$

(b) After 1 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-1/1} \right )\\ & = & 100 + 10\left (1 - e^{-1} \right )\\ & = & 100 + 10 (1 - 0.3679)\\ & = & 100 + 10(0.6321)\\ & = & 100 + 6.321\\ & = & 106.321\,^{\circ}F\\\end{array}$

2. Calculate the thermometer reading after 2.0 min

T₀ =106.321 °F

ΔT = 100 - 106.321 °F = -6.321 °F

t = t - 1, because the cooling starts 1 min late

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-(t - 1)/\tau} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-(2 - 1)/1} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-1} \right )\\ & = & 106.321 - 6.321 (1 - 0.3679)\\ & = & 106.321 - 6.321 (0.6321)\\ & = & 106.321 - 3.996\\ & = & 102.325\,^{\circ}F\\\end{array}$