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lord [1]
2 years ago
9

28. How many formula units are found in 0.250 moles of potassium nitrate?

Chemistry
1 answer:
Soloha48 [4]2 years ago
7 0

Answer:

\boxed {\boxed {\sf B. \ 1.5 *10^{23} \ formula \ units}}

Explanation:

1 mole of any substance contains the same number of particles. The particles can vary (atoms, molecules, formula units), but there are always 6.022*10²³ particles. In this case, the particles are formula units of potassium nitrate or KNO₃.

Let's create a ratio.

\frac {6.022*10^{23} \ formula \ units  \ KNO_3}{1 \ mol \ KNO_3}

Since we are trying to find the formula units in 0.250 moles, we multiply by that number.

0.250 \ mol \ KNO_3 *\frac {6.022*10^{23} \ formula \ units  \ KNO_3}{1 \ mol \ KNO_3}

The units of moles of potassium nitrate cancel.

0.250  *\frac {6.022*10^{23} \ formula \ units  \ KNO_3}{1 }

The denominator of 1 can be ignored, so we can make a simple multiplication problem.

0.250  *{6.022*10^{23} \ formula \ units  \ KNO_3}

1.5055 * 10^{23} \ formula \ units \ KNO_3

If we round to the nearest tenth, the 0 in the hundredth place tells us to leave the 5 in the tenth place.

1.5 *10^{23} \ formula \ units \ KNO_3

0.250 moles of potassium nitrate is approximately equal to 1.5*10²³ formula units of potassium nitrate and choice B is correct.

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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
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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}

3. Plot the temperature readings as a function of time.

The graphs are shown below.

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The answer is C. Life
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