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morpeh [17]
3 years ago
10

Florence often recalls the time when her children were young. While browsing through a store, she notices a cream-colored satin

dress, edged with hand-made lace. It reminds her of a dress her daughter used to wear when she was little. She decides to buy the dress as a gift for her granddaughter. This is an example of _____.
Chemistry
1 answer:
Ludmilka [50]3 years ago
5 0

Answer:

Autobiographical memory  

Explanation:

Autobiographical memory is the memory of specific events that you experienced personally earlier in your life.  

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How do you use nonmetal in a sentence?
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Here Is A Sentence You Could Possibly Use:

"Hydrogen and Nitrogen are two examples of non-metals."

(You can replace the two given elements with whatever two non-metallic elements you like.)

Hope This Helps!

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7 0
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Ammonia and hydrogen fluoride both have unusually high boiling points due to
nexus9112 [7]

Answer:

C. hydrogen bonding

Explanation:

Ammonia and hydrogen fluoride are both able to exhibit hydrogen bonding due to containing nitrogen (in ammonia) and fluoride (obviously in hydrogen fluoride). Remember the unique qualities of NOF. :)

5 0
3 years ago
Define What is boiling
kozerog [31]

Answer:

Boiling is the rapid vaporization of a liquid, which occurs when a liquid is heated to its boiling point, the temperature at which the vapour pressure of the liquid is equal to the pressure exerted on the liquid by the surrounding atmosphere.

Explanation:

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8 0
3 years ago
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}

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

The graphs are shown below.

6 0
3 years ago
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