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Alinara [238K]
3 years ago
8

Tell correctly I will mark you brainliest...

Chemistry
2 answers:
Oksi-84 [34.3K]3 years ago
8 0

Answer:

toothbrush- insulator

spoon - conductor

key - conductor

chain - conductor..

water- insulator

plastic ruler- insulator

metal wire- conductor..

pins n nails- conductor

Explanation:

katen-ka-za [31]3 years ago
4 0

Answer:

insulator :tooth brush ,plastic ruler,nails

conductor:spoon,metal wire,water,pin

key chain can be insulator or conductor becouse some of the are made by plastic and some of them are made by metal.

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Draw an example each of a hydrogen atom that can and can’t participate in hydrogen bonding.
frutty [35]

Answer:

A hydrogen bonding is a bond class that is produced from the attraction existing in a hydrogen atom and an oxygen, fluorine or nitrogen atom with a negative charge. This attraction, meanwhile, is known as dipole-dipole interaction and links the positive pole of one molecule with the negative pole of another.

Explanation:

The hydrogen atom, which has a positive charge, is known as the donor atom, while the oxygen, fluorine,  chlorine or nitrogen atom is the bond acceptor atom. In the substance in which they are most effective is in the water.

Hydrogen bonds have only one third of the strength of covalent bonds, but they have important effects on the properties of the substances in which they occur, especially in terms of melting and boiling points in crystal structures.

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.

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3 years ago
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Virty [35]
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Hope this helps

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8 0
3 years ago
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Valentin [98]
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