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Mademuasel [1]
3 years ago
13

Why are there wires in circuits?what is their job/function?

Chemistry
2 answers:
lisabon 2012 [21]3 years ago
8 0
Wires send electricity to point A to point B. In some cases, they send electricity both ways. So basiclly its a road taveling to Point B, Point A being the outlet
ra1l [238]3 years ago
3 0
To connect the parts of a circuit and transport electricity in it from one place to nother
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A 65.0-gram sample of some unknown metal at 100.0° C is added to 100.8 grams of water at 22.0° C. The temperature of the water r
vampirchik [111]
We do a heat balance to solve this:
(m cp ΔT)water = -(m cp ΔT)metal
100.8 (4.18) (27 - 22) = -65 (cp)(27-100)
cp = 100.8 (4.18) (27 - 22) / (-65 (27-100))
cp = 0.44 J/ (°C × g)

The specific heat of the metal is 0.44 J/ (°C × g)
5 0
3 years ago
When chemical, transport, or mechanical work is done by an organism, what happens to the heat generated?
dalvyx [7]

Explanation:

The heat generated when chemical, transport, or mechanical work is done by the organism is lost to the environment.

The the matter from one organism to the other is transferred via energy. The producer's produce and consumer's consume this energy according to 10% law. Most of the energy is lost (90% ) to the environment in form of heat.

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3 years ago
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3 years ago
The density of potassium, which has the BCC structure, is 0.855 g/cm3. The atomic weight of potassium is 39.09 g/mol. Calculate
Delicious77 [7]

<u>Answer:</u>

<u>For a:</u> The edge length of the crystal is 533.5 pm

<u>For b:</u> The atomic radius of potassium is 231.01 pm

<u>Explanation:</u>

  • <u>For a:</u>

To calculate the lattice parameter or edge length of the crystal, we use the equation:

\rho=\frac{Z\times M}{N_{A}\times a^{3}}

where,

\rho = density = 0.855g/cm^3

Z = number of atom in unit cell = 2  (BCC)

M = atomic mass of metal = 39.09 g/mol

N_{A} = Avogadro's number = 6.022\times 10^{23}

a = edge length of unit cell = ?

Putting values in above equation, we get:

0.855=\frac{2\times 39.09}{6.022\times 10^{23}\times (a)^3}\\\\\a=5.335\times 10^{-8}cm=533.5pm

<u>Conversion factor:</u>  1cm=10^{10}pm

Hence, the edge length of the crystal is 533.5 pm

  • <u>For b:</u>

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

R=\frac{\sqrt{3}a}{4}

where,

R = radius of the lattice = ?

a = edge length = 533.5 pm

Putting values in above equation, we get:

R=\frac{\sqrt{3}\times 533.5}{4}=231.01pm

Hence, the atomic radius of potassium is 231.01 pm

4 0
3 years ago
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