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Sonja [21]
3 years ago
5

Please help It’s due today

Chemistry
2 answers:
nikitadnepr [17]3 years ago
6 0

Answer:

Conductors

Explanation:

zubka84 [21]3 years ago
5 0

That would be B: conductors

(hope this helps ^^)

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As the temperature of a fixed volume of a gas increases the pressure wil
Sati [7]

Answer:

As the temperature of a fixed volume of a gas increase the pressure will increase.

Explanation:

According to the Gay- Lussac's Law,

" The pressure of given amount of gas is directly proportional to the temperature at a constant volume"

Mathematical expression:

           P ∝ T

          P = CT

          P / T = C

As the temperature increase, the pressure also increase.

The initial and final expression of volume and pressure can be written as,

P₁ / T₁  = P₂ / T₂

3 0
3 years ago
The diameter of a biscuit is approximately 51 millimeters (mm). An atom of bismuth (Bi) is approximately 320. picometers (pm) in
astraxan [27]

Answer:

1.5e+8 atoms of Bismuth.

Explanation:

We need to calculate the <em>ratio</em> of the diameter of a biscuit respect to the diameter of the atom of bismuth (Bi):

\\ \frac{diameter\;biscuit}{diameter\;atom(Bi)}

For this, it is necessary to know the values in meters for any of these diameters:

\\ 1m = 10^{3}mm = 1e+3mm

\\ 1m = 10^{12}pm = 1e+12pm

Having all this information, we can proceed to calculate the diameters for the biscuit and the atom in meters.

<h3>Diameter of an atom of Bismuth(Bi) in meters</h3>

1 atom of Bismuth = 320pm in diameter.

\\ 320pm*\frac{1m}{10^{12}pm} = 3.20*10^{-10}m

<h3>Diameter of a biscuit in meters</h3>

\\ 51mm*\frac{1}{10^{3}mm} = 51*10^{-3}m = 5.1*10^{-2}m

<h3>Resulting Ratio</h3>

How many times is the diameter of an atom of Bismuth contained in the diameter of the biscuit? The answer is the ratio described above, that is, the ratio of the diameter of the biscuit respect to the diameter of the atom of Bismuth:

\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1*10^{-2}m}{3.20*10^{-10}m}

\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1}{3.20}\frac{10^{-2}}{10^{-10}}\frac{m}{m}

\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1}{3.20}\frac{10^{-2}}{10^{-10}}\frac{m}{m}

\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{-2+10}

\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{8}=1.5e+8

In other words, there are 1.5e+8 diameters of atoms of Bismuth in the diameter of the biscuit in question or simply, it is needed to put 1.5e+8 atoms of Bismuth to span the diameter of a biscuit in a line.

6 0
3 years ago
Please help with question 4 and 5 please!!
ratelena [41]
Question four is : Gaseous state

Question five is : A rise in temperature increases the kinetic energy and speed of particles; it does not weaken the forces between them. The particles in solids vibrate about fixed positions; even at very low temperatures. Individual particles in liquids and gases have no fixed positions and move chaotically.

Hope this helps!

Have a great day!
8 0
3 years ago
Three friends were kicking soccer balls at a neighborhood field. Three soccer balls were rolling across the field. The black soc
BARSIC [14]

Answer: d :The blue and orange soccer balls; they have more mass than the black soccer ball, but changed speed by the same amount.

3 0
3 years ago
Read 2 more answers
Determine the number of grams of Carbon in 215 g of sucrose.
olya-2409 [2.1K]
<span>just find the percent mass of oxygen in sucrose again. and then multiply that by 50.00.</span>
6 0
3 years ago
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