Answer:
the individual atom in the molecule
Explanation:
In chemistry, the ball-and-stick model is a molecular model of a chemical substance. Invidual spheres there represent atoms in the molecule. The bigger atomic number the atom has, the larger diameter of the spheres this atom has in this model.
I hope this answer will help you. Have a nice day !
I believe that it would be Al1N1.
Problem One
You will use both m * c * deltaT and H = m * heat of fusion.
Givens
m = 12.4 grams
c = 0.1291
t1 = 26oC
t2 = 1204
heat of fusion (H_f) = 63.5 J/grams.
Equation
H = m * c * deltaT + m * H_f
Solution
H = 12.4 * 0.1291 * (1063 - 26) + 12.4 * 63.5
H = 1660.1 + 787.4
H = 2447.5 or 2447.47 is the exact answer. I have to leave the rounding to you. I have no idea where to round it although I suspect 2450 would be right for 3 sig digs.
Problem Two
Formula and Givens
t1 = 14.5
t2 = 50.0
E = 5680
c = 4.186
m = ??
E = m c * deltaT
Solution
5680 = m * 4.186 * (50 - 14.5)
5680 = m * 4.186 * (35.5)
5680 = m * 148.603 * m
m = 5680 / 148.603
m = 38.22 grams That isn't very much. Be very sure you are working in joules. You'd leave that many grams in the kettle after drying it thoroughly.
m = 38.2 to 3 sig digs.
Answer:
- 130.64°C.
Explanation:
- We can use the general law of ideal gas:<em> PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n and P are constant, and have two different values of V and T:
<em>V₁T₂ = V₂T₁</em>
<em></em>
V₁ = 634.0 L, T₁ = 21.0°C + 273 = 294.0 K.
V₂ = 307.0 L, T₂ = ??? K.
<em>∴ T₂ = V₂T₁/V₁ </em>= (307.0 L)(294.0 K)/(634.0 L) = <em>142.36 K.</em>
<em>∴ T₂(°C) = 142.36 K - 273 = - 130.64°C.</em>
Nuclear Fusion is the answer to the question who posted.
