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Natasha2012 [34]
3 years ago
14

What factors do not affect nuclear reactions, but do affect chemical reaction

Chemistry
1 answer:
Fudgin [204]3 years ago
7 0
Unlike nuclear reactions, nuclear reactions are not affected by changes in temperature,
pressure, of the presence of catalysts. Also nuclear reactions of given radioisotope cannot be slowed down, speeded up, or stopped.
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The molar mass of H2O is 18.01 g/mol. The molar mass of O2is 32.00 g/mol. What mass of H2O,in grams, must react to produce 50.00
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Answer is: mass of water is 56.28 grams.
Chemical reaction: 2H₂O → 2H₂ + O₂.
m(O₂) = 50.00 g.
n(O₂) = m(O₂) ÷ M(O₂).
n(O₂) = 50 g ÷ 32 g/mol.
n(O₂) = 1.5625 mol.
From chemical reaction: n(O₂) : n(H₂O) = 1 : 2.
n(H₂O) = 2 · 1.5625 mol.
n(H₂O) = 3.125 mol.
m(H₂O) = n(H₂O) · M(H₂O).
m(H₂O) = 3.125 mol · 18.01 g/mol.
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3 years ago
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Why dont the present shapes of the continents fit perfectly into a supercontinent
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Erosion? As time passes, the continents move? Some crumble? I don't know but I tried
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g Consider an ideal atomic gas in a cylinder. The upper part of the cylinder is a moveable piston of negligible weight. The heig
kumpel [21]

A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.

A cylindrical weight with a mass (m) of 3 kg is dropped, that is, its initial velocity (u) is 0 m/s and travels 10 m (s). Assuming the acceleration (a) is that of gravity (9.8 m/s²). We can calculate the velocity (v) of the weight in the instant prior to the collision with the piston using the following kinematic equation.

v^{2} = u^{2} + 2as = 2 (9.8 m/s^{2} ) (10m) \\\\v = 14 m/s

The object with a mass of 3 kg collides with the piston at 14 m/s, The kinetic energy (K) of the object at that moment is:

K = \frac{1}{2} m v^{2} = \frac{1}{2} (3kg) (14m/s)^{2} = 294 J

The kinetic energy of the weight is completely converted into heat transferred into the gas cylinder. Thus, Q = 294 J.

Given all the process is at 250 K (T), we can calculate the change of entropy of the gas using the following expression.

\Delta S_{gas} = \frac{Q}{T} = \frac{294 J}{250K} = 1.18 J/K

The change in the entropy of the environment, has the same value but opposite sign than the change in the entropy of the gas. Thus, \Delta S_{env} = -1.18 J/K

A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.

Learn more: brainly.com/question/22655760

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2 years ago
If a firework is light up is the kinetic energy increasing decreasing or the same
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Answer:

The kinetic energy is increasing, and the potential energy is decreasing.

Explanation:

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The atomic number and mass number do not change during this type of radioactive decay.
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In gamma decay, no change in proton number occurs, so the atom does not become a different element

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