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3 years ago

PLEASE HELP ME!!!!!!!!

Chemistry

2 answers:

ch4aika [34]3 years ago

7 0

So it has to be either the second or the third one, so let's focus on that. If the substance can make energy in a power plant, then it is most likely Uranium. Uranium is usually radioactive. Now the third one talks about fission, where the atoms that are unstable break apart into smaller atoms to become more stable. I'd say the second one, because it is obviously radioactive.

Mazyrski [523]3 years ago

6 0

I think the answer is c

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Calculate ΔrG∘ at 298 K for the following reactions.CO(g)+H2O(g)→H2(g)+CO2(g)2-Predict the effect on ΔrG∘ of lowering the temper

KonstantinChe [14]

Answer:

1) ΔG°r(298 K) = - 28.619 KJ/mol

2) ΔG°r will decrease with decreasing temperature

Explanation:

CO(g) + H2O(g) → H2(g) + CO2(g)

1) ΔG°r = ∑νiΔG°f,i

⇒ ΔG°r(298 K) = ΔG°CO2(g) + ΔG°H2(g) - ΔG°H2O(g) - ΔG°CO(g)

from literature, T = 298 K:

∴ ΔG°CO2(g) = - 394.359 KJ/mol

∴ ΔG°CO(g) = - 137.152 KJ/mol

∴ ΔG°H2(g) = 0 KJ/mol........pure substance

∴ ΔG°H2O(g) = - 228.588 KJ/mol

⇒ ΔG°r(298 K) = - 394.359 KJ/mol + 0 KJ/mol - ( - 228.588 KJ/mol ) - ( - 137.152 KJ7mol )

⇒ ΔG°r(298 K) = - 28.619 KJ/mol

2) K = e∧(-ΔG°/RT)

∴ R = 8.314 E-3 KJ/K.mol

∴ T = 298 K

⇒ K = e∧(-28.619/(8.314 E-3)(298) = 9.624 E-6

⇒ ΔG°r = - RTLnK

If T (↓) ⇒ ΔG°r (↓)

assuming T = 200 K

⇒ ΔG°r(200 K) = - (8.314 E-3)(200)Ln(9.624E-3)

⇒ ΔG°r (200K) = - 19.207 KJ/mol < ΔG°r(298 K) = - 28.619 KJ/mol

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3 years ago

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Answer:

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Explanation:

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2 years ago

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2 years ago

A steel beam that is 7.00 m long weighs 326 N. It rests on two supports, 3.00 m apart, with equal amounts of the beam extending

GrogVix [38]

Answer:

The answer to the question is

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Explanation:

To solve the question we list out the known variables first

Length of steel beam, L = 7.00 m

Weight of steal beam W = 326 N

Distance between supports = 3.00 m

Weight of Suki = 555 N

By moments theory, the sum of moments about a point = 0

and assuming the steal beam is uniform, the weight of the beam will act at the center of the beam

As described the supports are 2.00 m from each end of the beam thus

taking moments about one of the support with Suki between the end of the beam and the support we have

Distance from center of beam to support = 1.5 m

Distance of Suki to the nearest support = x m

therefore 555 × x + 326 × 1.5 = 0

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Therefore Suki is (2 - 0.881) m or 1.12 m close to the end before the beam begins to tip

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3 years ago