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2 years ago

What is one thing that could have prevented the Chernobyl tragedy?

Chemistry

2 answers:

irga5000 [103]2 years ago

3 0

Answer: if the staff at Chernobyl had been better trained.

Lilit [14]2 years ago

3 0

The Chernobyl series seems to suggest there were a number of ways the explosion could have been prevented. These include if the staff at Chernobyl had been better trained, if the Soviet government had learned from the lessons of the past and if they had not been so averse to spending money.

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Need help with 1 2 and 3

Pepsi [2]

1 would be the 3 one
Hope it’s correct

4 0

4 years ago

WILL GIVE BRAINLIEST Which effect is one likely result of a forest fire? a. extinction b. adaptation c. speciation d. forced mig

MAVERICK [17]

Answer:

d. forced migration

Explanation:

Certain hazardous occurrences affect living organisms in their natural habitat. One of those occurrences is forest fire. Forest fire or vegetation fire is an uncontrollable break out of fire in a vegetation, affecting the inhabitants of the area.

The occurrence of a forest fire will lead to a forced migration of organisms from their natural habitat. Animals and other mobile organisms will be forced to leave behind their devastating habitat and migrate to a less threatened area in order to survive.

4 0

3 years ago

Read 2 more answers

The orbital diagram shows the valence electron

dimulka [17.4K]

Answer:

its the first one in the 3p section mf

Explanation:

6 0

4 years ago

The equation represents the decomposition of a generic diatomic element in its standard state. 12X2(g)⟶X(g) Assume that the stan

aliina [53]

Answer:

$K^{2000K}=0.774\\\\K^{3000K}=12.56$

Explanation:

Hello,

In this case, considering the reaction, we can compute the Gibbs free energy of reaction at each temperature, taking into account that the Gibbs free energy for the diatomic element is 0 kJ/mol:

$\Delta _rG=\Delta _fG_{X}-\frac{1}{2} \Delta _fG_{X_2}=\Delta _fG_{X}$

Thus, at 2000 K:

$\Delta _rG=\Delta _fG_{X}^{2000K}=4.25kJ/mol$

And at 3000 K:

$\Delta _rG=\Delta _fG_{X}^{3000K}=-63.12kJ/mol$

Next, since the relationship between the equilibrium constant and the Gibbs free energy of reaction is:

$K=exp(-\frac{\Delta _rG}{RT} )$

Thus, at each temperature we obtain:

$K^{2000K}=exp(-\frac{4250J/mol}{8.314\frac{J}{mol\times K}*2000K} )=0.774\\\\K^{3000K}=exp(-\frac{-63120J/mol}{8.314\frac{J}{mol\times K}*3000K} )=12.56$

In such a way, we can also conclude that at 2000 K reaction is unfavorable (K<1) and at 3000 K reaction is favorable (K>1).

Best regards.

4 0

3 years ago

Increasing the temperature of a particular liquid from 298 K to 318 K causes its vapor pressure to double. What is the enthalpy

Delvig [45]

Answer:

27.3 kJ/mol

Explanation:

Step 1: Given data

Temperature 1 (T₁): 298 K

Vapor pressure 1 (P₁): P₁

Temperature 2 (T₂): 318 K

Vapor pressure 2 (P₂): 2 P₁

Step 2: Calculate the enthalpy of vaporization of this liquid

We will use the Clausius–Clapeyron equation.

ln (P₂/P₁) = -ΔHvap/R × (1/T₂ - 1/T₁)

ln 2 = -ΔHvap/(8.314 J/K.mol) × (1/318 K - 1/298 K)

ΔHvap = 2.73 × 10⁴ J/mol = 27.3 kJ/mol

5 0

3 years ago