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3 years ago

Why are car crashes more dangerous at higher speeds than at lower speeds

Chemistry

2 answers:

igomit [66]3 years ago

7 0

The impact of speed and velocity at greater speeds affects the car much greater than at slower speeds due to impact and timing and velocity and speed.

musickatia [10]3 years ago

4 0

When you go fast it hits harder so it causes more damage and less damage when slow,
Hope this helps xoxo

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What is the pH of a 2.5 x 10-5 M solution of HCI?

aalyn [17]

Answer:

The pH of the solution is 4.60.

Explanation:

The pH gives us an idea of the acidity or basicity of a solution. More precisely, it indicates the concentration of H30 + ions present in said solution. The pH scale ranges from 0 to 14: from 0 to 7 corresponds to acid solutions, 7 neutral solutions and between 7 and 14 basic solutions. It is calculated as:

pH = -log (H30 +)

pH= -log (2,5 x 10-5)

pH=4.60

7 0

3 years ago

When circumstances are right, most materials undergo a process called thermal expansion. Which statement best describes the caus

harina [27]

Answer: Matter is heated, and its particles spread out more

Explanation:

Thermal expansion occurs when there's an expansion of an object or material or when an object becomes bigger because of a rise in its temperature. This brings about the faster movement of the heated molecules and the atoms spreading out.

Therefore, the cause and effect of thermal expansion will be that when matter is heated, and its particles spread out more.

6 0

3 years ago

The protein lysozyme unfolds at a transition temperature of 75.5°C, and the standard enthalpy of transition is 509 kJ mol-1. Cal

spin [16.1K]

Answer:

0.4774 KJ/K.mol

Explanation:

We are told that the transition at 25.0°C occurs in three steps. Steps i, ii and iii.

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii

Now,

C_p,m(unfolded protein) = C_p,m(folded protein) + 6.28 kJ/K.mol

Now, for the first process, ΔS_i is given as;

ΔS_i = C_p,m × In(T2/T1)

We are given;

T1 = 25°C = 25 + 273.15K = 298.15 K

T2 = 75.5°C = 75.5 + 273.15 K=348.65 K

Thus;

ΔS_i = C_p,m × In(348.65/298.15)

Now, for the third process, ΔS_iii is given as;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(T1/T2)

Thus;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)

Now, we don't know C_pm. So, we have to find a way to eliminate it. We will do it by rewriting In(298.15/348.65) in such a way that when ΔS_iii is added to ΔS_i, C_p,m will cancel out. Thus;

In(298.15/348.65) can also be written as;

In(348.65/298.15)^(-1) or

- In(348.65/298.15)

Thus;

ΔS_iii = - [(C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)]

Now, let's add ΔS_iii to ΔS_i to get;

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] + [(-C_p,m - 6.28 kJ/K.mol) × In(348.65/298.15)]

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] - [C_p,m × In(348.65/298.15)] - [6.28In(348.65/298.15)]

First 2 terms will cancel out to give;

ΔS_i + ΔS_iii = -6.28In(348.65/298.15)

ΔS_i + ΔS_iii = -0.9826 KJ/K.mol

Now,for process ii;

ΔS_ii = standard enthalpy of transition/Transition Temperature

Thus;

ΔS_ii = (509 KJ/K.mol)/348.65

ΔS_ii = 1.46 KJ/K.mol

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii = -0.9826 + 1.46 = 0.4774 KJ/K.mol

5 0

3 years ago

When a given reaction is conducted in a calorimeter, energy is absorbed from the surrounding water and results in a decrease in

rjkz [21]

The potential energy of the products is higher than the potential energy of the reactants.

3 0

3 years ago

Read 2 more answers

Predict whether the changes in enthalpy, entropy, and free energy will be positive or negative for the boiling of water, and exp

Sedbober [7]

Answer:

ΔH > 0; ΔS >0; ΔG = 0

Not spontaneous when T < 100 °C;

Equilibrium when T = 100 °C

Spontaneous when T > 100 °C

Step-by-step explanation:

The process is

H₂O(ℓ) ⇌ H₂O(g)

ΔH > 0 (positive), because we must add heat to boil water

ΔS > 0 (positive), because changing from a liquid to a gas increases the disorder

ΔG = 0, because the liquid-vapour equilibrium process is at equilibrium at 100 °C

ΔG = ΔH – TΔS

Both ΔH and ΔS are positive.

If T = 100 °C, ΔG =0. ΔH = TΔS, and the system is at equilibrium.

If T < 100 °C, the ΔH term will predominate, because T has decreased below the equilibrium value.

ΔG > 0. The process is not spontaneous below 100 °C.

If T > 100 °C, the TΔS term will predominate, because T has increased above the equilibrium value.

ΔG < 0. The process is spontaneous above 100 °C.

4 0

3 years ago