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stira [4]
3 years ago
10

If a Kleenex box has a length of 8cm, width or 2cm, and a height of 4cm what is the volume of the Kleenex box?

Chemistry
1 answer:
Anettt [7]3 years ago
3 0

Answer:

The volume of the Kleenex box is 64 cm.

Explanation:

Volume = B x W x H

So:      V = 8cm x 2cm x 4cm = 64cm

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Explanation:

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3 years ago
It takes 2,500,00 Liters of Helium to fill the Goodyear Blimp. How many moles is this?
Valentin [98]

Answer:

102.26 moles of helium were required to Fill the Goodyear Blimp

Explanation:

To solve this question we need to use combined gas law:

PV = nRT

<em>Where P is pressure, V is volume of gas (2500L), n are moles of gas (Our incognite), R is gas constant (0.082atmL/molK) and T is absolute temperature</em>

<em />

Assuming atmospheric condition we can write P = 1atm and T = 25°C = 298.15K

Replacing:

PV/RT = n

1atm*2500L / 0.082atmL/molK*298.15K = n

<h3>102.26 moles of helium were required to Fill the Goodyear Blimp</h3>

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2 years ago
The scientific process is most similar to
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3 years ago
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alina1380 [7]

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Explanation:

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8 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
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