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vfiekz [6]
3 years ago
12

5. The density of a substance is 3.4 X 10^3 g/mL. Mixing it with another substance increases the density by 1.5 X 10^1 times. Wh

at is the density of the mixture?​
Chemistry
2 answers:
kaheart [24]3 years ago
7 0

Answer:

the total density of the mixture = 3.415 x 10^3 g/mL

Explanation:

i got that correct on exam

JulijaS [17]3 years ago
5 0

Answer:

the total density of the mixture = 3.415 x 10^3 g/mL

Explanation:

given:

density of a substance is 3.4 x 10^3 g/mL

Mixing it with another substance increases the density by 1.5 x 10^1 times.

find:

What is the density of the mixture?​

-------------------------------------------------

let density (a) = 3.4 x 10^3 g/mL

density(b) = 1.5 x 10^1 g/mL

since there is no specific density provided for the mixture, we then add both to increase the density,

total density = density(a) + density(b)

total density = 3.4 x 10^3 g/mL + 1.5 x 10^1 g/mL

total density = 3.415 x 10^3 g/mL

therefore,

the total density of the mixture = 3.415 x 10^3 g/mL

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So its temperature will not rise, since kinetic energy of molecules remains the same. The quantity of heat absorbed or released when a substance changes its physical phase at constant temperature (e g. From solid to liquid at melting point or from liquid to gas at boiling point) is termed as its latent heat.
4 0
2 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

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3 years ago
A scientist digs up sample of arctic ice that is 458,000 years old. He takes it to his lab and finds that it contains 1.675 gram
Fiesta28 [93]

Answer:

6.70 grams of krypton-81 was present when the ice first formed

Explanation:

Let use the below formula to find the amount of sample

N= N_0(\frac{1}{2})^n

where

n = \frac{t}{t_{\frac{1}{2}}}

here

t =  458,000 years

t_{\frac{1}{2}} = 229,000

\frac{t}{t_{\frac{1}{2}}} = \\frac{ 458,000}{229,000}

n = \frac{t}{t_{\frac{1}{2}}} = 2.000

Now substituting the values

1.675 = N_0(\frac{1}{2})^{2.000}}

1.675 = N_0\times (0.2500)

N_0= \frac{1.675}{0.2500}

N_0=6.70

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The _________ __________ is the bottom number on each element?
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The full chemical symbol for an element<span> shows its mass </span>number<span> at the top, and its atomic </span><span>number at the bottom</span>

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The recommended single dose for acetaminophen is 10.0 to 15.0 mg/kilograms of body weight for adults. Using this guideline calcu
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163 lb * 1 kg / 2.205 lb * 15.0 mg/kg = 1108.8 mg or about 1.11 g

6 0
3 years ago
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