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wlad13 [49]
3 years ago
15

Why does my dogs poop look green?

Chemistry
1 answer:
julsineya [31]3 years ago
5 0
Your dog may be eating a lot of vegetables. If not that, then they may be having stomach issues/something wrong with their diet.
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How many atoms are in 3.05 grams of copper?
vfiekz [6]

6.02 x10^23 atom  

3.5g x 1mol/63.55g Cu x 6.02 x 10^23/ 1mol=

3.32 x 10^22 atoms

5 0
3 years ago
Which of the following measurements is used to measure the enthalpy of combustion? A. None of these B. Measuring the change in t
DaniilM [7]

Answer: Option (C) is the correct answer.

Explanation:

In a substance, the total energy of its molecular motion is known as heat. Whereas when we measure the average energy of molecular motion of a substance then it is known as temperature.

So, any increase or decrease in temperature will lead to change in heat of a substance.

When one mole of a substance is burned then the amount of energy released  in the form of heat is known as heat of combustion.

Relation between heat and temperature is as follows.

                    q = m \times C \times \Delta T

Thus, we can conclude that to measure the enthalpy of combustion it cannot be measured, only calculated using the equation; q = mc \Delta T.

4 0
3 years ago
Do metals gain or lose electrons as they form ions
Anna71 [15]

Answer:

They usually lose electrons to form ions with 2 positive charges

Explanation:

6 0
3 years ago
Read 2 more answers
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
A bulb is added to a circuit. When you place your hand near the bulb, it feels warm. The bulb in the circuit would most accurate
alex41 [277]

Answer:

C energy source

Explanation:

I got this right

8 0
3 years ago
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