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Paul [167]
3 years ago
7

a solution is an acid or base and it doesn't react with metal. is it pH more likely to be 4 or 9? explain your answer

Chemistry
1 answer:
coldgirl [10]3 years ago
6 0
It is more likely 9. pH 4 is acidic and pH 9 is basic, and as the pH of a substance gets closer to 0 or 14, the substance becomes more corrosive or reactive. As 4 is closer to 0 than 9 is to 14, there is a much higher chance the solution has a pH of 9, because pH 4 is less neutral and therefore more corrosive/reactive than pH 9.
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How many oxygen molecules are in 22.4 liters of oxygen gas at 273 K and 101.3 kPa
Dmitriy789 [7]
This 273 K and 101.3 kPa is Standart conditions, so 22.4 L is 1 mol, 1 mol anything contains <span>6.023 x 10^23, 
AnswerA</span>
4 0
3 years ago
Read 2 more answers
5. Which coefficients correctly balance the formula equation
Tju [1.3M]

Answer:

which of these sentences contains a correctly formed possessive form of the name andres

8 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

5 0
3 years ago
How many moles of CaCl2 are contained in 0.448 L of a 0.85 M CaCl2 solution?
kati45 [8]

Answer:

0.3808

Explanation:

number of moles,n=Conc.XVol.

hence 0.85X0.448

3 0
3 years ago
If the freezing point of the solution had been incorrectly read 0.3 °C lower than the true freezing point, would the calculated
Dovator [93]

Answer : The molar mass of the solute would be low.

Explanation :

Formula used for depression in freezing point is:  

\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{w_b}{M_b}\times w_a}

where,

\Delta T_f = change in freezing point

\Delta T_s = freezing point of solution

\Delta T^o = freezing point of water

i = Van't Hoff factor

K_f = freezing point constant

m = molality

w_b = mass of solute

w_a = mass of solvent

M_b = molar mass of solute

From the formula we conclude that, when the freezing point of the solution read incorrectly that is freezing point of the solution is lower than the true freezing point then this means that change in freezing point would be high and the molar mass of the solute would be low.

Hence, the molar mass of the solute would be low.

6 0
3 years ago
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