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Alchen [17]
2 years ago
15

What is the first basic metal on the periodic table

Chemistry
1 answer:
kherson [118]2 years ago
4 0

The first basic metals on the periodic table are alkali metals.
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Routs<br> Which of the following is a physical change?
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Answer:

Explanation:

A physical change is a change in which the physical properties of matter are altered. These are properties are the forms and state.

Most physical changes are easily reversible and are pure state changes.

They  do not lead to the production of new compounds.

They involve no mass change and requires little to no energy.

Examples are melting, boiling, freezing, sublimation e.t.c

5 0
2 years ago
If the solubility of a N30 gas is 2.26g/l at 1.26atm of pressure, what is the solubility of
natali 33 [55]
The answer for your question is 0.27g/L
5 0
2 years ago
A balance is used to measure mass
FrozenT [24]

if this is true or false its true hope i helped

6 0
2 years ago
The formation of iodine is described by the following chemical equation:
olga nikolaevna [1]

Answer:

N2O2(g) +O2(g) ===> 2NO2(g)

Explanation:

For a nonelementary reaction, the reaction equation is described as the sum of all the steps involved. All these steps constitute the reaction mechanism. Each step in the mechanism is an elementary reaction. The rate law of the overall reaction involves the rate determining step (slowest step) in the reaction sequence.

Now look at the overall reaction 2NO(g) + O2(g) ---------> 2NO2(g)

The two steps in the mechanism are

2NO(g) --------->N2O2(g) (fast)

N2O2(g) +O2(g) -------> 2NO2(g) (slow)

Summing all the steps and cancelling out the intermediate N2O2(g), we obtain the reaction equation;

2NO(g) + O2(g) ---------> 2NO2(g)

Hence the answer.

7 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
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