An unknown solution has a ph of 2. How would you classify this solution

What is the periodic trend for electronegativity

soldier1979 [14.2K]

Answer:

mark me brainlest

Explanation:

Electronegativity values generally increase from left to right across the periodic table. Electronegativities generally decrease from top to bottom of a group. The highest electronegativity value is for fluorine.

3 0

2 years ago

Read 2 more answers

How many molecules are in 120 grams of Na2SO4

brilliants [131]

3.06 × 10^23 molecules

5 0

3 years ago

Write a balanced net ionic equation for the following reaction: SrCl2(aq) + H2CO3(aq) → SrCO3(s) + HCl (aq)

max2010maxim [7]

We will balance the equation in the following order: metals, amethals, carbon, hydrogen and oxygen (the most common order).
The metal present in the equation is Sr, which is already balanced (there are 1 on each side of the equation).
The amethal present in the equation is Cl. There is 2 Cl in the left side and only one in the right side. So, we will multiply the quantity of the molecule that contains Cl by 2. Doing this, we'll obtain:

$SrCl_2_{(aq)}+H_2CO_3_{(aq)}\to SrCO_3_{(s)}+2HCl_{(aq)}$

Looking at the equation, we can see that it is now fully balanced. Hence, a balanced equation of the reaction is:

$SrCl_2_{(aq)}+H_2CO_3_{(aq)}\to SrCO_3_{(s)}+2HCl_{(aq)}$

5 0

3 years ago

Which of the following are examples of conservable quantities? a. potential energy and length c. mechanical energy and mass b. m

Blababa [14]

Answer:

b

Explanation:

7 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