Find an answer to an inorganic chemistry fill in the blank question

This. Isn't that a special case of the first law of thermodynamics?

The three conditions are: closed system, no non-volumetric work, and constant pressure.

Where: h is called the enthalpy of heat, or enthalpy, which is a new function of state: a physical quantity that is consistent with the unit of energy; QP stands for the amount of heat absorbed by the system.

qp= h indicates that in the constant pressure reaction, all the heat qp absorbed by the system is used to change the enthalpy of the system.

The analysis is as follows: 1) Because the first law of thermodynamics is expressed as: a system changes from state i to state ii, in this process the system absorbs heat q, does work (volume work) w, and the amount of internal energy change of the system is expressed by u, then there is:

u = q – w ；

2) And because:

In the constant pressure reaction, p = 0,3) has: r u = q w = q p· v = q (pv).

So: q= r u + pv).

q= △r u +△pv)

u2 －u1) +p2v2 －p1v1)

u2 + p2v2) －u1 + p1v1)

u, p, v are all state functions, so u + pv is also a state function, so h = u + pv, then q = (u + pv) i.e.: r h = q

What does anhydrous copper sulphate do?

Summary. Inorganic chemistry is the science that studies the composition, properties, structure, and reactions of inorganic substances, and it is the oldest sub-discipline in chemistry. Inorganic substances include all chemical elements and their compounds, with the exception of most carbon compounds (except carbon dioxide, carbon monoxide, carbonic acid, carbon disulfide, carbonates).

Hello, inorganic chemistry question, inorganic chemistry, is a branch of chemistry that studies the first grade, file preparation, structure, properties, changes and applications of elements, elemental and inorganic compounds. It is of great significance for the comprehensive utilization of mineral resources, the production and research of inorganic raw materials and functional materials in modern technology.

Inorganic chemistry is the study of the composition, properties, structure and reactions of inorganic substances. Inorganic substances include all chemical elements and their compounds, with the exception of most carbon compounds (except carbon dioxide, carbon monoxide, carbonic acid, carbon disulfide, carbonates).

1. Equal, because the standard electrode electricity is only related to the pair itself, and has nothing to do with the reference electrode 2. The charge is balanced, there is a total of 3mol Cl-, and there is 1mol Cl- In the external world, the coordination number is 6, so there are 4 NH3, M=3, N=4

3. You can only make the roughest estimate of the conditions here, then the condition stability constant lgk'(my) = LGK(my)-LG y(h), the exact titration condition is LGK'(my)>=8, so the final measurement is the total amount of Al and Zn.

4. I won't do this either, I'm sorry.

5. Propane combustion: propane + 5 O2 = 3 CO2 + 4 H2O CHM (propane) = QV (propane) + NRT ( N=1) CHM (H2) = FHM (H2O).

RHM = CHM (propylene) + CHM (H2) - CHM (propane) substitution value can be solved.

In the case of isotherm of the fiber, p1v1=p2v2

For example, the partial pressure of hydrogen is imitated. p2=

Indeed, phosphoric acid is one of the most complex problems, there are too many equilibriums in it, and you also have the conservation of materials, which is the sum of all phosphorus concentrations, and the conservation of charges, the equality of positive and negative charges, and finally the formula that lists three equilibrium (three-step ionization of trinoric acids). where the product of acid and base = kw = 10 (-14), so you can calculate it, try it.

To give you an equation, let the equilibrium be that the concentrations of phosphorus are c, c1, c2, c3, hydrogen ions are c4, and hydroxide is kw c4

c+c1+c2+c3= ①

c1+2*c2+3*c3+kw/c4=c4 ②c1*c4/c=ka1 ③

c2*c4/c1=ka2 ④

c4*c3/c2=ka3 ⑤

Five equations and five unknowns can be solved equations, some of which can be simply calculated, such as: the two-step and three-step ionization of phosphoric acid can be basically ignored, that is, when calculating, the concentration of C2 and C3 is basically treated as 0 and the equation in , and C1=C4 can be obtained, so the following calculations are much simpler.

Solve it with chemical principles!

Solve using equilibrium constant expressions. Solve the unary quadratic equation [h+]=[h2po4-]; hpo42-]=ka2...

What's the pH? One less condition.

cds＋h2o→cd2+＋hs-＋oh-

ksp＝(a＋s)×s×s，a＝2×10-6 mol/l，ksp＝8×10^-27

s=[oh-]=，poh=

pH = that is, to make the concentration of CD2+ remaining in the solution not exceed 2 10-6 mol l, the pH should not be less.

Dissolution reaction: CaCO3(s) +2HAC Ca2+ +H2CO3 + 2AC-

Equilibrium concentration (mol l-1) x 2x

k = ca2+] h2co3] [ac-]2 / hac]2= [ca2+] co32-] h2co3] [h+]2 [ac-]2 / co32-] h+]2 [hac]2

ksp(caco3) ka(hac)^2 / ka(h2co3) ka(hco3-)

Check the table: ksp(caco3) =

ka(hac) =

ka(h2co3) =

ka(hco3-)

Substituting the k expression, we calculate: k =

k = x· =

Solution: x = mol l-1

i.e.: [Ca2+] mol L-1

AC-] 2 mol L-1 = mol L-1 dissolves CaCO3 in 1 L solution:

mol 100 g mol = 74 g The required acetic acid concentration is:

mol•l-1 = mol•l-1

You can't do the math here, you have to have all kinds of data. It's a matter of multiple balances. The specific method is to process the expressions of each constant, replace them with algebraic ideas, and finally obtain an expression, and then substitute it into the calculation.

Remember that you can't substitute data in the middle to calculate, otherwise you won't be able to calculate it!

To precipitate Zn2+, a minimum of C(S2-)=ksp(Zns) C(Zn2+)= is required

To precipitate Mn2+, a minimum of C(S2-)=ksp(Mns) C(Mn2+)= is required

It can be seen that ZNS precipitates first.

To make Zns precipitate completely(C(Zn2+)<1 10 (-5)), C(S2-)>Ksp(Zns) C(Zn2+)=, and at the same time not to allow Mn2+ precipitation, C(S2-)H2S = 2H+ +S2-,K=K1K2,C(H2S) when H2S is saturated, C(H2S)=[K1K2·C(H2S) C(S2-)]1 2).

It can be calculated< c(h+) "i.e. < ph<< p>