What is the molar mass of chcl3?

"It is not the union that organizes the employees; it is management."

blsea [12.9K]

The statement "it is not the union that organizes the employees; it is management " means that :

The management creates the climate which allows the union to convince employees to accept union representation.

This excerpt shows that although the union helps in the organization of employees for better and easy management of staffs by the management, the management creates an enabling environment by creating stipulations which allows the u ions to convince staffs to join and accept union representation in staff welfare matters.

Hence we can conclude that, the management creates the climate which allows the union to convince employees to accept union representation.

Learn more about union representation : brainly.com/question/10659490

3 0

3 years ago

2KClO3(s) => 2KCl(s) + 3O2(g) is there an increase or decrease of entropy?

ValentinkaMS [17]

Answer:

Entropy increases

Explanation:

Entropy (S) is a measure of the degree of disorder. For a given substance - say water - across phases the following is true ...

S(ice) < S(water) << S(steam)

For a chemical process, entropy changes can be related to increasing or decreasing molar volumes of gas from reactant side of equation to product side of equation. That is ...

if molar volumes of gas increase, then entropy increases, and

if molar volumes of gas decrease, then entropy decreases.

For the reaction 2KClO₃(s) => 2KCl(s) + 3O₂(g)

molar volumes of gas => 0Vm* 0Vm 3Vm

*molar volumes (Vm) apply only to gas phase substances. Solids and liquids do not have molar volume.

Since the reaction produces 3 molar volumes of O₂(g) product vs 0 molar volumes of reactant, then the reaction is showing an increase in molar volumes of gas phase substances and its entropy is therefore increasing.

3 0

3 years ago

At equilibrium, the concentrations of the products and reactants for the reaction, H2 (g) + I2 (g)  2 HI (g), are [H2] = 0.106

lana [24]

Answer:

The new equilibrium concentration of HI: [HI] = 3.589 M

Explanation:

Given: Initial concentrations at original equilibrium- [H₂] = 0.106 M; [I₂] = 0.022 M; [HI] = 1.29 M

Final concentrations at new equilibrium- [H₂] = 0.95 M; [I₂] = 0.019 M; [HI] = ? M

Given chemical reaction: H₂(g) + I₂(g) → 2 HI(g)

The equilibrium constant ( $K_{c}$ ) for the given chemical reaction, is given by the equation:

$K_{c} = \frac {[HI]^{2}}{[H_{2}]\: [I_{2}]}$

At the original equilibrium state:

$K_{c} = \frac {(1.29\: M)^{2}}{(0.106\: M) \times (0.022\: M)}$

$K_{c} = \frac {1.6641}{0.002332} = 713.59$

Therefore, at the new equilibrium state:

$K_{c} = \frac {[HI]^{2}}{(0.95\: M) \times (0.019\: M)}$

$\Rightarrow K_{c} = 713.59 = \frac {[HI]^{2}}{0.01805}$

$\Rightarrow [HI]^{2} = 713.59 \times 0.01805 = 12.88$

$\Rightarrow [HI] = \sqrt {12.88} = 3.589 M$

Therefore, the new equilibrium concentration of HI: [HI] = 3.589 M

6 0

3 years ago

Which best describes how a fetus exchanges gases while in the womb?

vagabundo [1.1K]

Answer:

C. Gas exchange occurs through the placenta.

Explanation:

The fetus is a developing baby in the womb or uterus. The fetus is incapable of performing certain life processes itself and hence, depends on the mother to do them. One of these processes is GAS EXCHANGE. Gas exchange occurs in the fetus via a structure called PLACENTA.

Placenta is a structure that forms in the uterus during pregnancy. It helps the developing fetus supply oxygen and nutrients from the mother and also remove wastes from the fetus back to the mother. The oxygen is taken into the fetus via the PLACENTA while the carbon dioxide (waste product of respiration) is removed via the PLACENTA. Hence, Gas exchange occurs through the placenta.

5 0

3 years ago

In the compound h2s hydrogen has an oxidation number of 1+ and sulfur has an oxidation number of

elena-14-01-66 [18.8K]

Oxidation number is defined as the total number of electrons that are gained or lost by the atom to form a chemical bond.
the oxidation number of the compound H₂S is 0.
the sum of the oxidation numbers of the individual elements should add up to the oxidation number of the compound.

(oxidation number of H x 2 H ions) + oxidation number of S = 0
since we know the oxidation number of H, lets name the oxidation number of S = x
(+1 * 2 )+ (x) = 0
2 + x = 0
x = -2
oxidation number of S is -2

3 0

4 years ago

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