when 25 ml of 0.12 m aqueous ammonia is titrated with 0.12 m hydrobromic acid, what is the ph at the equivalence point?

Why does thermal energy decreases when potassium is dissolved in water

Levart [38]

Answer:

The process of dissolving can be endothermic (temperature goes down) or exothermic (temperature goes up).

When water dissolves a substance, the water molecules attract and “bond” to the particles (molecules or ions) of the substance causing the particles to separate from each other.

The “bond” that a water molecule makes is not a covalent or ionic bond. It is a strong attraction caused by water’s polarity.

It takes energy to break the bonds between the molecules or ions of the solute.

Energy is released when water molecules bond to the solute molecules or ions.

If it takes more energy to separate the particles of the solute than is released when the water molecules bond to the particles, then the temperature goes down (endothermic).

If it takes less energy to separate the particles of the solute than is released when the water molecules bond to the particles, then the temperature goes up (exothermic).

Explanation:

8 0

3 years ago

How many nitrogen atoms are in 2.79 mo.of NH3

alexandr1967 [171]

1 mol of any particles has 6.02 * 10 ²³ particles.

If we look at 1 NH3 (1 mol NH3 or 1 molecule NH3), we can see that 1 molecule NH3 has 1 atom of N and 3 atoms of H; also 1 mole of NH3 has 1 mole of N atoms and 3 moles of H atoms.

So, 1 mol of NH3 has 1 mol of N atoms,
and 2.79 mol NH3 have 2.79 mol of N atoms.

2.79 mol of N atoms* 6.02 * 10 ²³ N atoms/ 1 mol N atoms = 1.68*10²⁴ N-atoms

Answer is 1.68*10²⁴ N-atoms.

5 0

3 years ago

If element R has 2 valence electron and element Q has 6 valence electrons, what is the likely compound that will form:

kvv77 [185]

Explanation:

Sydney has been the work that is a great place 56 to be able and a day and day for a week or two types and we will have a very nice time of course but 6yy6667 has 7575665yy44y

4 0

2 years ago

Reliable results are results that can be...<br> A: communicated<br> B: evaluated<br> C: duplicated

insens350 [35]

Reliable results are results that can be... A: Communicated

4 0

3 years ago

Read 2 more answers

The equilibrium constant Kc for the reaction PCl3(g) + Cl2(g) ⇌ PCl5(g) is 49 at 230°C. If 0.70 mol of PCl3 is added to 0.70 mol

Ymorist [56]

Answer : The correct option is, (B) 0.11 M

Solution :

First we have to calculate the concentration $PCl_3$ and $Cl_2$ .

$\text{Concentration of }PCl_3=\frac{\text{Moles of }PCl_3}{\text{Volume of solution}}$

$\text{Concentration of }PCl_3=\frac{0.70moles}{1.0L}=0.70M$

$\text{Concentration of }Cl_2=\frac{\text{Moles of }Cl_2}{\text{Volume of solution}}$

$\text{Concentration of }Cl_2=\frac{0.70moles}{1.0L}=0.70M$

The given equilibrium reaction is,

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

Initially 0.70 0.70 0

At equilibrium (0.70-x) (0.70-x) x

The expression of $K_c$ will be,

$K_c=\frac{[PCl_5]}{[PCl_3][Cl_2]}$

$K_c=\frac{(x)}{(0.70-x)\times (0.70-x)}$

Now put all the given values in the above expression, we get:

$49=\frac{(x)}{(0.70-x)\times (0.70-x)}$

By solving the term x, we get

$x=0.59\text{ and }0.83$

From the values of 'x' we conclude that, x = 0.83 can not more than initial concentration. So, the value of 'x' which is equal to 0.83 is not consider.

Thus, the concentration of $PCl_3$ at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of $Cl_2$ at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of $PCl_5$ at equilibrium = x = 0.59 M

Therefore, the concentration of $PCl_3$ at equilibrium is 0.11 M

3 0

3 years ago