CaF2(s)⇄Ca2+(aq)+2F−(aq) Ksp=3.9×10−11

PLEASE ANSWEER THIS HURRY HURRRRRRRRRYY

klemol [59]

The first one would be it

6 0

4 years ago

Help this mirrorball, please

grin007 [14]

Answer:

Option B

Explanation:

Avagadro's hypothesis showed that at constant temperature and pressure equal volume of all gases contains equal no of molecules.

Avagadro's constant is known as 6.022×10^23

5 0

3 years ago

Read 2 more answers

Please help! :)

Reika [66]

Answer:

There will be 525.2 grams of K3N produced

Explanation:

Step 1: Data given

Number of moles of potassium oxide ( K2O) = 6 moles

Magnesium nitride (Mg3N) = in excess

Molar mass of K3N = 131.3 g/mol

Step 2: The balanced equation

Mg3N2 + 3K2O → 3MgO + 2K3N

Step 3: Calculate moles of K3N

The limiting reactant is K2O.

For 1 mol Mg3N2 consumed, we need 3 moles of K2O to produce 3 moles of MgO and 2 moles of K3N

For 6 moles K2O we'll have 2/3 * 6 = 4 moles of K3N

Step 4: Calculate mass of K3N

Mass of K3N = moles K3N * molar mass K3N

Mass of K3N = 4 moles * 131.3 g/mol

Mass of K3N = 525.2 grams

There will be 525.2 grams of K3N produced

8 0

3 years ago

Find the enthalpy of neutralization of HCl and NaOH. 137 cm3 of 2.6 mol dm-3 hydrochloric acid was neutralized by 137 cm3 of 2.6

liraira [26]

Answer : The correct option is, (D) 89.39 KJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.3562 mole of HCl neutralizes by 0.3562 mole of NaOH

Thus, the number of neutralized moles = 0.3562 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $137ml+137ml=274ml$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 274ml=274g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 274 g

$T_{final}$ = final temperature of water = 325.8 K

$T_{initial}$ = initial temperature of metal = 298 K

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

$q=274g\times 4.18J/g^oC\times (325.8-298)K$

$q=31839.896J=31.84KJ$

Thus, the heat released during the neutralization = -31.84 KJ

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -31.84 KJ

n = number of moles used in neutralization = 0.3562 mole

$\Delta H=\frac{-31.84KJ}{0.3562mole}=-89.39KJ/mole$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 89.39 KJ/mole

3 0

4 years ago

Which method would increase the solubility of a gas

ch4aika [34]

Increasing the partial pressure of the gas over the liquid or lowering the temperature of the liquid can increase the solubility of a gas in a liquid. I hope this. Let me know if anything is unclear or if you want any further explanation.

3 0

3 years ago

Read 2 more answers