Please help me solve this

Tính [H] và [OH] của dung dịch HC1 0,1M? Tính pH của dung dịch?

Brrunno [24]

Answer:

po jakiemu to jest XD

Explanation:

po jakiemu to jest XD

3 0

3 years ago

Ammonia NH3 may react with oxygen to form nitrogen gas and water.4NH3 (aq) + 3O2 (g) \rightarrow 2 N2 (g) + 6H2O (l)If 2.15g of

bagirrra123 [75]

Answer:

NH3 is the limiting reactant

The % yield is 36.1 %

Explanation:

Step 1: Data given

Mass of NH3 = 2.15 grams

Mass of O2 = 3.23 grams

Molar mass of NH3 = 17.03 g/mol

Molar mass of O2 = 32 g/mol

volume of N2 produced = 0.550 L

Temperature = 295 K

Pressure = 1.00 atm

Step 2: The balanced equation:

4NH3 (aq) + 3O2 (g) → 2 N2 (g) + 6H2O (l)

Step 3: Calculate moles of NH3

Moles NH3 = Mass NH3 / Molar Mass NH3

Moles NH3 = 2.15 grams / 17.03 g/mol

Moles NH3 = 0.126 moles

Step 4: Calculate moles of O2

Moles O2 = 3.23 grams / 32 g/mol

Moles O2 = 0.101 moles

Step 5: Calculate the limiting reactant

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

NH3 is the limiting reactant. It will completely be consumed ( 0.126 moles).

O2 is in excess, there will be 3/4 * 0.126 = 0.0945 moles consumed

There will remain 0.101 - 0.945 = 0.0065 moles of O2

Step 6: Calculate moles of N2

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

For 4 moles NH3 , we'll have 2 moles of N2 produced

For 0.126 moles NH3 consumed, we'll have 0.063 moles of N2 produced.

Step 7: Calculate volume of N2 produced

p*V = n*R*T

⇒ with p = the pressure of the gas = 1.00 atm

⇒ with V = the volume = TO BE DETERMINED

⇒ with n = the number of moles N2 = 0.063 moles

⇒ with R = the gasconstant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 295

V = (nRT)/p

V = (0.063*0.08206*295)/1

V = 1.525 L = theoretical yield

Step 8: Calculate the % yield

% yield = actual yield / theoretical yield

% yield = (0.550 L / 1.525 L)*100%

% yield = 36.1 %

4 0

3 years ago

If the half-life of a radioisotope is 10,000 years, the amount remaining after 20,000 years is

Cloud [144]

Answer:

Considering the half-life of 10,000 years, after 20,000 years we will have a fourth of the remaining amount.

Explanation:

The half-time is the time a radioisotope takes to decay and lose half of its mass. Therefore, we can make the following scheme to know the amount remaining after a period of time:

Time_________________ Amount

t=0_____________________x

t=10,000 years____________x/2

t=20,000 years___________x/4

During the first 10,000 years the radioisotope lost half of its mass. After 10,000 years more (which means 2 half-lives), the remaining amount also lost half of its mass. Therefore, after 20,000 years, the we will have a fourth of the initial amount.

6 0

3 years ago

Write the chemical equation for fot the reaction between mgcl2 and the soap you preapred

In-s [12.5K]

Soap is the sodium or potassium salt of long chain of fatty acid. Fatty acids when treated with NaOH or KOH forms Soap. This process is called as Saponification. Examples of Soap are as follow,

1. Sodium Stearate C₁₇H₃₅COONa

2. Potassium Oleate C₁₇H₃₃COOK

Reaction of Soap with MgCl₂;

When Soap is treated with MgCl₂ or CaCl₂ it forms insoluble precipitate called S.C.U.M. The reactions with MgCl₂ are as follow,

2C₁₇H₃₅COONa + MgCl₂ --------> 2C₁₇H₃₅COOMg + 2 NaCl

2C₁₇H₃₃COOK + MgCl₂ --------> 2C₁₇H₃₅COOMg + 2 KCl

These reaction are often found in hard water. And this reaction decreases the effectiveness of soap.

5 0

3 years ago

What amount of sucrose (C12H22O11) should be added to 5.83 mol water to lower the vapor pressure of water at 50 °C to 72.0 torr?

Triss [41]

Answer:

The amount of sucrose that must be added is 1.66 moles

Explanation:

Colligative property of lowering vapor pressure has this formula:

Vapor pressure of pure solvent (P°) - Vapor pressure of solution = P° . Xm

We have both vapor pressure (pure solvent and solution9, so let's determine the ΔP

ΔP = 92.6 Torr - 72 Torr = 20.6 Torr

Let's add the data in the formula

20.6 Torr = 92.6 Torr . Xm

Xm = Mole fraction of solute → (mol of solute/ mol of solute + mol of solvent)

Mol of solvent = 5.83 mol (data from the problem)

Therefore Xm = 20.6 Torr / 92.6 Torr → 0.222

Let's find out the moles of solute (our unknown value)

0.22 = moles of solute / moles of solute + 5.83 moles of solvent

0.222 (moles of solute + 5.83 moles of solvent) = moles of solute

0.222 moles of solute + 1.29 moles of solvent = moles of solute

1.29 moles of solvent = moles of solute - 0.222 moles of solute

1.29 moles = 0.778 moles of solute

1.29 / 0.778 = moles of solute → 1.66 moles

3 0

3 years ago