All categories

3 years ago

In a titration, what is the name of the substance that is being determined?

1 answer:

8 0

In a titration, the substance that is unknown and being identified is called analyte. A titration is where a known solution or concentration called the titrant is used to identify and measure an unknown substance which is the analyte.

You might be interested in

A certain liquid X has a normal boiling point of 108.30 °C and a boiling point elevation constant Kb=1.07 °C kg/mol. A solution

Fynjy0 [20]

Answer:

34,6g of (NH₄)₂SO₄

Explanation:

The boiling-point elevation describes the phenomenon in which the boiling point of a liquid increases with the addition of a compound. The formula is:

ΔT = kb×m

Where ΔT is Tsolution - T solvent; kb is ebullioscopic constant and m is molality of ions in solution.

For the problem:

ΔT = 109,7°C-108,3°C = 1,4°C

kb = 1.07 °C kg/mol

Solving:

m = 1,31 mol/kg

As mass of X = 600g = 0,600kg:

1,31mol/kg×0,600kg = 0,785 moles of ions. As (NH₄)₂SO₄ has three ions:

0,785 moles of ions× $\frac{1(NH_{4})_{2}SO_{4}}{3Ions}$ = 0,262 moles of (NH₄)₂SO₄

As molar mass of (NH₄)₂SO₄ is 132,14g/mol:

0,262 moles of (NH₄)₂SO₄× $\frac{132,14g}{1mol}$ = 34,6g of (NH₄)₂SO₄

I hope it helps!

7 0

4 years ago

A 33.0 mL sample of 1.15 M KBr and a 59.0 mL sample of 0.660 M KBr are mixed. The solution is then heated to evaporate water unt

Oksi-84 [34.3K]

Answer:

We need 13.06 grams of silver nitrate to precipitate out silver bromide in the final solution

Explanation:

Step 1: Data given

Sample 1: The 1.15 M sample has a volume of 33.O mL

Sample 2: The 0.660 M sample has a volume of 59.0 mL

Molar mass of KBr = 119 g/mol

Molar mass of AgNO3 = 169.87 g/mol

Step 2: Calculate number of moles for both samples

Number of moles = Molarity * Volume

Sample 1: 1.15 M * 33 *10^-3 L = 0.03795 moles

Sample 2: 0.660 M *59*10^-3 L = 0.03894 moles

Total mol KBr = 0.03795 + 0.03894 = 0.07689 moles

Step 3: Calculate total mass

mass = Number of moles * Molar mass

mass = 0.07689 moles * 119 g/moles = 9.15 grams ( in 55mL)

Step 4: Calculate moles of AgBr

AgNO3 reacts with KBr

KBr(aq) + AgNO3(aq) → AgBr(s) + KNO3(aq)

1 mole of KBr consumed, needs 1 mole of AgNO3 to produce 1 mole of AgBr and 1 mole of KNO3

So 0.07689 moles of KBr wll need 0.07689 moles of AgNO3

Step 5: Calculate mass of silver nitrate

mass of AgNO3 = Moles of AgNO3 * Molar mass of AgNO3

mass of AgNO3 = 0.07689 moles * 169.87 g/mol = 13.06 grams

We need 13.06 grams of silver nitrate to precipitate out silver bromide in the final solution

8 0

3 years ago

A saturated solution of manganese(II) hydroxide was prepared, and an acid–base titration was performed to determine its KspKsp a

trapecia [35]

Answer:

10.945 x 10^-4

Explanation:

Balanced equation:

Mn(OH)2 + 2 HCl --> MnCl2 + H2O

it takes 2 moles HCL for each mole Mn(OH)2

Next find the molarity of the Mn(OH)2 solution

= (1 mole Mn(OH)2 / 2 mole HCl) X (0.0020 mole HCl / 1000ml) X (4.86 ml)

= 4.86 x 10^-3 mole

this is now dissolved in (70 + 4.86) = 74.86 ml or 0.07486 L

thus [Mn(OH)2] = 4.86 x 10^-3 mole / 0.07486 L = 0.064921 M

Ksp = [Mn2+][OH-]^2 = 4x^3 = 4(0.064921)^3 = 10.945 x 10^-4

6 0

3 years ago

Both protons and neutrons (and their anti-particles) froze out:

VladimirAG [237]

I think Both protons and neutrons (and their anti-particles) froze out at 1013 K, about 0.0001 seconds after the Big Bang. Protons and neutrons are sub atomic particles of an atom that are found in the nucleus of an atom. Proton is the positively charge particle while the neutron has no charge. The proton positive charge accounts for the positive nuclear charge.

6 0

4 years ago

nswer the following questions relating to HCl, CH3Cl, and CH3Br.HCl(g), can be prepared by the reaction of concentrated H2SO4(ag

alexdok [17]

Answer:

It is an example of double displacement reaction.

4.8 g of NaCl is needed to react.

Explanation:

Balanced reaction: $H_{2}SO_{4}(aq.)+2NaCl(s)\rightarrow 2HCl(g)+Na_{2}SO_{4}(aq.)$

Here, oxidation states of H, S, O, Na and Cl do not change during reaction. Hence it is not a redox reaction.

In this reaction, cations and anions of the reactants interchange their partners during reaction. Hence, it is an example of double displacement reaction.

As $H_{2}SO_{4}(aq.)$ remain in excess amount therefore NaCl (s) is the limiting reagent. Hence production of HCl entirely depends on amount of NaCl used.

Molar mass of HCl = 36.46 g/mol

So, 3.0 g of HCl = $\frac{3.0}{36.46}$ mol of HCl = 0.082 mol of HCl

According to balanced equation-

2 moles of HCl are produced from 2 moles of NaCl

So, 0.082 moles of HCl are produced from 0.082 moles of NaCl

Molar mass of NaCl = 58.44 g/mol

So, mass of 0.082 moles of NaCl = $(0.082\times 58.44)$ g = 4.8 g

Hence 4.8 g of NaCl is needed to react.

4 0

3 years ago