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matrenka [14]
2 years ago
14

The point in a titration at which the indicator changes is called the

Chemistry
1 answer:
irakobra [83]2 years ago
8 0

Answer:

End point

Explanation:

The point at which the indicator changes color is called the endpoint. So the addition of an indicator to the analyte solution helps us to visually spot the equivalence point in an acid-base titration

#correct me if I'm wrong

keep safe and study hard

brainliest please thank you

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Explain,using your own example ,why you must always give a unit when reporting a measurement
Sphinxa [80]

Answer : Unit of any measurement can be defined as a definite magnitude of a given quantity, which is defined and adopted by convention or by law, that is widely used as a standard for measurement of the same kind of quantity. If the unit system was not given to the measurements we would haphazardly complicate the calculations.


Example : If someone wrote 500 in the paper as a answer to a question. But forgot to mention the units of the measurements. It can be misinterpreted by anyone as there are no units written besides the number (unit less). Complications grow. One cannot simply predict is it 500 m? 500 g? 500 mL? 500 km? 500 s? or 500 M? anything can be guessed. To be specific about the answer to any problem we should always use units. Which will help to narrow down the approach of any problem and give a measurable related quantity about the associated problem.


4 0
3 years ago
Four (4.0) liters of helium gas are stored at 125 atm pressure. If the temperature and number of particles do not change, what w
seropon [69]

Answer:

10 L

Explanation:

The only variables are pressure and volume, so we can use Boyle's Law:

p1V1 = p2V2

Data:

p1 = 125 atm; V1  = 4.0 L  

p2 =  50 atm; V2 = ?

Calculation:

125 × 4.0 = 50V2

       500 = 50 V2

        V2 = 500/50 = 10 L

The new volume will be 10 L.

5 0
3 years ago
When 78.6 g of urea CH4N2O are dissolved in 700. g of a certain mystery liquid X, the freezing point of the solution is 4.9 °C
iogann1982 [59]

Answer:

The van't Hoff factor of NaCl in liquid X is 1.69

Explanation:

Step 1: Data given

Mass of urea = 78.6 grams

Molar mass of urea = 60.06 g/mol

Mass of liquid X = 700 grams = 0.700 kg

he freezing point of the solution is 4.9°C lower than the freezing point of pure X

When 78.6 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 8.5°C lower than the freezing point of pure X.

Molar mass of NaCl = 58.44 g/mol

Step 2: Calculate moles

Moles urea = mass / molar mass

Moles urea = 78.6 grams / 60.06 g/mol

Moles urea = 1.31 moles

Moles NaCl = 78.9 grams / 58.44 g/mol

Moles NaCl = 1.35 moles

Step 3: Calculate molality

Molality = moles / mass of liquid

Molality urea = 1.31 moles / 0.700 kg

Molality = 1.87 molal

Molality NaCl = 1.35 moles / 0.700 kg

Molality NaCl = 1.92 molal

Step 4: Calculate the freezing point depression constant of X

ΔT = i*Kf*m

⇒with ΔT = the freezing point depression = 4.9 °C

⇒with i = the van't hoff factor of urea = 1

⇒with Kf =the freezing point depression consant of X = TO BE DETERMINED

⇒with m = the molality of urea solution = 1.87 molal

4.9 °C = 1 * Kf * 1.87 molal

Kf == 4.9 / 1.87

Kf = 2.62 °C/m

Step 5: Calculate the van't Hoff facotr of NaCl in X

ΔT = i*Kf*m

⇒with ΔT = the freezing point depression = 8.5 °C

⇒with i = the van't hoff factor of urea = TO BE DETERMINED

⇒with Kf =the freezing point depression consant of X = 2.62 °C/m

⇒with m = the molality of urea solution = 1.92 molal

8.5 °C = i * 2.62 °C/m * 1.92 m

i = 8.5 / (2.62 * 1.92)

i = 1.69

The van't Hoff factor of NaCl in liquid X is 1.69

5 0
2 years ago
What mass of silver chloride can be produced from 1.27L of a 0.156M solution of silver nitrate?
pentagon [3]

Answer:

m AgCl = 28.395 g

Explanation:

  • AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

∴ [AgNO3] = 0.156 mol/L

∴ V = 1.27L

⇒ mol AgNO3 = 0.156 mol/L * 1.27 L = 0.19812 mol AgNO3

mass AgCL:

⇒ m AgCl = 0.19812 mol AgNO3 * mol AgCl/molAgNO3 * 143.32gAgCl/molAgCl

⇒ m AgCl = 28.395 g

4 0
2 years ago
Charge on monatomic ions: a) I b) Sr c) K d) N e) S f) In
wlad13 [49]
A) I = -1, B) Sr = +2, C) K = +1, D) N = -3, E) S = -2, F) In = +3
4 0
2 years ago
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