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3 years ago

The point in a titration when the added amount of standard reagent is equal to the amount of analyte being titrated.

Chemistry

1 answer:

creativ13 [48]3 years ago

8 0

That point is called as EQUILIBRIUM, and then you have to note that value to compare with others

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A compound with the empirical formula ch2 has a molar mass of 98 g/mol what is the molecular formula for this compound ?

madam [21]

Approximate molecular masses:
Molecular mass of C = 12
Molecular mass of H = 1

Let n = moles required for CH₂.
Then
nCH₂ = 98
n(12 + 2*1) = 98
14n = 98
n = 7

Answer: The molecular formula is 7CH₂

7 0

2 years ago

What is the complete ionic equation for

murzikaleks [220]

Answer:

what is the net ionic equation

H2SO4(aq) + Cal2(aq) → CaSO4(s) + 2Hl(aq)?

A. H++ SO42- + Ca2+ + 21 → CaSO4 + H+ +1-

B. 2H+ + S042- + Ca2+ + 21° → Ca2+ + SO42- + 2H+ + 21

C. S042- + Ca2+ → CaSO4,

D. 2H+ + SO42- + Ca2+ + 2I- → CaSO4 + 2H+ + 2I-

cancel the spectator ion that is the ions which does not take place in the reaction

for this case is 2 H^+ and 2 i^-

5 0

3 years ago

The substances in a chemical reaction that are combined or separated to form new substances are the_______

djyliett [7]

The answer is reactants.

5 0

3 years ago

H2 + O2 ---> H2O<br> Classifying reaction

Degger [83]

Answer:

hydrogen + oxygen = water

3 0

3 years ago

Read 2 more answers

A sailor on a trans-Pacific solo voyage notices one day that if he puts 735.mL of fresh water into a plastic cup weighing 25.0g,

Gennadij [26K]

Answer:

Amount of salt in 1 L seawater = 34 g

Explanation:

According to Archimedes' principle, mass of freshwater and cup = mass of equal volume of seawater

mass of freshwater = density * volume

1 cm³ = 1 mL

mass of freshwater = 0.999 g/cm³ * 735 cm³ = 734.265 g

mass of freshwater + cup = 734.265 + 25 = 759.265 g

Therefore, mass of equal volume of seawater = 759.265 g

Volume of seawater displaced = 735 mL = 0.735 L (assuming the cup volume is negligible)

1 liter = 1000 cm³ = 1000 mL;

Density of seawater = mass / volume

Density of seawater = 759.265 g / 0.735 L = 1033.01 g/L

Density of freshwater in g/L = 0.999 g/ (1/1000) L = 999 g/L

mass of 1 Liter seawater = 1033.01 g

mass of 1 Liter freshwater = 999 g

mass of salt dissolved in 1 L of seawater = 1033.01 g - 999 g = 34.01 g

Therefore, amount of salt in 1 L seawater = 34 g

4 0

3 years ago