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3 years ago

What is the molarity of a solution that contains 40.

Chemistry

1 answer:

Allushta [10]3 years ago

8 0

Answer:

Explanation:

because it be an the only answer there

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Helpe pls I will mark brainlest

olya-2409 [2.1K]

Answer:

What is the question needing to be answered?

Explanation:

Usually with a scientific hypothesis, you ask the question in an if, then statement. For example: If it rains outside, then the ground will be wet. It's kind of like a cause and effect statement.

4 0

3 years ago

A chemistry student needs to standardize a fresh solution of sodium hydroxide. She carefully weighs out 385.mg of oxalic acid H2

Vlada [557]

Answer:

The molarity of the sodium hydroxide solution is 0.0692 M

Explanation:

Step 1: Data given

Mass of H2C2O4 = 385 mg = 0.385 grams

volume = 250 mL = 0.250 L

Volume of NaOH = 123.7 mL = 0.1237 L

Molar mass H2C2O4 = 90.03 g/mol

Step 2: The balanced equation

2NaOH + H2C2O4 → Na2C2O4 + 2H2O

Step 3: Calculate moles H2C2O4

Moles H2C2O4 = mass H2C2O4 / molar mass H2C2O4

Moles H2C2O4 = 0.385 grams / 90.03 g/mol

Moles H2C2O4 = 0.00428 moles

Step 4: Calculate molarity of H2C2O4

Molarity H2C2O4 = moles / volume

Molarity H2C2O4 = 0.00428 moles / 0.250 L

Molarity H2C2O4 = 0.01712 M

Step 5: Calculate molarity of NaOH

2*Ca*Va = n*Cb*Vb

⇒ with Ca = Molarity of H2C2O4 = 0.01712 M

⇒ Va = volume of H2C2O4 = 0.250 L

⇒Cb = molarity of NaOH = TO BE DETERMINED

⇒ Vb = volume of NaOH = 0.1237 L

Cb = (2*0.01712*0.250)/0.1237

Cb = 0.0691 M

The molarity of the sodium hydroxide solution is 0.0692 M

4 0

3 years ago

List 5 spheres and give an example of something you would find in each sphere . Ill give 50 points

Brilliant_brown [7]

Hydrosphere, atmosphere, geosphere, biosphere, anthrosphere.

Hydrosphere: Liquid, vapour or ice.

Atmosphere: Oxygen.

Geosphere: Minerals.

Biosphere: Ecosystem.

Anthrosphere: Human habitats

3 0

2 years ago

Read 2 more answers

Write the cell notation for an electrochemical cell consisting of an anode where Mn (s) is oxidized to Mn2 (aq) and a cathode wh

morpeh [17]

Answer:

Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)

Explanation:

In writing the cell notation for an electrochemical cell, the anode is written on the left hand side while the cathode is written on the right hand side. The two half cells are separated by two thick lines which represents the salt bridge.

For the cell discussed in the question; the Mn(s)/Mn^2+(aq) is the anode while the Co^2+(aq)/Co(s) half cell is the cathode.

Hence I can write; Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)

8 0

3 years ago

A colorless liquid has a molar mass of 60.01 g/mol. When the liquid was analyzed, it was 46.7% nitrogen and 53.3% oxygen. What i

MAXImum [283]

first we have to find the empirical formula of the compound

empirical formula is the simplest ratio of whole numbers of components making up a compound

for 100 g of the compound

N O

mass 46.7 g 53.3 g

number of 46.7 g/ 14 g/mol 53.3 g/ 16 g/mol

moles = 3.34 mol = 3.33 mol

divide by the least number of moles

3.34/3.33 = 1.00 3.33/ 3,33 = 1.00

therefore number of atoms are

N - 1

O - 1

empirical formula is - NO

mass of empirical unit - 14 g/mol + 16 g/mol = 30 g

molecular formula is actual composition of elements in the compound

molecular mass - 60.01 g/mol

number of empirical units = molecular mass / empirical unit mass

= 60.01 g/mol / 30 g = 2

there are 2 empirical units

2(NO)

molecular formula = N₂O₂

6 0

3 years ago