A single paper clip weighs about one gram. So 10 paper clips would weigh?

GoOd MoRnInG EvErY OnE BlA BlA BlA BlA

azamat

Answer:

you too ^^

Explanation:

4 0

2 years ago

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Alkenes ______

Vlada [557]

Answer:

B

Explanation:

alkene is no polar compounds

and insoluble in water

6 0

2 years ago

Which of the following represents the correct temperature and pressure at STP?

melamori03 [73]

B is the correct answer if I remember

3 0

3 years ago

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Which of the following aqueous solutions would have the highest boiling point? 3.0 M solution of molecular compound sucrose (C12

Ne4ueva [31]

Answer: option b) 3.0 M solution of ionic compound aluminum chloride (AlCl₃)

Explanation:

1) The solutions given are:

a) 3.0 M solution of molecular compound sucrose (C₁₂H₂₂O₁₁)

b) 3.0 M solution of ionic compound aluminum chloride (AlCl₃)

c) 3.0 M solution of ionic compound lithium bromide (LiBr)

d) 3.0 M solution of ionic compound calcium fluoride (CaF₂)

2) The elevation of the boiling point of a solution is a colligative property.

3) That means that the elevation of the boiling point depends on the number of solute particles in the solvent and not the identity per se of the solute particles.

4) The formula of the elevation of the boiling point, ΔTb is:

ΔTb = i × m × Kb.

Where:

i) i is the Van't Hoof constant and is equal to the number of particles for each molecule or unit formula.

If the compound is a molecule (covalent bond) i = 1, if the compound is ionic, i is the number of ions product of the dissociation (I will show you how to determine it for each of the compounds given).

ii) m is the molality of the solution (moles of solute per kg of solvent)

iii) Kb is the molality boiling constant. It is a constant for every solvent. It does not change with the solute.

4) For our four compounds, m and kb are egual, so you must analyze the i factor for each solution.

a) 3.0 M solution of molecular compound sucrose (C₁₂H₂₂O₁₁)

Since it is a molecular compound it does not dissociate and i = 1

b) 3.0 M solution of ionic compound aluminum chloride (AlCl₃)

Assume the ionic compound dissociates 100%. The for eveary unit of AlCl₃ there will be 4 ions. Those are 4 particles and means that i = 4.

c) 3.0 M solution of ionic compound lithium bromide (LiBr)

Assuming also 100% dissociation of this ionic compound, since every LiBr unit dissociates into two ions, i = 2.

d) 3.0 M solution of ionic compound calcium fluoride (CaF₂)

Following same reasoning, i = 3.

5) Conclusion: since the greatest i factor is 4, and given all the other factors equal, the 3.0 M solution of ionic compound aluminum chloride (AlCl₃), option b, would be the aqueous solutions with the highest boiling point,

8 0

2 years ago

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Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2N

Gnesinka [82]

Answer:

NH3 is the limiting reactant

The theoretical yield is 216.0 kg urea

The % for this reaction is 78.8 %

Explanation:

Step 1: Data given

Mass of ammonia = 122.5 kg

Mass of carbon dioxide = 211.4 kg

Mass of urea produced = 170.3 kg

Molar mass of ammnoia = 17.031 g/mol

Molar mass of carbon dioxide = 44.01 g/mol

Moalr mass of urea = 60.06 g/mol

Step 2: The balanced equation

2NH3(aq) + CO2(aq) --> CH4N2O(aq) + H2O(l)

Step 3: Calculate moles of NH3

Number of moles = mass / Molar mass

Moles NH3 = 122500 grams / 17.031 g/mol

Moles NH3 = 7192.77 moles

Step 4: Calculate moles of CO2

Moles CO2 = 211400 / 44.01 g/mol

Moles CO2 = 4803.45 moles

Step 5: Calculate limiting reactant

For 2 moles NH3 consumed, we need 1 moles of CO2 to produce 1 mole urea and 1 mole H2O

NH3 is the limiting reactant. It will completely be consumed (7192.77 moles).

CO2 is in excess. There will be consumed 7192.77/2 = 3596.4 moles

There will remain 4803.45 - 3596.4 = 1207.05 moles of CO2

Step 6: Calculate moles of urea produced:

For 2 moles NH3 consumed, we need 1 moles of CO2 to produce 1 mole urea and 1 mole H2O

For 7192.77 moles of NH3, we have 3596.4 moles of urea produced

Step 7: Calculate mass of urea

Mass urea = moles urea * molar mass urea

Mass urea = 3596.4 moles * 60.06 g/mol

Mass urea = 216000 grams = 216 kg = theoretical yield

Step 8: Calculate % yield

% yield = (actual yield / theoretical yield)*100%

% yield = (170.3 / 216) *100% = 78.8%

The % for this reaction is 78.8 %

3 0

2 years ago