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2 years ago

In a neutralization reaction, bases donate.

Chemistry

1 answer:

Sedaia [141]2 years ago

4 0

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Answer: I think it is hydroxide ion but im not sure

Explanation:

I hope this helped!

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- Zack Slocum

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Most particles would travel____from their source to a screen that lit up when struck.

quester [9]

Answer: in a straight path!

Explanation: hope this helps

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2 years ago

How many grams of hydrogen gas are needed to react completely with 54.0 g of oxygen gas, given the following balanced chemical r

Shkiper50 [21]

Answer:

6.75g of hydrogen will completely react with 54g of oxygen

Explanation:

6 0

2 years ago

Calculate the concentration in mol/L, M, of an aqueous sugar solution with a concentration of 23.5% (w/w) and density of 1.005 g

Temka [501]

Answer:

The concentration in mol/L is 0.683M

Explanation:

23.5% (w/w)

This data means that in 100 g of solution, we have 23.5 grams of solute.

From this point, we can calculate the moles of sugar.

Moles = Mass / Molar mass

Moles = 23.5 g /342.30 g/m

Moles = 0.068 moles

Density data make us know, the volume of our solution.

solution δ = solution mass / solution volume

δ = 1.005 g/mL = 100 g / solution volume

solution volume = 100g / 1.005 g/ml

solution volume = 99.5 mL

In conclusion, 0.068 moles are in 99.5 mL

Molarity (M) is mol/L

Let's convert 99.5 mL in L

99.5 mL / 1000 = 0.0995 L

0.068 m / 0.0995L = 0.683

If we convert moles in mmoles, we can also get Molarity (M)

mmoles / mL = M

0.068 moles . 1000 = 68 mmoles

68 mmmoles / 99.5mL = 0.683

8 0

3 years ago

A sample of Francium-212 will decay to one sixteenth of its original amount after 80 minutes. What is the half-life of Francium-

lbvjy [14]

20 minutes.

The sample would lose one half the quantity of francium in each half-life.

$1/16 = 1/2^{4} = 1/2 \times1/2 \times1/2 \times1/2$

Thus a mass decrease by a factor of 16 would correspond to a period of four half-lives. It took 80 minutes for the sample to lose all these francium, therefore

$t_{1/2} = 80/4 = 20$ minutes.

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3 years ago

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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