experiment 2: suppose you added 0.5 g of fp sample 1 instead of 2.0 g, what would happen to the freezing point temperature of the solution?

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Answer 1

If you added 0.5 g of sample 1 instead of 2.0 g, the freezing point temperature of the solution would decrease.

When a solute is added to a solvent, it disrupts the formation of the solvent's crystal lattice structure, lowering the freezing point of the solution. The extent to which the freezing point is lowered depends on the concentration of the solute particles in the solution. In this case, by reducing the amount of sample 1 from 2.0 g to 0.5 g, the concentration of solute particles in the solution would decrease.

Since the freezing point depression is directly proportional to the concentration of solute particles, a decrease in the amount of sample 1 would result in a smaller decrease in the freezing point temperature compared to if 2.0 g were added. In other words, the solution would experience a less significant decrease in freezing point temperature with only 0.5 g of sample 1.

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predict whether the hcl, clo- is conjugate acid/base pair or not. group of answer choices yes no

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Yes, HCl is a strong acid and thus it does not have a conjugate base.

But, when HCl gets dissolved in water, it gives H+ and Cl- ions as its products. Here, Cl- acts as the conjugate base of HCl. Thus, HCl and Cl- form a conjugate acid-base pair. Therefore, the answer is: yes, HCl and Cl- form a conjugate acid-base pair.HCl is a strong acid and thus it does not have a conjugate base. But, when HCl gets dissolved in water, it gives H+ and Cl- ions as its products. Here, Cl- acts as the conjugate base of HCl. Thus, HCl and Cl- form a conjugate acid-base pair. Therefore, the answer is: yes, HCl and Cl- form a conjugate acid-base pair.

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ammonia is produced using the haber process. calculate the mass of ammonia produced when 35.0g of nitrogen reacts with 12.5 g of hydrogen

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The balanced chemical equation of the Haber process is:

N2 + 3H2 → 2NH3

To calculate the mass of ammonia produced when 35.0g of nitrogen reacts with 12.5 g of hydrogen using the Haber process, we need to find the limiting reactant first.

Limiting reactant is the reactant which gets completely consumed in a chemical reaction, limiting the amount of product produced. Therefore, we must calculate the moles of each reactant using their molar masses and compare them to find the limiting reactant.

For nitrogen, the molar mass = 28 g/mol

Number of moles of nitrogen = 35.0 g / 28 g/mol = 1.25 mol

For hydrogen, the molar mass = 2 g/mol

Number of moles of hydrogen = 12.5 g / 2 g/mol = 6.25 mol

From the above calculations, it can be observed that hydrogen is in excess as it produces more moles of NH3. Thus, nitrogen is the limiting reactant.

Using the balanced chemical equation, the number of moles of NH3 produced can be calculated.

Number of moles of NH3 = (1.25 mol N2) × (2 mol NH3/1 mol N2) = 2.50 mol NH3Now,

to find the mass of NH3 produced, we can use its molar mass which is 17 g/mol.Mass of NH3 produced = (2.50 mol NH3) × (17 g/mol) = 42.5 g

Therefore, the mass of ammonia produced when 35.0g of nitrogen reacts with 12.5 g of hydrogen using the Haber process is 42.5 g.

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the basal rate of consumption of o2 by a 70-kg person is 16 mol o2 per day. this will oxidize food and then be reduced to water, providing energy for the person according to: o2 4h 4e- 2h2o a) the current (in amperes, c/s) corresponding to this rate of

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The current corresponding to the basal rate of oxygen consumption of a 70-kg person, which is 16 mol O2 per day, is approximately 0.19 Amperes.

To calculate the current, we need to convert the number of moles of oxygen consumed to the number of electrons involved in the reduction of oxygen.

From the balanced equation: O2 + 4H+ + 4e- → 2H2O, we can see that for every 4 moles of oxygen consumed, 4 moles of electrons are involved.

Therefore, the number of moles of electrons involved in the reduction of oxygen is also 16 mol.

To calculate the charge in coulombs (C), we use Faraday's constant (F) which is equal to 96485 C/mol.

Charge (C) = moles of electrons × Faraday's constant

Charge = 16 mol × 96485 C/mol

Charge ≈ 1543760 C

Finally, to calculate the current (I) in Amperes (A), we divide the charge by the time in seconds. Assuming a day consists of 24 hours (86400 seconds), we have:

Current (A) = Charge (C) / Time (s)

Current ≈ 1543760 C / 86400 s

Current ≈ 17.86 A

Therefore, the current corresponding to the basal rate of oxygen consumption of a 70-kg person is approximately 0.19 Amperes.

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