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3 years ago

5. Base your answer to the question on the particle diagrams

Chemistry

1 answer:

Luden [163]3 years ago

7 0

Answer:sample 3

Explanation:because if only have one kind of substance which makes it pure

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Which compound can act as both a Brønsted-Lowry acid and a Brønsted-Lowry base?

allsm [11]

Water.
In certain circumstances, water acts as a proton donor.
e.g NH3(aq) + H2O(l) --> NH4+(aq) + OH-(aq)

In other circumstances, water can act as a proton acceptor.
e.g HNO3(aq) + H2O(l) --> H3O+(aq) + NO3-(aq)

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3 years ago

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An aqueous solution of 10.03 g of catalase, an enzyme found in the liver, has a volume of 1.05 L at 27°C. The solution's osmotic

-Dominant- [34]

Answer : The molar mass of catalase is, $2.40\times 10^5g/mol$

Explanation :

Formula used :

$\pi =CRT\\\\\pi=\frac{w}{M\times V}RT$

where,

$\pi$ = osmotic pressure = 0.745 torr = 0.000980 atm (1 atm = 760 torr)

C = concentration

R = solution constant = 0.0821 L.atm/mol.K

T = temperature = $27^oC=273+27=300K$

w = mass of catalase = 10.03 g

M = molar mass of catalase = ?

V = volume of solution = 1.05 L

Now put all the given values in the above formula, we get:

$0.000980atm=\frac{10.03g}{M\times 1.05L}\times (0.0821L.atm/mole.K)\times (300K)$

$M=2.40\times 10^5g/mol$

Therefore, the molar mass of catalase is, $2.40\times 10^5g/mol$

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3 years ago

Calculate the freezing point of a solution 1.25 g benzene (C6H6) in 125 g of chloroform (CHCl3).

posledela

Answer:

The freezing point for the solution is -64.09°C

Explanation:

This problem can be solved, by the freezing point depression. This colligative problem shows, that the freezing point of a solution is lower than the freezing point of pure solvent.

ΔT = Kf. m

ΔT = T° freezing pure solvent - T° freezing solution

Kf = Cryscopic constant, for chloroform is 4.68

T°freezing pure solvent = -63.5°C

m is mol/kg of solvent → molality

Let's determine the moles of benzene

1.25 g / 78 g/mol = 0.0160 mol

Let's convert the mass of solvent to kg

125 g . 1kg / 1000 g = 0.125 kg

m = 0.0160 mol / 0.125 kg → 0.128 m

Let's go to the formula to replace the data

-63.5°C - T° freezing solution = 4.68 °C/m . 0.128 m

T° freezing solution = - (4.68 °C/m . 0.128 m + 63.5°C)

T° freezing solution = - 64.09°C

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3 years ago

Why is a high temperature needed to melt potassium iodide?

goldfiish [28.3K]

Potassium Iodide have a lot of strong bonds which requires a very high temperature (high energy) to break them and change into a different state of matter.

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A biologist determines that a lizard's body temperature is 29 degrees C. Which of the following correctly describes the biologis

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Answer:

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