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3 years ago

* Do you think odor is a property? Explain your ideas.

Chemistry

1 answer:

damaskus [11]3 years ago

7 0

Answer:

This might be simply as matter of convenience. It is now well known that different molecules have one of several smells. The sensation of smell is produced when the odor-causing molecule binds to certain sites (called receptors) high up in the nose.

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A 100. mL volume of 0.800 M calcium chloride is mixed with 400. mL of water to make 500. mL of solution. What is the final molar

strojnjashka [21]

Answer:

0.99mol/L

Explanation:

Molarity ( M) = # of moles of solute / volume of solution (L)

Volume of Solution = 100 mL or 0.1 L

Molarity ( M) = # of moles of solute / volume of solution (L)

Molarity (M) = 0.099 mol / 0.1 L = 0.99mol/ L

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3 years ago

Indigestion can be cured by antacid

gayaneshka [121]

Answer: ummmm

Explanation:

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3 years ago

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the elements carbon hydrogen and oxygen are all part of the same blank on the periodic table........ i i think GROUP diangol row

Mrrafil [7]

The elements Carbon, Hydrogen, and Oxygen are all part of Non-Metals.

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3 years ago

The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures

QveST [7]

Answer:

For A: The $K_p$ for the given reaction is $4.0\times 10^1$

For B: The $K_c$ for the given reaction is 1642.

Explanation:

The given chemical reaction follows:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

For A:

The expression of $K_p$ for the above reaction follows:

$K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}$

We are given:

$p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm$

Putting values in above equation, we get:

$K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1$

Hence, the $K_p$ for the given reaction is $4.0\times 10^1$

For B:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = $4.0\times 10^1$

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642$

Hence, the $K_c$ for the given reaction is 1642.

7 0

3 years ago

This question has multiple parts. Work all the parts to get the most points. An aqueous antifreeze solution is 39.0% ethylene gl

nlexa [21]

Answer:

Molality = 10.300 m

Molarity = 6.5970 M

mole fraction = 0.156549

Explanation:

39.0 % = ethylene glycol

61.0 % = water

imagine the total mass = 100g

39.0% ethylene glycol = 39g

61.0 % water = 61g

1) Molality = number of moles / mass of solvent (kg)

Molar mass of ethylene glycol = 62.07g/mole

mole = 39g / 62.07g/mole = 0.6283 moles

Molality = moles / mass of solvent = 0.6283 moles / 0.061kg = 10.300 m

2) Molarity = number of moles / volume of solution

Since we know the density of the solution = 1.05g /ml

⇒ volume = 100g / 1.05g /mL = 95.24 mL = 0.09524 L

Molarity = 0.6283 moles / 0.09524 L = 6.5970 M

3) Mole fraction

moles water = 61g / 18.02g/mole = 3.38513 moles

Total number of moles = moles of ethylene glycol + moles of water = 0.6283 + 3.38513 = 4.0134276 moles

Mole fraction = 0.6283/ 4.0134276 = 0.156549

3 0

3 years ago