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3 years ago

Which units express heat capacity

Chemistry

2 answers:

vladimir1956 [14]3 years ago

7 0

Answer:

Heat capacity units: J/K

Explanation:

Heat capacity (C) is the amount of heat (Q) required to raise the temperature(T) of a substance by a unit degree.

$C = \frac{Q}{T}$

It is usually expressed in terms of J/K

Heat capacity is an intensive property i.e. it is not dependent on the amount of the substance. i.e, the amount of heat required to raise the temperature of 1 g of copper per kelvin would be the same as that corresponding to 100 g.

sveta [45]3 years ago

6 0

The heat capacity of a defined system is the amount of heat (usually expressed in calories, kilocalories, or joules) needed to raise the system's temperature by one degree (usually expressed in Celsius or Kelvin). It is expressed in units of thermal energy per degree temperature. To aid in the analysis of systems having certain specific dimensions, molar heat capacity and specific heat capacity can be used. To measure the heat capacity of a reaction, a calorimeter must be used. Bomb calorimeters are used for constant volume heat capacities, although a coffee-cup calorimeter is sufficient for a constant pressure heat capacity.

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The compound trimethylamine, (CH3)3N, is a weak base when dissolved in water. Write the Kb expression for the weak base equilibr

GREYUIT [131]

Answer:

The $K_b$ expression for the weak base equilibrium is:

$K_b=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}$

Explanation:

$(CH_3)_3N(aq)+H_2O(l)\rightlefharpoons (CH_3)_3NH^++OH^-(aq)$

The expression of the equilibrium constant of base $K_c$ can be given as:

$K_c=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N][H_2O]}$

] $K_b=K_c\times [H_2O]=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}$

As we know, water is pure solvent, we can put $[H_2O]=1$

$K_b=K_c\times 1=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}$

So, the the $K_b$ expression for the weak base equilibrium is:

$K_b=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}$

6 0

3 years ago

Let the ED50 of a recreational drug be defined as the amount required for 50% of a test group to feel high or get a buzz. If the

Anastaziya [24]

Explanation:

1) The ED50 value of ethanol = 470 mg per kg of person body = 470 mg/kg

Mass of person = 70 kg

Amount of ethanol required by person of 70 kg to get buzz:

$70 kg\times 470 mg/kg=32,900 mg$

A 70 kg party goer need to quickly consume in order to have a 50% chance of getting a buzz is 32,900 mg of ethanol.

Mass of ethanol in beer bottle = 13.9 g

Amount of ethanol required by person of 70 kg to get buzz:

$70 kg\times 470 mg/kg=32,900 mg =32.9 g$

1 mg = 0.001 mg

Bottles of beer would a 70 kg party goer need to consume:

$=\frac{32.9 g}{13.9 g}=2.367\approx 2$

Bottles of beer would a 70 kg party goer need to consume a 50% chance of getting a buzz is 2.

2)The LD50 value of ethanol = 4,700 mg per kg of person body = 4,700 mg/kg

Mass of person = 70 kg

Amount of ethanol required by person of 70 kg to have a 50% chance of dying:

$70 kg\times 4,700 mg/kg=329,000 mg$

A 70 kg party goer need to quickly consume in order to have a 50% chance of dying is 329,000 mg of ethanol.

Mass of ethanol in Jack Daniels bottle = 356 g

Amount of ethanol required by person of 70 kg to have a 50% chance of dying:

$70 kg\times 4,700 mg/kg=329.000 mg =329 g$

1 mg = 0.001 mg

Bottles of Jack Daniels would a 70 kg party goer need to consume to have a 50% chance of dying:

$=\frac{329 g}{356 g}=0.923\approx 1$

Bottles of Jack Daniels would a 70 kg party goer need to consume to have a 50% chance of dying is 1.

4 0

3 years ago

The primary forces of attraction between water molecules in h2o(l) are

finlep [7]

Hydrogen bonds 100% sure Have a great evening

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3 years ago

The value for Kw is 1.0 x 10-14 at 298K, but this value is temperature dependent. Given what you know about Kw and acid/base che

Papessa [141]

Answer:

The correct answer is option b.

Explanation:

$H_2O\rightleftharpoons H^++OH^-$

The ionic product of water : $K_w$

$K_w=[H^+][OH^-]$

A pure water has equal concentration of hydrogen ions and hydroxide ions, hence neutral.

$K_w=[H^+][H^+]=[H^+]^2$

The ionic product of water at 283 K = $K_w=0.29\times 10^{-14}$

$K_w=[H^+]^2$

$0.29\times 10^{-14}=[H^+]^2$

$[H^+]=5.385\times 10^{-8} M$

The pH of the water at 283 k;

$pH=-\log [H^+]$

$=-\log[5.385\times 10^{-8} M]=7.26$

A pure water has equal concentration of hydrogen ions and hydroxide ions,So water will e neutral at this temperature also.

The correct answer is option b.

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3 years ago

One sentence describing the difference between pure substances and mixtures.

Alja [10]

Answer:

A pure substance is made up of only one ingredient or combination. A mixture is made up of two or maybe more separate components that are not chemically linked.

Explanation:

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3 years ago

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