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patriot [66]
4 years ago
6

20 Systems in nature tend to undergo changes toward(1) lower energy and higher entropy(2) lower energy and lower entropy(3) high

er energy and higher entropy(4) higher energy and lower entropy
Chemistry
2 answers:
astraxan [27]4 years ago
7 0

Answer is: (1) lower energy and higher entropy.

Entropy is the measure of the molecular disorder and it is system’s thermal energy per unit temperature that is unavailable for doing useful work.

For a reaction to be spontaneous under standard conditions at all temperatures, the signs of ΔH° and ΔS° must be negative and positive, respectively.

Gibbs free energy (G) determines if reaction will proceed spontaneously, if ΔG is negative, reaction is spontaneous.

ΔG = ΔH - T·ΔS.

ΔG - changes in Gibbs free energy.

ΔH - changes in enthalpy.

ΔS - changes in entropy.

T is temperature in Kelvins.

morpeh [17]4 years ago
5 0
Systems in nature tend to undergo changes toward Lower energy and higher entropy.
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PLEASE HELP CHEM BABES I HAVE BEEN CRYING FOR A WHILE NOW
slavikrds [6]

Answer: See below

Explanation:

1. To calculate the mass, you know you can convert by using molar mass. Since mass is in grams, we can use molar mass to convert moles to grams. This calls for the Ideal Gas Law.

Ideal Gas Law: PV=nRT

We manipulate the equation so that we are solving for moles, then convert moles to grams.

n=PV/RT

P= 100 kPa

V= 0.831 L

R= 8.31 kPa*L/mol*K

T= 27°C+273= 300 K

Now that we have our values listed, we can plug in to find moles.

n=\frac{(100kPa)(0.831L)}{(8.31kPa*l/molK)(300K)}

n=0.033mol

We use the molar mass of NO₂ to find grams.

0.033mol*\frac{46.005g}{1mol }=1.52 g

The mass is 1.52 g.

2. To calculate the temperature, we need to use the Ideal Gas Law.

Ideal Gas Law: PV=nRT

We can manipulate the equation so that we are solving for temperature.

T=PV/nR

P= 700.0 kPa

V= 33.2 L

R= 8.31 kPa*L/mol*K

n= 70 mol

Now that we have our values, we can plug in and solve for temperature.

T=\frac{(700kPa)(33.2L)}{(70mol)(8.31 kPa*L/molK)}

T=40K

The temperature is 40 K.

3 0
3 years ago
What environmental factor could be altered in order to increase the initial reaction rate in experiment one if the initial conce
nadezda [96]
Idk lolololololololololol
7 0
3 years ago
To squeeze a gas into a smaller space
Ray Of Light [21]
Compressing the gas. It will increase the pressure.
6 0
3 years ago
BRAINLIESTT ASAP! PLEASE HELP :)
Nadya [2.5K]

Hello, and this is my own words Periodic trends are specific patterns in the properties of chemical elements that are revealed in the periodic table of elements. Major periodic trends include electronegativity, ionization energy, electron affinity, atomic radii, ionic radius, metallic character, and chemical reactivity.

Hope This Helps. and have good day!

3 0
4 years ago
a 50.00-mL sample of bleach solution contains 0.214 M HClO and 0.667 M NaClO. The Ka of hypochlorous acid is 3.0 ✕ 10−8. Find th
amm1812

Answer:

A) pH = 8.11

B) pH = 7.96

Explanation:

The pH of buffer is calculated using Henderson Hassalbalch's equatio, which is

pH=pKa+log\frac{[salt]}{[acid]}

pKa = -logKa = -log(3.0 ✕ 10⁻⁸)

pKa = 7.5

[salt] = 0.667

[Acid] = 0.214

pH = 7.5 + log\frac{0.667}{0.214}=7.99

A) If the solution is halved it means the concentration of both salt and acid will be the same.

The volume will change

Number of moles of acid = molarity X volume = 0.214 X 25 = 5.35 mmol

Number of moles of salt = molarity X volume = 0.667 X 25 = 16.68 mmol

moles of NaOH added = molarity X volume = 0.100 X 10 = 1 mmol

These moles of NaOH will react with acid to give same amount of salt and thus the moles of acid will decrease

New moles of acid = 5.35 - 1 = 4.35

New moles of salt = 16.68 + 1 = 17.68

the new pH will be

pH = 7.5 + log\frac{17.68}{4.35}=8.11

B)

moles of HCl added = molarity X volume = 0.100 X 10 = 1 mmol

These moles of HCl will react with salt to give same amount of acid and thus the moles of salt will decrease

New moles of acid = 5.35  + 1 = 6.35

New moles of salt = 16.68 - 1 = 15.68

the new pH will be

pH = 7.5 + log\frac{15.68}{5.35}=7.97

3 0
3 years ago
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