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3 years ago

Y’all what’s this I don’t feel like doing it and I’ll give brainlist too

Chemistry

1 answer:

maw [93]3 years ago

3 0

Answer:

2.Closed System: A closed system is a thermodynamic system where energy can be exchanged with its surrounding but not matter. Open System: Open systems can exchange matter with the surrounding. Closed System: Closed systems cannot exchange matter with the surrounding. Open System: The mass of the system will vary with time in open systems.

Explanation:

1.The particle model is the name for the diagrams used to draw solids, liquids and gases. In the model, the particles are shown as circles or spheres. However, the particles in ice, liquid water and steam look the same because they are all water, but in different states of matter. Why do we use the particle model in science?

3.An open system can be defined as a system which can exchange both matter and energy with the surrounding. For example, the earth can be recognized as an open system. In this case, the earth is the system, and space is the surrounding. Sunlight can reach the earth surface and we can send rockets to space.

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Can the ratio of hydrogen to oxygen change if a solute is dissolved in the water

coldgirl [10]

Answer:

Water molecules feature the combinations of hydrogen and oxygen atoms in a 2:1 ratio. Since they are present in a fixed ratio of mass, water molecules obey the law of constant proportions. Water is formed when two molecules of the diatomic hydrogen gas, combine with one molecule of the diatomic oxygen gas to produce two molecules of water

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3 years ago

Could someone help me with this too? Much appreciated! I am a bit stuck.

Mashcka [7]

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3 years ago

What happens when a catalyst is used in a chemical reaction? The reaction rate increases. The reaction rate decreases. The react

topjm [15]

Catalysts speed up a reaction so it would be the reaction rate increases.

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3 years ago

Read 2 more answers

There are two common oxides of copper; one is 80% copper and the other is 89% copper by weight. Calculate the formula and name e

irakobra [83]

80% copper (Cu)

Cu: 80 : 63.546 = 1.259

O: 20 : 16 = 1.25

Cu:O = 1 : 1

the formula: CuO

89% copper (Cu)

Cu: 89 : 63.546 = 1.4

O: 11 : 16 = 0.6875

Cu:O = 2:1

the formula: Cu₂O

8 0

2 years ago

Learn how changes in binding free energy affect binding and the ratio of unbound and bound molecules.

Delvig [45]

C)[D]/[ED] = 5.20

D)[D]/[ED] = 5.20

E)[D']_T = 1.495* 10 ^-7 M

F)[D'] / [ED'] = 0.0579

Explanation:

E = 250 nM =2.5* 10 ^-7 mol/L , T=298.15 K

Dissociation constant of K_D = 1.3 μM (1.3 *10 ^-6 M)

E + D ⇄ ED → K_a = [ED] / [D][E] (association constant)

ED ⇄ E + D → K_D = [E][D] / [ED] (dissociation constant)

[E] =2.5*10^-7 mol/L

K_D = 1.3* 10^-6 M

K_D = [E][D] / [ED] → [D]/[ED] = K_D / [E]

= [D]/[ED] = 1.3* 10 ^-6 / 2.5 *10^-7

= 13/25 * 10

=130/25 = 5.20

[D]/[ED] = 5.20

ΔG =RTln Kd

ΔG_2 for E and D = 1.987 * 298.15 * ln 1.3*10^-6

ΔG_2 592.454 * [ln 1.3 +ln 10^-6]

ΔG_1 = 592.424 [0.2623 - 13.8155]

ΔG_2 = -592.424 * 13.553

ΔG_1 = -8184.633 cal/ mol

ΔG_1 = -8184.633 * 4.18 J/mol = -34244.508 J?mol

ΔG_1 = -34.245 KL/mol

so, ΔG_2 = ΔG_1 - 10.5 KJ/mol

ΔG_2 = -34.245 - 10.5

ΔG_2 = -44.745KJ / mol

ΔG_2 =RT ln K_D

-44.745 *10^3

=8.314 *298.15 lnK_D

lnK_D' = - 44745 / 2478.81 g

ln K_D' = -18.051

K_D' = -18.051

K_D' = e^-18.051

[D]/[ED] = 5.20

[E] = 2.5* 10 ^-7 mol/ L = a

K_D' = [E][D] / [ED'] E +D' → ED'

K_D' = a/2(x-(a/2) / (a/2)

KD' = x - a/2

=2.447 *10^-8 = (2.5/2) * 10^-7

x=2.447 * 10^-8 + 1.25 * 10^-7

x = 2.447 *10^-8 + 1.25 * 10 ^-7

x= 10^-7 [1.25 + 0.2447]

x = 1.4947 * 10^-7

[D']_T = 1.495* 10 ^-7 M

K_D' = [E][D'] / [ED']

[D'] / [ED'] = KD' / [E]

[D'] / [ED'] = 1.447 *10^-8 / 2.5* 10^-7

[D'] / [ED'] = 0.5788 * 10^-1

[D'] / [ED'] = 0.0579

5 0

3 years ago