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slega [8]
3 years ago
12

Choose the statement that correctly describes the diagram. A. Side B has a higher concentration of particles than side A. B. The

re are two types of particles with varying amounts. C. Side A has a higher concentration of particles than side B. D. Side A and side B have equal numbers of particles. Mark for review (Will be highlighted on the review page)
Chemistry
2 answers:
Luda [366]3 years ago
8 0
A. Side B has higher concentration of particles than side A
umka21 [38]3 years ago
3 0

Answer:

I would also go with A

Explanation:

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A reaction is non-spontaneous at any temperature when
slavikrds [6]

Answer:

A reaction is non-spontaneous at any temperature when the Gibbs free energy > 0.

Explanation:

There is a state function, that determines if a reaction is sponaneous or non spontaneous:

ΔG = Gibbs free energy

A reaction is non spontaneous when it does require energy to produce that reaction. It will be spontaneous, when the reaction does not require energy to be occured.

The formula is: ΔG = ΔH - T.ΔS

ΔH → Enthalpy → Energy gained or realeased as heat.

ΔH < 0 → <em>Exothermic reaction. Spontaneity is favored </em>

T  → Temperature

ΔS → Entropy →  Degree of disorder of a system.

When the system has a considered disorder ΔS > 0, disorder increases.

When the system is more ordered, ΔS < 0, disorder decreases.

The reaction will be non spontaneous if, the enthalpy is positive (endothermic reaction) and the ΔS < 0 (disorder decreases). It will not occur if we do not give energy.

ΔG < 0 → Spontaneous reaction

ΔG > 0 → Non spontaneous reaction

ΔG = 0 → System in equilibrium

8 0
3 years ago
When trying to clean muddy, dirty river water, which method would work
Llana [10]
When trying to clean muddy, dirty river water, FILTRATION would work best
6 0
3 years ago
Convert the following temperatures to Kelvin:
maxonik [38]

Explanation:

A. 100°C to Kelvins

T(K)=T(^oC)+273.15

T(K)=100(^oC)+273.15=373.15 K

B 600°R to Kelvins

(T)^oK=((T)^oR)\times 1.8

(T)^oK=600\times 1.8 K = 1080 K

C. 98°F to Kelvins

(T(K)-273.15)=(T(^oF)-32)\times \frac{5}{9}

(T(K))=(98(^oF)-32)\times \frac{5}{9}+273.15=309.81K

D. 77.4°F to degree Celsius

((T)^oC)=((T)^oF-32)\times \frac{5}{9}

(T)^oC =(77.4^oF-32)\times \frac{5}{9}=25.22^oC

E. 77.4 K to degree Celsius

T(^oC)=T(^K)-273.15

T(^oC)=77.4(K)-273.15=-195.75^oC

F. 77.4°R to degree Celsius

(T)^oC=((T)^oR-491.67)\times \frac{5}{9}

(T)^oC=((77.4)^oR-491.67)\times \frac{5}{9}=-230.15 ^oC

7 0
3 years ago
When a species population no longer has traits adapted to its environment, the species will
MatroZZZ [7]

I think that A or C is the answer.

3 0
2 years ago
Read 2 more answers
Calculate the hydrogen ion concentration in a 3.4 x 10-3 M<br> solution of KOH.
notsponge [240]

Explanation:

KOH→K {}^{ + }  + OH {}^{ - }  \\ [OH {}^{ - } ] = 3.4 \times  {10}^{ - 3}  \\ k _{w} = [OH {}^{ - } ] [H {}^{  + } ]  \\ 1 \times  {10}^{ - 14}  = 3.4 \times  {10}^{ - 3}  \times [H {}^{  +  } ]  \\ [H {}^{  +  } ]  = 2.94 \times  {10}^{ - 12}  \: M

7 0
2 years ago
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