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4 years ago

What is a ideal solution?

2 answers:

8 0

In chemistry, an ideal solution or idealmixture is a solution with thermodynamic properties analogous to those of a mixture of ideal gases. ... The vapor pressure of the solutionobeys Raoult's law, and the activity coefficient of each component (which measures deviation from ideality) is equal to one.

xenn [34]4 years ago

5 0

Solution with thermodynamic properties analogous to those of a mixture with ideal gasses

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What is the capillary rise of ethanol in a glass tube with a 0.1 mm radius if the surface tension of ethanol is 0.032Jm2 and the

iragen [17]

Answer:

Explanation:

Expression for capillary rise is as follows

h = 2T / ρ g r where T is surface tension , ρ is density of liquid and r is radius of capillary tube.

T = .032 J m⁻²

ρ = .71 g / cm³

= 7100 kg / m³

r = .1 x 10⁻³ m = 10⁻⁴ m

h = 2 x .032 / (7100 x 9.8 x 10⁻⁴ )

h = .0092 m

= 9.2 mm .

5 0

3 years ago

The same reaction is begun with an initial concentration of 0.05 M O3 and 0.02 M NO. Under these conditions, the reaction reache

Ivahew [28]

The balanced equation of the reaction is:

O3(g) + NO (g) → O2 (g) + NO2 (g)

Then the ratios of reaction is 1 mol O3 : 1 mol NO : 1 mol O2 : 1 mol NO2

If you have initially 0.05 M of O3 and 0.02 M of NO, the reaction will end when all the NO is consumed.

The by the stoichiometry 0.02 mol of O3 will be consumed in 8 seconds.

And the rate of reaction is change in concetration divided by the time.

The change in concentration in O3 is 0.02 M

Then, the rate respect O3 is 0.02 M / 8 seconds = 0.0025 M/s

8 0

3 years ago

enzyme‑catalyzed, single‑substrate reaction E + S − ⇀ ↽ − ES ⟶ E + P . The model can be more readily understood when comparing t

laila [671]

Complete Question

The complete question is shown on the first uploaded image

Answer:

[S]<<KM | [S]=KM | [S]>>KM | Not true

____________ | Half of the active | Reaction rate is | Increasing

$[E_{free}]$ is about | sites are filled of | independent of | $[E_{Total}]$ will

equal to $[E_{total}]$ . | | [S] | lower KM

_____________________________________________|____________

[ES] is much | | Almost all active

lower than | | sites are filled

$[E_{free}]$ | |

Explanation:

Generally the combined enzyme $[ES]$ is mathematically represented as

$[ES] = \frac{[E_{total}][S]}{K_M + [S]}----(1)$

for Michaelis-Menten equation

Where $[S]$ is the substrate concentration and $K_M$ is the Michaelis constant

Considering the statement $[S] < < K_M$

Looking at the equation $[S]$ is denominator so it can be ignored(it is far too small compared to $K_M$ ) hence the above equation becomes

$[ES] = \frac{[E_{total}][S]}{K_M}$

Since $[S]$ is less than $K_M$ it means that $\frac{[S]}{K_M} < < 1$

so it means that $[ES] < < [E_{total}]$

What this means is that the number of combined enzymes $[ES]$ i.e the number of occupied site is very small compared to the the total sites $[E_{total}]$ i.e the total enzymes concentration which means that the free sites [ $E_{free}]$ i.e the concentration of free enzymes is almost equal to $[E_{total}]$