The equation for density, d, is

4 mol of water is made up of 2.408×10^24 molecules of water.

zhenek [66]

It would be 20 give me bravest

6 0

3 years ago

Identify some other substances (besides KCl) that might give a positive test for chloride upon addition of AgNO3. do you think i

nlexa [21]

Answer:

-The other substances that give a positive test with AgNO3 are other chlorides present, iodides and bromide.

-It is reasonable to exclude iodides and bromides but it is not reasonable to exclude other chlorides

Explanation:

In the qualitative determination of halogen ions, silver nitrate solution(AgNO3) is usually used. Now, various halide ions will give various colours of precipitate when mixed with with silver nitrate. For example, chlorides(Cl-) normally yield a white precipitate, bromides(Br-) normally yield a cream precipitate while iodides (I-) normally yield a yellow precipitate. Thus, all these ions or some of them may be present in the system.

With that being said, if other chlorides are present, they will also yield a white precipitate just like KCl leading to a false positive test for KCl. However, since other halogen ions yield precipitates of different colours, they don't lead to a false test for KCl. Thus, we can exclude other halides from the tendency to give us a false positive test for KCl but not other chlorides.

5 0

3 years ago

Calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting tota

Ghella [55]

Answer:

(a) Δ $S_{sys}$ = 2.881 J/K; Δ $S_{sur}$ = -2.881 J/K; total change in entropy = 0

(b)Δ $S_{sys}$ = 2.881 J/K; Δ $S_{sur}$ = 0 ; total change in entropy = 2.881 J/K

(c) Δ $S_{sys}$ = 0 ; Δ $S_{sur}$ = 0 ; total change in entropy = 0

Explanation:

In the given problem, we need to calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting total change in entropy, when a sample of nitrogen gas of mass 14 g at 298 K and 1.00 bar doubles its volume. We have the following variable:

mass (m) = 14 g

Temperature = 298 K

Pressure = 1.00 bar

Initial volume = $V_{1}$

Final volume = $V_{2}$ = $2V_{1}$

(a) Change in entropy of the system Δ $S_{sys}$ = $nRIn\frac{V_{2} }{V_{1} }$

where R = 8.314 J/(mol*K)

n = number of moles = mass/molar mass = 14/ 28 = 0.5 moles

Δ $S_{sys}$ = 0.5*8.314*ln2 = 2.881 J/K

Change in entropy of the surrounding Δ $S_{sur}$ = -2.881 J/K

Therefore, for a reversible process, the total change in entropy = Δ $S_{sys}$ +Δ $S_{sur}$ = 2.881 - 2.881 = 0

(b) Because entropy is a state function, we use the same procedure as in part (a). Thus, Δ $S_{sys}$ = 2.881 J/K

Since surrounding does not change in this process Δ $S_{sur}$ = 0.

total change in entropy = Δ $S_{sys}$ +Δ $S_{sur}$ = 2.881 - 0 = 2.88 J/K

(c) For an adiabatic reversible expansion, q(rev) = 0, thus:

Δ $S_{sys}$ = 0

Since heat energy is not transferred from the system to the surrounding

Δ $S_{sur}$ = 0

total change in entropy = Δ $S_{sys}$ +Δ $S_{sur}$ = 0

6 0

3 years ago

Donna and Ellis have been married for over 30 years which of the following is likely to be most important in their relationship

OleMash [197]

30 because it is the highest number of marriage

8 0

3 years ago

A 111.6 gram sample of iron (MW=55.8) was heated from 0 degrees C to 20 degrees C. It absorbed 1004 Joules of energy. What is th

qwelly [4]

Use the equation q=ncΔT.
q= heat absorbed our released (in this case 1004J)
n= number of moles of sample ( in this case 2.08 mol)
c=molar heat capacity
ΔT=change in temperature (in this case 20°C)
You have to rewrite the equation for c.
c=q/nΔT
c=1004J/(2.08mol x 20°C)
c=24.1 J/mol°C

I hope this helps

7 0

3 years ago