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Arte-miy333 [17]
4 years ago
13

What is an example of a colloid? A. Soda B.Fog C.Jewelry D.Silverware

Chemistry
2 answers:
TEA [102]4 years ago
5 0

Answer:

B. fog

Explanation:

This is because a liquid (water) is dispersed throughout the air (gas)

Olegator [25]4 years ago
4 0

Answer:

the current answer is

silverware

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Geothermal energy helps to
disa [49]

Answer:A

Explanation:

4 0
4 years ago
The daily value of phosphorous is 800 mg, how many grams of phosphorous are recommended​
Lady_Fox [76]

.8 grams of phosphorus are recommended.

7 0
4 years ago
A sample consisting of 2 moles He is expanded isothermally at 0 degrees from 5.0dm3 to 20.0dm3. Calculate w, q and deltaU for ea
FinnZ [79.3K]

Answer:

i) \Delta U=0 w=-6293 J q=6293 J

ii) \Delta U=0 w=-3404,52 J q=3404,52 J

ii) \Delta U=0 w=0 J q=0 J

Explanation:

As the initial and final states of the sample are the same, the ΔU of the sample is, for the three cases

\Delta U=n.C_{V}.\Delta T=0 since \Delta T=0

i)Reversibly P_{ext} =P_{sys} so w can be calculated by  

w=-n.R.T.ln(\frac{V_{f}}{V_{i}})=-2 \times 8.314\frac{J}{mol K} \times 273,15K \times ln(\frac{}{5dm^{3}})=-6293 J

and because of the first law of thermodynamics

q=-w=6293 J

ii)Irreversibly with P_{ext} =P_{f}

we can calculate P_{f} by the law of ideal gases

P_{f} =\frac{n\times R\times T}{V_{f}} =\frac{2\times 0.082\frac{dm^{3}atm}{mol K}\times 273,15K}{20dm^{3}} =2,24 atm

then w can be calculated by

w=-P_{ext} \times \Delta V=-2,24 atm \times (20-5) dm^{3} \times frac{101.325J }{atm dm^{3}=-3404,52J

and  

q=-w=3404,52J

iii)a free expansion P_{ext} so w=0 (there's no work at vaccum) and q=-w=0

7 0
3 years ago
What do you notice about seasons in countries positioned along the Equator? Why do you think this is the case?​
never [62]

Answer:

Explanation:

Places near the Equator experience little seasonal variation. They have about the same amount of daylight and darkness throughout the year. These places remain warm year-round. Near the Equator, regions typically have alternating rainy and dry seasons.

3 0
3 years ago
Read 2 more answers
The carbon-oxygen bond in CO has a higher bond dissociation enthalpy than a carbon-oxygen bond in CO2. Which is the best explana
kkurt [141]

Answer:

CO HAS A TRIPLE BOND WHILE C-O BOND IN CO2 IS A DOUBLE BOND

CO HAS A LONE PAIR ON CARBON WHILE CO2 DOES NOT

Explanation:

Bond dissociation bond enthalpy or energy is the energy needed to break 1 mole of a divalent molecule into separate atoms mostly in the gaseous state.

The carbon and oxygen in carbon monoxide form a triple bond as carbon monoxide has 10 electrons in their outermost shell which results into six shared electrons in 3 bonding orbitals as against the double bond formed by other carbon compounds. Four electrons come from oxygen and the remaining two from carbon and due to this, two electrons from oxygen will occupy one orbital and this forms a dative bond. Also because of the triple bond, carbon monoxide is often regarded as a more stable compound than carbon dioxide with a double bond. This gives it its higher bond dissociation enthalpy value and more energy is needed to break it into its separate atoms. This is in conjunction with a larger bond length similar to the bong length in a triple bond. This makes it more stronger than the bond dissociation enthalpy of carbon dioxide having a double bond.

7 0
4 years ago
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