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sleet_krkn [62]
3 years ago
11

2. Write the symbol proposed by Dalton for silver, sulphur. oxygen and gold.​

Chemistry
1 answer:
slega [8]3 years ago
6 0

Answer:

check the pdf hope it helps :)

Explanation:

Download pdf
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The small bags of silica gel you often see in a new shoe box are placed there to control humidity. Despite its name, silica gel
Zinaida [17]

So we look equation for the free Gibbs free energy (ΔG) which depends on entalpy (ΔH), temperature (T) and entropy (ΔS):

ΔG = ΔH - TΔS

ΔG is negative (-) because the water absorption on the silica gel surface is a spontaneous process.

ΔH is negative (-) because the water absorption on the silica gel surface is a exothermic process (it releases heat and if you want to desorb the water form the silica gen you need to add heat which is a endothermic process).

ΔS is negative (-) because the water is adsorbed, so from disorderly state you take the water molecules and put them in a orderly state and by doing that you decrease the entropy.

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3 years ago
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What advantage does a heterogeneous catalyst provide over a homogeneous catalyst in industrial processes?
irga5000 [103]
A heterogeneous catalyst can be easily separated from reactants.
8 0
3 years ago
6)
MrMuchimi

Answer:

(4) Water, stirring, and filtering  

Explanation:

The added water will dissolve the sugar but not the sand. When you filter the mixture, the sand will be trapped in the filter paper and the dissolved sugar will pass through the pores of the paper.

(1) and (2) are wrong. You don't separate a mixture of sugar and sand by adding more sand. Furthermore, neither substance will boil at 100 °C.

(3) is wrong. You can dissolve the sugar in water but, if you boil the water away, the sugar and sand will still be together.

7 0
4 years ago
When 131.0 mL of water at 26.0°C is mixed with 81.0 mL of water at 85.0°C, what is the final temperature? (Assume that no heat i
JulsSmile [24]

Answer:

63.52°C is the final temperature

Explanation:

1) 131.0 mL of water at 26.0°C

Mass of water = m

Volume of the water =131.0 mL

Density of the water = 1.00 g/mL

Density=1.00 g/mL=\frac{m}{131.0 mL}

m = 131.0 g

Initial temperature of the water = T_i = 26.0°C

Final temperature of the water = T_f

Change in temperature ,\Delta T=T_f-T_i

Heat absorbed 131.0 g of water = Q

Q=m\times c\times \Delta T

2) 81.0 mL of water at 85.0°C

Mass of water = m'

Volume of the water =81.0 mL

Density of the water = 1.00 g/mL

Density=1.00 g/mL=\frac{m'}{81.0 mL}

m' = 81.0 g

Initial temperature of the water = T_i' = 85.0°C

Final temperature of the water = T_f'

Change in temperature ,\Delta T'=T_f'-T_i'

Heat lost by 81.0 g of water = Q'

Q'=m'\times c\times \Delta T'

After mixing both liquids the final temperature will become equal fro both liquids.

T_f=T_f'

Since, heat lost by the water at higher temperature will be equal to heat absorbed by the water at lower temperature.

Q=-Q' (Law of conservation of energy.)

Let the specific heat of water be c

m\times c\times \Delta T=m'\times c\times \Delta T'

131.0 g\times c(T_f-26^oC)=-(81.0.0 g\times c(T_f-85^oC))

T_f=63.52^oC

63.52°C is the final temperature

4 0
4 years ago
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