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kykrilka [37]
3 years ago
6

What are the 7 methods of separating mixtures?​

Chemistry
2 answers:
omeli [17]3 years ago
5 0

Answer:

wanna friend

Explanation:

pls

Gennadij [26K]3 years ago
3 0

Methods Of Separating Mixtures

Handpicking.

Threshing.

Winnowing.

Sieving.

Evaporation.

Distillation.

Filtration or Sedimentation.

Separating Funnel.

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In a certain industrial process involving a heterogeneous catalyst, the volume of the catalyst (in the shape of a sphere) is 10.
olga nikolaevna [1]

Answer:

a. The second run will be faster.

d. The second run has twice the surface area.

Explanation:

The rate of a reaction is proportional to the surface area of a catalyst. Given the volume (V) of a sphere, we can find its surface area (A) using the following expression.

A=\pi ^{1/3} (6V)^{2/3}

The area of the 10.0 cm³-sphere is:

A=\pi ^{1/3} (6.10.0)^{2/3}=22.4cm^{2}

The area of each 1.25 cm³-sphere is:

A=\pi ^{1/3} (6. 1.25)^{2/3}=5.61cm^{2}

The total area of the 8 1.25cm³-spheres is 8 × 5.61 cm² = 44.9 cm²

The ratio of  8 1.25cm³-sphere to 10.0 cm³-sphere is 44.9 cm²/22.4 cm² = 2.00

Since the surface area is doubled, the second run will be faster.

6 0
3 years ago
Nvm i have to answer lol
DIA [1.3K]

Answer:

i will do a joke... hmmm...

Explanation:

why did the nurse need a red pen?

To draw out blood. lol

5 0
2 years ago
Read 2 more answers
A substance added in The final stages to remove sulphur from coal is
Deffense [45]

Answer:

calcium oxide is injected into the final stage of the scubber, wich then reacts with the sulfur dioxide to form calcium sulfite.

Explanation:

8 0
3 years ago
Find the "density" of a copper sample whose mass is 134.4 grams and whose volume is 15.0mL
Sonja [21]

Answer:

8.96 g/mL

Explanation:

density = mass / volume

density = 134.3g / 15.0 mL

density = 8.96 g/mL

3 0
2 years ago
Given the standard heats of reaction
ANTONII [103]

Answer:

Explanation:

M(s) → M (g ) + 20.1 kJ --- ( 1 )

X₂ ( g ) → 2X (g ) + 327.3 kJ ---- ( 2 )

M( s) + 2 X₂(g) → M X₄ (g ) - 98.7 kJ ----- ( 3 )

( 3 ) - 2 x ( 2 ) - ( 1 )

M( s) + 2 X₂(g) - 2 X₂ ( g ) - M(s)  → M X₄ (g ) - 98.7 kJ -  2 [ 2X (g ) + 327.3 kJ ] - M (g ) - 20.1 kJ

0 = M X₄ (g ) - 4 X (g ) - M (g ) - 773.4 kJ

4 X (g ) +  M (g ) =  M X₄ (g ) - 773.4kJ

heat of formation of M X₄ (g ) is - 773.4 kJ

Bond energy of one M - X bond =  773.4 / 4 =  193.4 kJ / mole

6 0
3 years ago
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