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BigorU [14]
3 years ago
5

PLEASE NEED HELP ASAP

Chemistry
1 answer:
Zigmanuir [339]3 years ago
8 0

I would love to help but I cant see the picture

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why mixture of stone pebbles And water is concerned considered suspension you are provided with a mixture of sand salt and camph
baherus [9]

Answer:

camphor sublimates salt is soluble in water while sand does not sublime and does not dissolve in water you first heat the mixture in a beaker covered with a watch glass camphor will then accumulate on the watch glass then you dissolve the remaining mixture of sand and salt salt will dissolve forming a salt solution then you filter using a filter paper and a beaker the residue on the filter paper is sand while the filtrate is salt solution you then heat the salt solution so that it can evaporate leaving salt particles thus you will have obtained salt sand and camphor

6 0
3 years ago
Balance each of these equations.
Oksanka [162]

Answer:

A. 2NO + O2 -> 2NO2

B. 4Co + 3O2 -> 2Co2O3

C. 2Al + 3Cl2 -> Al2Cl6

D. 2C2H6 + 7O2 -> 4CO2 + 6H2O

E. TiCl4 + 4Na -> Ti + 4NaCl

8 0
3 years ago
How does one calculate the mass of the sample?
antoniya [11.8K]

Answer:

HUH I NEED A PIC of the question

Explanation:

4 0
3 years ago
Identify the property of the matter described below.
ZanzabumX [31]

Answer: C.

Explanation: Alcohol floats on oil and water sinks in oil. Water, alcohol, and oil layer well because of their densities, but also because the oil layer does not dissolve in either liquid. The oil keeps the water and alcohol separated so that they do not dissolve in one another. ... Water sinks because it is more dense than oil.

7 0
3 years ago
A 225-g sample of aluminum was heated to 125.5 oc, then placed into 500.0 g water at 22.5 oc. (the specific heat of aluminum is
Sidana [21]
when  heat gained = heat lost 

when AL is lost heat and water gain heat

∴ (M*C*ΔT)AL = (M*C*ΔT) water

when M(Al) is the mass of Al= 225g 

C(Al) is the specific heat of Al = 0.9 

ΔT(Al) = (125.5 - Tf) 

and Mw is mass of water  = 500g

Cw is the specific heat of water = 4.81 

ΔT = (Tf - 22.5) 

so by substitution:

∴225* 0.9 * ( 125.5 - Tf) = 500 * 4.81 * (Tf-22.5)

∴Tf = 30.5 °C
7 0
3 years ago
Read 2 more answers
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