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slega [8]
3 years ago
7

4 methods that can be used to separate mixtures are

Chemistry
2 answers:
TiliK225 [7]3 years ago
7 0
Coffee Filter
By Hand
Boiling
Magnetism
Evaporation
a Funnel

or what you might be asking


Distillation
Crystallization
<span>Chromatography
</span><span>Filtration</span>
RUDIKE [14]3 years ago
6 0
Filtration
distillation
crystallization
chromatography
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How would you seprate a mixture of sand and salt?
lys-0071 [83]
First, you mix the salt and sand with water, so the salt dissolves. Next, you filter the sand out, so you have the slat water and sand separated. Then, you evaporate the water, leaving the salt behind.
3 0
3 years ago
Read 2 more answers
A solution of 0.0027 M K2CrO4 was diluted from 3.00 mL to 100. mL. What is the molarity of the new solution?
Stells [14]

Answer:

[K₂CrO₄] → 8.1×10⁻⁵ M

Explanation:

First of all, you may know that if you dilute, molarity must decrease.

In the first solution we need to calculate the mmoles:

M = mmol/mL

mL . M = mmol

0.0027 mmol/mL . 3mL = 0.0081 mmoles

These mmoles  of potassium chromate are in 3 mL but, it stays in 100 mL too.

New molarity is:

0.0081 mmoles / 100mL = 8.1×10⁻⁵ M

4 0
3 years ago
List the ions in solution in the electrolysis of conc.sodium chloride<br>​
Licemer1 [7]
Chloride ions Cl –(aq) (from the dissolved sodium chloride) are discharged at the positive electrode as chlorine gas, Cl 2(g) sodium ions Na +(aq) (from the dissolved sodium chloride) and hydroxide ions OH –(aq) (from the water) stay behind - they form sodium hydroxide solution, NaOH(aq)
5 0
3 years ago
Given the equation representing a reaction at equilibrium: N₂ + 3H₂ ⇄ 2NH₃ What occurs when the concentration of H₂ is increased
ValentinkaMS [17]
That would cause the equation to shift right, and make more NH3 and decrease the amount of N2
4 0
3 years ago
Heat is transferred directly from a heat reservoir at 200 C to another heat reservoir at 5C. If the amount of heat transferred i
Daniel [21]

Answer:

ΔS=0.148  KJ/K

Explanation:

Given that

Q = 100 KJ

T₁=200°C

T₁=200+273 = 437 K

T₂=5°C

T₂=5 + 273 = 278 K

Reservoir 1 is rejecting heat that is why it taken as negative while the reservoir 2 is gaining the heat that is why it is taken as positive.

So the total change in entropy given as

ΔS=  - Q/T₁ + Q/T₂

ΔS=  - 100/473 + 100/278  KJ/K

ΔS=0.148  KJ/K

4 0
3 years ago
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