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Sergio [31]
3 years ago
6

What does this demonstration show?

Chemistry
2 answers:
Rina8888 [55]3 years ago
7 0
There is nothing there....
alexgriva [62]3 years ago
3 0
Huh there’s nothing there ?
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State two substances that can be used as a filter in water treatment.​
Alchen [17]

Answer:

1. pure chlorine

2.chloramine

Explanation:

7 0
2 years ago
When 711 J of heat is added to a 41.1 g of an unknown metal, the temperature of the metal increases from 22oC to 94oC. Calculate
alekssr [168]
The   specific  heat  capacity  of   the  metal  is   calculated  as follows

  heat energy =  Mc delta  T
heat  energy (Q) = (711 j  )
m=mass    (41.1 g)
c=  specific  heat  capacity  =?
delta  t =  change  in  temperature(  94 -22=72)

c=  Q/m  delta  T
  c=711 j  /  (41.1 g  x72)  =  0.240  j/g /k
5 0
4 years ago
Liquid A and liquid B form a solution that behaves ideally according to Raoult's law. The vapor pressures of the pure substances
Rama09 [41]

Answer:

Vapor pressure of solution → 151.1 Torr

Option 2.

Explanation:

Raoult's Law is relationed to colligative property about vapor pressure. A determined solute, can make, the vapor pressure of solution decreases.

ΔP = P° . Xm

where Xm is the mole fraction of solute, P° (vapor pressure of pure solvent)

and ΔP = Vapor pressure of pure solvent - Vapor pressure of solution.

In order to determine the vapor pressure of solution, we need to determine, the vapor pressure of B and A in the solution

B's pressure = P° B . Xm

When we add A to B, A works as the solute and B, as the solvent.

Vapor pressure of pure B is 135 torr. (P° B)

In order to determine, the Xm, we use the moles of A and B

Xm = 5.3 mol of B / (1.28 + 5.3) → 0.806

B's pressure = 135 Torr . 0.806 → 108.81 Torr

If mole fraction of B is 0.806, mole fraction for A (solute) will be (1 - 0.806)

A's pressure = 218 Torr . 0.194 → 42.3 Torr

Vapor pressure of solution is sum of vapor pressures of solute + solvent.

Vapor pressure of solution = 42.3 Torr + 108.81 Torr → 151.1 Torr

6 0
3 years ago
What rules should you follow when multiplying and dividing? Check all that apply.
xeze [42]

Answer:

exponents

Explanation:

The exponents in scientific notation should be added when multiplying, and subtracted when dividing. The exponents need to be the same so the leading numbers can be multiplied or divided. The result has the same number of decimal places as the least precise number.

EZ

7 0
3 years ago
Determine the ΔH for the following reaction 2NH3 + 5/2O2 = 2NO(g) + 3 H2O(g)
beks73 [17]

The enthalpy change, ΔH for the following reaction 2\:NH_{3}(g) + \frac{5}{2}\:O_{2}(g)\rightarrow 2\:NO (g) + 3\:H_{2}O (g) is -452.76 kJ.

<h3>What is enthalpy change, ΔH, of a reaction?</h3>

The enthalpy change of a reaction is the heat changes that occurs when a reaction proceeds to formation of products.

  • Enthalpy change, ΔH = ΔH of products - ΔH of reactants

The equation of the reaction is given below

2\:NH_{3}(g) + \frac{5}{2}\:O_{2}(g)\rightarrow 2\:NO (g) + 3\:H_{2}O (g)

\Delta{H_{f}\:of\:NO = 90.25 kJ; \Delta{H_{f}\:of\:H_{2}O =-241.82kJ;  \Delta{H_{f}\:of\:NH_{3} =-46.1 kJ;  \Delta{H_{f}\:of\:O_{2} =0

\Delta{H_{f}\:of\:rxn = (90.25*2)+(-241.82*3)-( -46.1*2)= -452.76\:kJ

Therefore, the enthalpy change, ΔH for the following reaction 2\:NH_{3}(g) + \frac{5}{2}\:O_{2}(g)\rightarrow 2\:NO (g) + 3\:H_{2}O (g) is -452.76 kJ.

Learn more about enthalpy change at: brainly.com/question/14047927

#SPJ1

4 0
2 years ago
Read 2 more answers
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