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4 years ago

14

An aqueous solution containing 35.5 g of an unknown molecular (non-electrolyte) compound in 151.0 g of water was found to have a

freezing point of -1.3 ?C

1 answer:

Tasya [4]4 years ago

8 0

ccccccccccccccccccccccccccccccccccc

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When does an object move<br><br>

natulia [17]

Answer:

When a force pushes or pulls the object, the object will move in the direction of the force. The bigger the force, and the lighter the object, the greater the acceleration. It can also make something slow down, speed up or change direction.

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3 years ago

Which radioactive waste can be stored for decay and then safely released into the environment?

Natalka [10]

The radioactive waste can be stored for decay and then safely released into the environment is N-16.

<h3>What are the types of decay?</h3>

Decays can be of three types: alpha, beta and gamma. Each of them corresponds to a different radioactive particle, which changes the nucleus of the emitting atom according to its characteristics.

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2 years ago

The density of an unknown metal was determined to be 2.85 g/ml. the actual density was 2.70 g/ml. what is the percent error in t

Tasya [4]

% error = $\frac{|experimental - theoretical|}{theoretical}$ x 100%

Experimental: 2.85
Actual (theoretical): 2.70

% error = $\frac{2.85-2.70}{2.70}$ x 100% = .055555 x 100% = 5.56%

7 0

3 years ago

algol13

Distilling ocean water would be a good solution to meet this community's water needs because it will prevent some water borne diseases such as cholera, diarrhea, typhoid etc.

<h3>Importance of water distillation</h3>

Water Distillation Systems has several advantages and some of the advantages include;

It removes waterborne biological contaminants such as bacteria, viruses, organic and inorganic chemicals, heavy metals, etc
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Thus, distilling ocean water would be a good solution to meet this community's water needs because it will prevent some water borne diseases such as cholera, diarrhea, typhoid etc.

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6 0

2 years ago

7. A 2.0 L container had 0.40 mol of He(g) and 0.60 mol of Ar(g) at 25°C.

Finger [1]

Answer:

a) Ek Ar > Ek He

b) v Ar < v He

c) If v Ar = 431 m/s ⇒ v He = 1710.44 m/s

d) Pt = 12.218 atm

e) P He = 4.887 atm and P Ar = 7.33 atm

Explanation:

container:

∴ V = 2.0 L

∴ n He = 0.4 mol

∴ n Ar = 0.6 mol

∴ T = 25°C ≅ 298 K

a) Internal energy (U) :

∴ U = Ek + Ep = kinetic energy + potential energy

∴ Ep: the potential interaction energy is neglected, assuming ideal gas mixture

⇒ U = Ek = N(1/2mv²)= 3/2 NKT

∴ N = nNo ....number of moleculas

∴ K = 1.380 E-23 J/K....Boltzmann's constant

∴ No = 6.022 E23 molec/mol....Avogadro's number

for He:

⇒ N = (0.4)(6.022 E23) = 2.4088 E23 molec

⇒ Ek = (3/2)(2.4088 E23)(1.380 E-23 J/K)(298) = 1485.892 J

for Ar:

⇒ N = (0.6)(6.022 E23) = 3.6132 E 23 molec

⇒ Ek = (3/2)(3.6132 E23)(1.380 E-23 J/K)(298) = 2228.838 J

** Ar gas has a greater average kinetic energy

b) He:

∴ N(1/2)mv² = (3/2)NKT

⇒ mv² = 3KT

⇒ v² = 3KT/m

⇒ v = √3KT/m

∴ m He = (0.4 mol)(4.0026 g/mol) = 1.601 g He = 1.601 E-3 Kg He

⇒ v = √(3(1.380 E-23)(298)/(1.601 E-3)) = 2.776 E-9 m/s He

Ar:

∴ m Ar = (0.6)(39.948 g/mol) = 23.969 g = 0.0239 Kg Ar

⇒ v = 6.99 E-10 m/s

** v Ar < v He

c) r = V Ar / v He = (6.99 E-10 m/s)/(2.776 E-9 m/s) = 0.252

∴ If v Ar = 431 m/s

⇒ v He = v Ar/0.252 = 431 m/s / 0.252 = 1710.44 m/s

d) Pt = ntRT / V

∴ nt = 0.4 + 0.6 = 1 mol

⇒ Pt = (1mol)(0.082 atm.L/K.mol)(298 K)/(2.00 L) = 12.218 atm

e) P He = nRT/V = (0.4)(0.082)(298)/2 = 4.8872 atm

⇒ P Ar = Pt - PHe = 12.218 - 4.8872 = 7.33 atm

3 0

3 years ago