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vesna_86 [32]
3 years ago
14

An object has a mass of 55 grams and a density of 3.23g/ml ? what is the volume

Chemistry
1 answer:
lara [203]3 years ago
8 0
Volume= mass divided by density
V= m/d
55/3.23
= 17.03
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Iteru [2.4K]

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6. Color Change.

Production of an odor.

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What is the strongest type of intermolecular force between solve and solvent in Cu(s) in Ag(s)?
maks197457 [2]

The strongest intermolecular forces are in ion-ion bonds which happen when a metal bonds to another metal. 2. The next strongest forces are ion-dipole bonds which happen when metals bond to nonmetals. 3.

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In the second step of this reaction, isocyanic acid reacts to form melamine and carbon dioxide: hnco(l)→c3n3(nh2)3(l)+co2(g) bal
slava [35]

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7 0
3 years ago
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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
Utilizing dimensional analysis find the number of inches in a kilometer given that 1 inch equals 2.54 cm
hodyreva [135]
To solve this problem, separate it into chunks that you know. You know that there are 2.54 centimeters in 1 inch. You know that there are 100 centimeters in 1 meter. You know that there are 1000 meters in a kilometer. Therefore, we'll convert in this order: 1) from kilometers to meters, 2) from meters to centimeters, and 3) from centimeters to inches.
1) 1km × 1000m/1km
= 1000m
2) 1000m × 100cm/1m
= 100000cm
3) 100000cm × 1in/2.54cm
≈ 39370in
So, there are approximately 39370 inches in a kilometer.
4 0
4 years ago
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