Ask question

Ask question

All categories

3 years ago

15

Imagine you could observe the individual atoms that make up a piece of matter and that you observe the motion of the atoms becom

ing more orderly. what can you assume about the system

1 answer:

Evgen [1.6K]3 years ago

8 0

It is cooling down or solidifying. I think.

You might be interested in

Considering 0.10 m solutions of each substance, which contains the smallest concentration of ions

mihalych1998 [28]

Question options:

A) K2SO4

B) FeCl₃

C) NaOH

D) NH₃

E) KCl

Answer:

D. NH₃

Explanation:

K2SO4 = 2 K+ + SO42-

[K+]= 2 x 1.0 = 2.0 M ; [SO42-] = 1.0 M

total concentrations of ions = 2.0 + 1.0 = <em>3.0 M</em>

FeCl3 = Fe3+ + 3Cl-

[Fe3+] = 1.0 M ; [Cl-] = 3 x 1.0 = 3.0

total concentration ions = 1.0 + 3.0 =<em> 4.0 M</em>

NaOH = Na+ + OH-

[Na+] = [OH-] = 1.0 M

total concentration ions = 1.0 + 1.0 = <em>2.0 M</em>

<u>NH3 is a weak acid so the concentration of NH4+ and OH- </u><u><em>< 2.0</em></u>

KCl = K+ + Cl-

[K+] = [Cl-] = 1.0 M

total concentration ions = 1.0 + 1.0 =<em> 2.0 M</em>

7 0

3 years ago

Given the Henry’s law constant for O2(4.34*109Pa) at 25° C, calculate the molar concentration of oxygen in air-saturated and O2s

Flauer [41]

This is an incomplete question, here is a complete question.

The Henry's law constant for oxygen dissolved in water is 4.34 × 10⁹ g/L.Pa at 25⁰C.If the partial pressure of oxygen in air is 0.2 atm, under atmospheric conditions, calculate the molar concentration of oxygen in air-saturated and oxygen saturated water.

Answer : The molar concentration of oxygen is, $2.67\times 10^2mol/L$

Explanation :

As we know that,

$C_{O_2}=k_H\times p_{O_2}$

where,

$C_{O_2}$ = molar solubility of $O_2$ = ?

$p_{O_2}$ = partial pressure of $O_2$ = 0.2 atm = 1.97×10⁻⁶ Pa

$k_H$ = Henry's law constant = 4.34 × 10⁹ g/L.Pa

Now put all the given values in the above formula, we get:

$C_{O_2}=(4.34\times 10^9g/L.Pa)\times (1.97\times 10^{-6}Pa)$

$C_{O_2}=8.55\times 10^3g/L$

Now we have to molar concentration of oxygen.

Molar concentration of oxygen = $\frac{8.55\times 10^3g/L}{32g/mol}=2.67\times 10^2mol/L$

Therefore, the molar concentration of oxygen is, $2.67\times 10^2mol/L$

8 0

3 years ago

A test tube containing CaCO₃ is heated until the entire compound decomposes. If the test tube plus calcium carbonate originally

Marysya12 [62]

Find the molar mass of CaCO3 then subtract the molar mass what it originally weighed and the loss of mass. Hopefully this works!

8 0

3 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

Ainat [17]

Answer:

The correct answer is 4

Explanation:

Boron trifluoride (BF₃) has a molecular geometry (as shown in the image in the question) referred to as trigonal planar; this is because each of the the fluorine atoms/molecules (bonded to the central boron atom) is placed in such a way that they form the three "end points"/"domains" of an equilateral triangle. Hence, the correct option is the last option.

7 0

3 years ago