Calculate the moles of argon gas at STP in 33 L. Round answer to 2 significant figures.

How would you prepare 100 ml of 0.4 M MgSO4 from a stock solution of 2 M MgSO4?

miss Akunina [59]

OK, so to answer this question, you will simply use the molality equation which is as follows:
M1V1 = M2V2
In the givens you have:
M1 = 2M
V1 is the unknown
M2 = 0.4M
V2 = 100 ml

plug in the givens in the above equation:
2 x V1 = 0.4 x 100
therefore:
V1 = 20 ml

Based on this: you should take 20 ml of the 2 M solution and make volume exactly 100 ml in a volumetric flask by diluting in water.

7 0

2 years ago

A solution of 2-propanol and 1-octanol behaves ideally. Calculate the chemical potential of 2-propanol in solution relative to t

andrew-mc [135]

Answer:

The chemical potential of 2-propanol in solution relative to that of pure 2-propanol is lower by 2.63x10⁻³.

Explanation:

The chemical potential of 2-propanol in solution relative to that of pure 2-propanol can be calculated using the following equation:

$\mu (l) = \mu ^{\circ} (l) + R*T*ln(x)$

Where:

μ (l): is the chemical potential of 2-propanol in solution

μ° (l): is the chemical potential of pure 2-propanol

R: is the gas constant = 8.314 J K⁻¹ mol⁻¹

T: is the temperature = 82.3 °C = 355.3 K

x: is the mole fraction of 2-propanol = 0.41

$\mu (l) = \mu ^{\circ} (l) + 8.314 \frac{J}{K*mol}*355.3 K*ln(0.41)$

$\mu (l) = \mu ^{\circ} (l) - 2.63 \cdot 10^{3} J*mol^{-1}$

$\mu (l) - \mu ^{\circ} (l) = - 2.63 \cdot 10^{3} J*mol^{-1}$

Therefore, the chemical potential of 2-propanol in solution relative to that of pure 2-propanol is lower by 2.63x10⁻³.

I hope it helps you!

8 0

3 years ago

In an experiment to study the photoelectric effect, a scientist measures the kinetic energy of ejected electrons as afunction of

crimeas [40]

Answer:

a) v₀ = 4.41 × 10¹⁴ s⁻¹

b) W₀ = 176 KJ/mol of ejected electrons

c) From the graph, light of frequency less than v₀ will not cause electrons to break free from the surface of the metal. Electron kinetic energy remains at zero as long as the frequency of incident light is less than v₀.

d) When frequency of the light exceeds v₀, there is an increase of electron kinetic energy from zero steadily upwards with a constant slope. This is because, once light frequency exceeds, v₀, its energy too exceeds the work function of the metal and the electrons instantaneously gain the energy of incident light and convert this energy to kinetic energy by breaking free and going into motion. The energy keeps increasing as the energy and frequency of incident light increases and electrons gain more speed.

e) The slope of the line segment gives the Planck's constant. Explanation is in the section below.

Explanation:

The plot for this question which is attached to this solution has Electron kinetic energy on the y-axis and frequency of incident light on the x-axis.

a) Wavelength, λ = 680 nm = 680 × 10⁻⁹ m

Speed of light = 3 × 10⁸ m/s

The frequency of the light, v₀ = ?

Frequency = speed of light/wavelength

v₀ = (3 × 10⁸)/(680 × 10⁻⁹) = 4.41 × 10¹⁴ s⁻¹

b) Work function, W₀ = energy of the light photons with the wavelength of v₀ = E = hv₀

h = Planck's constant = 6.63 × 10⁻³⁴ J.s

E = 6.63 × 10⁻³⁴ × 4.41 × 10¹⁴ = 2.92 × 10⁻¹⁹J

E in J/mol of ejected electrons

Ecalculated × Avogadros constant

= 2.92 × 10⁻¹⁹ × 6.023 × 10²³

= 1.76 × 10⁵ J/mol of ejected electrons = 176 KJ/mol of ejected electrons

c) Light of frequency less than v₀ does not possess enough energy to cause electrons to break free from the metal surface. The energy of light with frequency less than v₀ is less than the work function of the metal (which is the minimum amount of energy of light required to excite electrons on metal surface enough to break free).

As evident from the graph, electron kinetic energy remains at zero as long as the frequency of incident light is less than v₀.

d) When frequency of the light exceeds v₀, there is an increase of electron kinetic energy from zero steadily upwards with a constant slope. This is because, once light frequency exceeds, v₀, its energy too exceeds the work function of the metal and the electrons instantaneously gain the energy of incident light and convert this energy to kinetic energy by breaking free and going into motion. The energy keeps increasing as the energy and frequency of incident light increases and electrons gain more speed.

e) The slope of the line segment gives the Planck's constant. From the mathematical relationship, E = hv₀,

And the slope of the line segment is Energy of ejected electrons/frequency of incident light, E/v₀, which adequately matches the Planck's constant, h = 6.63 × 10⁻³⁴ J.s

Hope this Helps!!!

5 0

3 years ago

Which of the following statements about electronic configurations of atoms and mononuclear ions, and atomic orbital quantum numb

ruslelena [56]

Answer:

a. The electronic configuration of the hydride anion is 1s2. TRUE

b. The valence electronic configuration of strontium is 4d2. FALSE

c. For a given value of l the number of possible values of ml is 2l + 1. TRUE

d. Cu+ has the same electronic configuration as Ni. TRUE

e. The magnetic quantum number is never larger than the principle quantum number (for a given orbital). TRUE

Explanation:

a. The electronic configuration of the hydride anion is 1s² is true since the hydriden anion is the hdrogen atom which has gained an electron and we will add that electron to the 1s¹ configuration of H.

b. The valence electronic configuration of strontium is 4d2 is false since Sr is an element of period 5 , therefore its valece electrons are in in period five and it has 2 electrons because Sr belongs to group 2.

c. For a given value of l the number of possible values of ml is 2l+1 is true since this number gives the magnetic orientation for the sublevel. Thus for s there is only one orientation, then ml = 2 (0 ) +1 . Por p with l equal to 1 we have three possible orientations : 2(1) + 1 =3. The d and f sublevels have 10 and 14 orientations.

d. Cu⁺ has the the same electronic configuration as Ni is true since Cu, atomic number 29, has one more electron than its neighbor Ni with an atomic number of 28. If we remove one electron from copper we are gong to have the same 28 electrons niquel has in its neutral state.

e. The magnetic quantum number is never larger than the principal quantum number for a given orbital is true since l, the magnetic quantum number can have values up to n-1, the principal quantum number.

4 0

2 years ago

When 28.0 g of acetylene reacts with hydrogen, 24.5 g of ethane is produced. What is the percent yield of C2H6 for the reaction?

Afina-wow [57]

Answer:

$Y=75.6\%$

Explanation:

Hello.

In this case, since no information about the reacting hydrogen is given, we can assume that it completely react with the 28.0 g of acetylene to yield ethane. In such a way, via the 1:1 mole ratio between acetylene (molar mass = 26 g/mol) and ethane (molar mass = 30 g/mol), we compute the yielded grams, or the theoretical yield of ethane as shown below:

$m_{C_2H_6}^{theoretical}=28.0gC_2H_2*\frac{1molC_2H_2}{26gC_2H_2}*\frac{1molC_2H_6}{1molC_2H_2} *\frac{30gC_2H_6}{1molC_2H_6}\\ \\m_{C_2H_6}^{theoretical}=32.3gC_2H_6$

Hence, by knowing that the percent yield is computed via the actual yield (24.5 g) over the theoretical yield, we obtain:

$Y=\frac{24.5g}{32.3g}*100\%\\ \\Y=75.6\%$

Best regards.

3 0

2 years ago