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blagie [28]
3 years ago
8

The successive ionization energies of a certain element are I1 = 589.5 kJ/mol, I2 =1145 kJ/mol, I3= 4900 kJ/mol, I4 = 6500 kJ/mo

l, and I5 = 8100 kJ/mol. This pattern of ionization energies suggests that the unknown element is:_________
a. K
b. Si
c. As
d. Ca
e. S
Chemistry
2 answers:
stepladder [879]3 years ago
8 0

Answer:

=D

Explanation:

Calcium has 2 Valence electrons which are easy to remove. From the ionization energy values, I1 = 589.5 kJ/mol, I2 =1145 kJ/mol, I3= 4900 kJ/mol, I4 = 6500 kJ/mol, and I5 = 8100 kJ/mol, the energy difference between the 2nd and 3rd electron is the highest because the energy required to remove the 3rd electron is geeatest which implies that the 3rd electron must be a completed octet-orbital electron which is more difficult to remove than other electrons

Gnom [1K]3 years ago
7 0

Answer:

As

Explanation:

For any element to exhibit the pattern of ionization energy shown in the question, it must possess five electrons in its outermost shell. These five electrons are not lost at once. They are lost progressively until the valence shell becomes empty. The ionization energy increases steadily as more electrons are lost from the valence shell.

The only pentavalent element among the options in arsenic, hence the answer.

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For each of the following pairs of complexes, identify which one you would predict to have the larger Δo value, and explain why.
mash [69]

Answer:

a) [Fe(H2O)6]3+

b) [Fe(CN)6]3−

c) [Ru(CN)6]3-

Explanation:

. [Mn(H2O)6]2+ or [Fe(H2O)6]3+

The both complexes are d5 complexes with the same ligand , water. Water is a weak ligand and note that Mn^2+ often have a crystal field stabilization energy of zero hence

[Fe(H2O)6]3+ will possess a greater ∆o value.

The splitting of d orbitals according to the crystal field theory depends on the;

i)geometry of the complex

ii) nature of the metal ion,

iii)charge on the metal ion,

iv) ligands that surround the metal ion.

When the geometry and the ligands are held constant, the order of crystal field splitting is as follows;

Pt4+ > Ir3+ > Rh3+ > Co3+ > Cr3+ > Fe3+ > Fe2+ > Co2+ > Ni2+ > Mn2+

[Fe(H2O)6]3+ or [Fe(CN)6]3−

[Fe(CN)6]3− will have a greater ∆o because the cyanide ion is a strong field ligand compared to water. A strong field ligand causes a greater splitting of the octahedral crystal field compared to a weak field ligand.

. [Fe(CN)6]3− or [Ru(CN)6]3-

[Ru(CN)6]3- will exhibit a greater crystal field splitting. Crystal field splitting increases with the second and third row transition elements when compared to the crystal field splitting of the first row transition elements. Note that, there is an increase of approximately 30%–50% in Δo on going from a first-row transition metal to a second-row metal and another 30%–50% increase on going from a second-row to a third-row metal when they have the same geometry and oxidation state.

4 0
3 years ago
Snape grows tired of these conceptual questions and thinks it's time for a problem. What is the retention factor if the distance
MAVERICK [17]

Answer:

The Retention factor (rf) value is = 0.2

Explanation:

  • Retention factor (Rf) is factor used substances that could be separated using Chromatography. Retention factor determines how fast the component can move on the chromatogram (stationary phase) after elution. Elution occurs when mobile phase (solvent) moves across the stationary phase when the solute has been spotted on the origin.
  • Retention factor (Rf) ranges from value between 0 and  1. The closer the value to 1, the faster it can move upon elution. Rf can be calculated.
  • Rf value = distance moved by the solute / distance moved by the solvent

             = 0.40cm / 2.00cm

             = 0.2

6 0
3 years ago
An aqueous solution in a 55 gallon (208 l drum), characterized by minimal buffering capacity, received 4kg of phenol and 1.5 kg
lianna [129]

An aqueous solution in a 55 gallon (208 l drum), characterized by minimal buffering capacity, received 4kg of phenol and 1.5 kg of sodium phenate. What is the ph of the solution. The pka of phenol = 9.98. Mw of phenol and sodium phenate are 94 g/mol and 116 g/mol, respectively.


Volume of solution = 55 gallons = 208.2 L [ 1 gallon = 3.78 L]

moles of phenol = mass / molar mass = 4000 g / 94 = 42.55 moles

moles of sodium phenate = mass / molar mass = 1500 / 116 = 12.93 moles

pKa of phenol = 9.98

We know that the pH of buffer is calculated using Hendersen Hassalbalch's equation

pH = pKa + log [salt] / [acid]

volume is same for both the sodium phenate and phenol has we can directly take the moles of each in the formula

pH = 9.98 + log [12.93 / 42.55] = 9.46



4 0
3 years ago
Mount Everest rises to a height of 8,850 m above sea level. At a base camp on the mountain the atmospheric pressure is measured
yan [13]

Answer:

Water will boil at 76^{0}\textrm{C}.

Explanation:

According to clausius-clapeyron equation for liquid-vapor equilibrium:

                                         ln(\frac{P_{2}}{P_{1}})=\frac{-\Delta H_{vap}^{0}}{R}[\frac{1}{T_{2}}-\frac{1}{T_{1}}]

where, P_{1} and P_{2} are vapor pressures of liquid at T_{1} (in kelvin) and T_{2} (in kelvin) temperatures respectively.

Here, P_{1} = 760.0 mm Hg, T_{1} = 373 K, P_{2} = 314.0 mm Hg

Plug-in all the given values in the above equation:

                          ln(\frac{314.0}{760.0})=\frac{-40.7\times 10^{3}\frac{J}{mol}}{8.314\frac{J}{mol.K}}\times [\frac{1}{T_{2}}-\frac{1}{373K}]

                       or, T_{2}=349 K

So, T_{2}=349K=(349-273)^{0}\textrm{C}=76^{0}\textrm{C}

Hence, at base camp, water will boil at 76^{0}\textrm{C}

6 0
3 years ago
What is the mass of 1.25 L of ammonia gas at STP
Alex73 [517]

Answer:

mass 1.25 Liters NH₃(gas) = 0.949 grams (3 sig-figs)

Explanation:

At STP (Standard Temperature-Pressure conditions => 0°C(=273K) and 1atm pressure,  1 mole <u>any</u> gas will occupy 22.4 Liters.

So, given 1.25 Liters ammonia gas at STP, convert to moles then multiply by formula wt. (17g/mole gives mass of NH₃.

moles NH₃(gas) = 1.25L NH₃(gas)/22.4L NH₄(gas)· NH₃(gas)mole⁻¹ = 0.0558 mole NH₃(gas).  

Converting to grams NH₃(gas) = 0.0558 mole NH₃(gas) x 17 g·mol⁻¹ = 0.949 grams NH₃(gas).

7 0
3 years ago
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