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gladu [14]
3 years ago
11

Write the symbol for every chemical element that has atomic number greater than 55 and atomic mass less than 144.0 u

Chemistry
1 answer:
laila [671]3 years ago
3 0
You will need a periodic table to help you answer this problem. The atomic numbers are arrange from lowest to highest in the periodic table. You can locate element number 55 to be Cesium with an atomic weight of 132.905 amu. So, you start from element 56. The following elements are:

56     Barium       137.328 amu
57     Lanthanium   138.905 amu
58     Cerium       140.116 amu
59     <span>Praseodymium    140.908 amu
60     Neodymium   144.243 amu

Neodymium is already greater than 144 amu. Therefore, these elements only include Barium, Lanthanium, Cerium and Praseodymium.</span>
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8. Sulfur has a first ionization energy of 1000 kJ/mol. Photons of what frequency are required to ionize one mole of Sulfur?​
Lynna [10]

Answer:

the frequency of photons v = 1.509\times10^{39}Hz

Explanation:

Given:  first ionization energy of 1000 kJ/mol.

No. of moles of sulfur = 1 mole

\Delta E_1 = 1000KJ/mol

We know that plank's constant

h = 6.626\times10^{-34} Js

Let the frequency of photons be ν

Also we know that ΔE = hν

this implies ν = ΔE/h

= \frac{10^6J}{6.626\times10^{-34} Js}

v = 1.509\times10^{39}Hz

Hence, the frequency of photons v = 1.509\times10^{39}Hz

6 0
3 years ago
What is the total number of moles of oxygen atoms in 1 mole of N2O3
Rama09 [41]
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7 0
3 years ago
Explain how creativity can play a role in the construction of scientific questions and hypotheses.
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6 0
3 years ago
Assuming complete dissociation, what is the ph of a 3.24 mg/l ba(oh)2 solution?
weqwewe [10]
We need to first find the molarity of Ba(OH₂) solution.
A mass of 3.24 mg is dissolved in 1 L solution.
Ba(OH)₂ moles dissolved - 3.24 x 10⁻³ g/171.3 g/mol = 1.90 x 10⁻⁵ mol
dissociaton of Ba(OH)₂ is as follows;
Ba(OH)₂ --> Ba²⁺ + 2OH⁻
1 mol of Ba(OH)₂ dissociates to form 2OH⁻ ions.
Therefore [OH⁻] = (1.90 x 10⁻⁵)x2  = 3.8 x 10⁻⁵ M
pOH = -log[OH⁻]
pOH = -log (3.8 x 10⁻⁵)
pOH = 4.42
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6 0
3 years ago
The equilibrium constant (Kp) for the decomposition of phosphorus pentachloride (PCl5) to phosphorus trichloride (PCl3) and mole
Oksanka [162]

Answer: The partial pressure of Cl_2 is 1.86 atm

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_c

The given balanced equilibrium reaction is,

                            PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)

Pressure at eqm.     0.973 atm                 0.548atm      x atm

The expression for equilibrium constant for this reaction will be,

K_c=\frac{(p_{PCl_3}\times (p_{Cl_2})}{(p_{PCl_5})}

Now put all the given values in this expression, we get :

1.05=\frac{(0.548)\times (x)}{(0.973)}

By solving the term 'x', we get :

x = 1.86 atm

Thus, the partial pressure of Cl_2 is 1.86 atm

4 0
4 years ago
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