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OLga [1]
2 years ago
4

Which statement is true about the properties of the majority of the elements in the far left column of the periodic table? A. Th

ey are highly reactive. B. They have no luster. C. They have low electrical conductivity. D. They have low heat conductivity.
Chemistry
2 answers:
Degger [83]2 years ago
4 0

<u>Answer: </u>

Option: A. The statement alkali metals are highly reactive is true.

<u>Explanation: </u>

The far left column metals are named as Alkali earth metals. Since their ionization energies are low and have larger atomic radii, they are known for their high reactivity. These metals are malleable, ductile. Alkali metals are great conductors of electricity and heat. The most reactive alkali metals in this group are Cesium and Francium. They are shiny, silvery and soft (lustrous).

vladimir1956 [14]2 years ago
3 0

Answer: A

Explanation:

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A 0.529-g sample of gas occupies 125 ml at 60. cm of hg and 25°c. what is the molar mass of the gas?
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<span>Let's </span>assume that the gas has ideal gas behavior. <span>
Then we can use ideal gas formula,
PV = nRT<span>

</span><span>Where, P is the pressure of the gas (Pa), V is the volume of the gas (m³), n is the number of moles of gas (mol), R is the universal gas constant ( 8.314 J mol</span></span>⁻¹ K⁻¹) and T is temperature in Kelvin.<span>
<span>
</span>P = 60 cm Hg = 79993.4 Pa
V = </span>125  mL = 125 x 10⁻⁶ m³

n = ?

<span> R = 8.314 J mol</span>⁻¹ K⁻¹<span>
T = 25 °C = 298 K
<span>
By substitution,
</span></span>79993.4 Pa<span> x </span>125 x 10⁻⁶ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 298 K<span>
                                          n = 4.0359 x 10</span>⁻³ mol

<span>
Hence, moles of the gas</span> = 4.0359 x 10⁻³ mol<span>

Moles = mass / molar mass

</span>Mass of the gas  = 0.529 g 

<span>Molar mass of the gas</span> = mass / number of moles<span>
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<span>                                    = </span>131.07 g mol</span>⁻¹<span>

Hence, the molar mass of the given gas is </span>131.07 g mol⁻¹

4 0
2 years ago
A sealed, insulated calorimeter contains water at 310 K. The surrounding air temperature is 298 K, and the water inside the calo
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2 years ago
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If the mass percentage composition of a compound is 72.1% Mn and 27.9% O, its empirical formula is
mojhsa [17]

Answer:

MnO- Manganese Oxide

Explanation:

Empirical formula: This is the formula that shows the ratio of elements

present in a  

compound.

   

How to determine Empirical formula

1. First arrange the symbols of the elements present in the compound

alphabetically to  determine the real empirical formula. Although, there

are exceptions to this rule, E.g H2So4

2. Divide the percentage composition by the mass number.

3. Then divide through by the smallest number.

4. The resulting answer is the ratio attached to the elements present in

a compound.

           

                                                                              Mn                         O    

                         

% composition                                                      72.1                      27.9    

                       

Divide by mass number                                       54.94                     16  

                                 

                                                                               1.31                      1.74    

                       

Divide by the smallest number                         1.31                      1.31                          

                                                                               1                    1.3

                                                 

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8 0
2 years ago
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STatiana [176]

Answer: The longest wavelength of light that will produce free chlorine atoms in solution is 493 nm.

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1 mole = 6.022\times 10^{23} molecules

For 6.022\times 10^{23} molecules = 242,800 Joules

For one molecule of chlorine gas =  \frac{242800 Joules/mol}{6.022\times 10^{23} mol^{-1}}=40,318.83\times 10^{-23}Joules

According to photoelectric equation:

E=h\nu=\frac{hc}{\Lambda }

E = Energy of the photon of light used to produce free chlorine atoms

\nu= frequency of the light used to produce free chlorine atoms

h = Planck's constant =6.626\times 10^{-34}J.s, c = speed of light=3\times 10^8 m/s

\lambda = wavelength of the light used to produce free chlorine atoms

40,318.83\times 10^{-23}J=\frac{hc}{\Lambda }=\frac{6.626\times 10^{-34} J.s\times 3\times 10^8 m/s}{\lambda }

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The longest wavelength of light that will produce free chlorine atoms in solution is 493 nm.

5 0
2 years ago
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