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svp [43]
3 years ago
13

¿Cuáles son las características de los materiales de laboratorio? Por ejemplo: exactitud, resistencia a la temperatura, etc.

Chemistry
1 answer:
boyakko [2]3 years ago
7 0

Answer:

lo siento, no sé punto libre. :p

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[tex]\purple{\rule{45pt}{7pt}}\purple{\rule{45pt}{999999pt}}[tex]

8 0
2 years ago
How do I balance <br>H2O ⇾ H2 + O2
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I change the coefficient 2 to 2 times 2 that will give me 4 so at this point on each side of the equation I have the same number for each type of atom that means this equation is balanced.
6 0
3 years ago
Which atom in the ground state requires the least amount of energy to remove its valence electron?(1) lithium atom (2) potassium
Fudgin [204]
<h3>Answer:</h3>

                   Rubidium (Rb)

<h3>Explanation:</h3>

                           Ionization Energy is defined as, "the minimum energy required to knock out or remove the valence electron from valence shell of an atom".

<h3>Trends in Periodic table:</h3>

               Along Periods:

                                        Ionization Energy increases from left to right along the periods because moving from left to right in the same period the number of protons (atomic number) increases but the number of shells remain constant hence, resulting in strong nuclear interactions and electrons are more attracted to nucleus hence, requires more energy to knock them out.

              Along Groups:

                                        Ionization energy decreases from top to bottom along the groups because the number of shells increases and the distance between nucleus and valence electrons also increases along with increase in shielding effect provided by core electrons. Therefore, the valence electrons experience less nuclear attraction and are easily removed.

<h3>Conclusion:</h3>

                   Given elements belong to same group hence, Rubidium present at the bottom of remaining elements will have least ionization energy due to facts explained in trends of groups above.

5 0
3 years ago
UCI Chemistry researchers, Prof. F. Sherwood Rowland and Dr. Mario Molina werefirst to discovered in 1973 that chlorofluorocarbo
Olin [163]

Answer:

15.27895 x 10⁶kg of chlorine radical is added to the atmosphere in a year due to 100 million MVACs                                                                                              

Explanation:

Chlorofluorocarbons (CF₂Cl₂) from refrigerants produce chlorine radicals according to the following equation

         CF₂Cl₂ → CF2Cl  + Cl ⁻ .........(1)

From equation 1, one mole of CF₂Cl₂ produces one mole of Chlorine radical

From the question,

The emission rate of CF₂Cl₂ is 59.5mg/hour/MVAC

In one day the emission rate would be 59.5 x 24hours

                                                                  = 1428mg/day

In one year, the emission rate would be 1428mg/day x 365days

                                                                 =  521220mg/year

                                                                   = 521.220g/year/MVAC

Therefore the emission rate for 100 million MVAC using CF₂CL₂ in a year is

                                                                    = 52122 x 10⁶g/year/MVAC

                                                                    = 52122 x 10³kg/year/MVAC

The molar mass of CF₂CL₂                       = 120.913g/mol

No of moles  of CF₂CL₂                              = mass/ molar mass

                                                                     = 52122 x 10⁶g / 120.913g/mol

                                                                      = 431 x 10⁶ moles of CF₂Cl₂

From equation 1,  since one mole of CF₂Cl₂ produces one mole of Chlorine radical, it implies that

431 x 10⁶ moles of CF₂Cl₂ would produce 431 x 10⁶ moles of chlorine radical,

Therefore, to find the mass of chlorine radical produced, we use the formula

No of moles of chlorine radical  = mass/ molar mass

431 x10⁶ moles = mass of chlorine radical /molar mass of chlorine radical

431 x 10⁶ moles = mass/ 35.45g/mol

mass of chlorine =  431 x 10⁶ moles x 35.45 g/mol

                            =    15278.95 x 10⁶ g

In Kg, the mass    =   15,278.95 x 10³kg of cholrine radical

                             =   15.27895 x 10⁶ Kg of chlorine radical

                                                                                                 

6 0
3 years ago
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