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Pachacha [2.7K]
3 years ago
10

List six elements with only 1 electron in an s orbital

Chemistry
1 answer:
Verdich [7]3 years ago
6 0

Answer:

Lithium=Li2s^1\\Sodium=Na3s^1\\Potassium=K4s^1\\Rubidium=Rb5s^1\\Cesium=Cs6s^1\\Francium=Fr7s^1

Explanation:

When it comes to electron configuration and orbitals, it's important to first identify what exactly we are trying to identify. Below is a given example:

He1s^2

He=element

1=level

s=orbital

(exponent)^2=electrons

Looking at the periodic table, identify the alkali metal family on the periodic table, or group one elements:

Lithium=Li2s^1\\Sodium=Na3s^1\\Potassium=K4s^1\\Rubidium=Rb5s^1\\Cesium=Cs6s^1\\Francium=Fr7s^1

Notice how each configuration has an exponent of 1, representative of a single electron in their s-orbital.

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2.What two factors must be held constant for density to be a constant ratio?
irina1246 [14]

Answer:

Temperature and Pressure

Explanation:

Temperature and pressure cause change in volume.

So any change in volume will alter the ratio of density as given by equation of density.

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2 years ago
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What is the empirical formula of a compound containing 5.03 grams carbon, 0.42 grams hydrogen, and 44.5 grams chlorine?
djverab [1.8K]

The empirical formula of the compound is CHCl₃.

<h3>Calculation:</h3>

Given,

Mass of carbon = 5.03 g

Mass of hydrogen = 0.42 g

Mass of chlorine = 44.5 g

Molecular weight of carbon = 12 g

Molecular weight of hydrogen = 1 g

Molecular weight of chlorine = 35.4 g

First, calculate the moles of each element,

                      Moles = given mass/ molecular weight

Moles of carbon = 5.03/12 = 0.42

Moles of hydrogen = 0.42/1 = 0.42

Moles of chlorine = 44.5/ 35.4 = 1.26

Divide the moles of each element by the smallest number of moles,

0.42 mol of C/ 0.42 = 1 C

0.42 mol of H/ 0.42 = 1 H

1.26 mol of Cl/0.42 = 3 Cl

The ratio of elements is 1:1:3

Therefore the empirical formula of the compound will be CHCl₃.

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2 years ago
Why do some objects fall through the air at a different rate then others (this is for science)
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Answer:

the amount of air resistance depends upon the speed of the object, more massive objects fall faster than less massive objects because they are acted upon by a larger force of gravity; for this reason, they accelerate to higher speeds until the air resistance force equals the gravity force.

4 0
3 years ago
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 64.0 mL of 0.0600 M EDTA . Titration of the excess unreacted EDTA req
tigry1 [53]

Answer:

the concentration of Cd^{2+}  in the original solution= 0.0088 M

the concentration of Mn^{2+} in the original solution = 0.058 M

Explanation:

Given that:

The volume of the sample  containing Cd2+ and Mn2+ =  50.0 mL; &

was treated with 64.0 mL of 0.0600 M EDTA

Titration of the excess unreacted EDTA required 16.1 mL of 0.0310 M Ca2+

i.e the strength of the Ca2+ = 0.0310 M

Titration of the newly freed EDTA required 14.2 mL of 0.0310 M Ca2+

To determine the concentrations of Cd2+ and Mn2+ in the original solution; we have the following :

Volume of newly freed EDTA = \frac{Volume\ of \ Ca^{2+}* Sample \ of \ strength }{Strength \ of EDTA}

= \frac{14.2*0.0310}{0.0600}

= 7.3367 mL

concentration of  Cd^{2+} = \frac{volume \ of \  newly  \ freed \ EDTA * strength \ of \ EDTA }{volume \ of \ sample}

= \frac{7.3367*0.0600}{50}

= 0.0088 M

Thus the concentration of Cd^{2+} in the original solution = 0.0088 M

Volume of excess unreacted EDTA = \frac{volume \ of \ Ca^{2+} \ * strength \ of Ca^{2+} }{Strength \ of \ EDTA}

= \frac{16.1*0.0310}{0.0600}

= 8.318 mL

Volume of EDTA required for sample containing Cd^{2+}   and  Mn^{2+}  = (64.0 - 8.318) mL

= 55.682 mL

Volume of EDTA required for Mn^{2+}  = Volume of EDTA required for

                                                                sample containing  Cd^{2+}   and  

                                                             Mn^{2+} --  Volume of newly freed EDTA

Volume of EDTA required for Mn^{2+}  = 55.682 - 7.3367

= 48.3453 mL

Concentration  of Mn^{2+} = \frac{Volume \ of EDTA \ required \ for Mn^{2+} * strength \ of \ EDTA}{volume \ of \ sample}

Concentration  of Mn^{2+} =  \frac{48.3453*0.0600}{50}

Concentration  of Mn^{2+}  in the original solution=   0.058 M

Thus the concentration of Mn^{2+} = 0.058 M

6 0
3 years ago
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