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navik [9.2K]
3 years ago
15

Compared to the nonmetals in Period 2, the metals in Period 2 generally have larger

Chemistry
1 answer:
konstantin123 [22]3 years ago
6 0
The answer is atomic radii; the size or radii of an atom increases from left to right, versus the ionization energies and electronegativities of atoms which increase from right to left.
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The distance between Earth and its moon is 384000000 meters. Express this distance in kilometers.
Andreyy89
384000000m

to change from m to km you just need to divide that give unit in Meters by 1000

which equals,
384000

so,

384000000 meters = 384000 kilometers
3 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
What happens when an atom is exposed to an energy source such as electricity?
yarga [219]
It moves to a higher energy level<span />
7 0
2 years ago
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If you have 3cm of each<br> substance, which would<br> have the MOST MASS
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5 0
3 years ago
Solid sodium metal reacts violently with water, producing heat, hydrogen gas, and sodium hydroxide. How many molecules of hydrog
asambeis [7]

Answer:

4.6 × 10²³ molecules:

Step-by-step solution

You will need a balanced equation with masses, moles, and molar masses, so let's gather the information in one place:

M_r:  22.99

          2Na + 2H₂O ⟶ 2NaOH + H₂

m/g:    35

1. Calculate the <em>moles of Na </em>

Moles of Na = 35 g Na × (1 mol Na/22.99 g Na)

Moles of Na = 1.52 mol Na

2. Calculate the <em>moles of H₂ </em>

Moles of H₂ = 1.52 mol Na × (1 mol H₂/2 mol Na)

Moles of H₂= 0.761 mol H₂

3. Calculate the molecules of H₂

6.022 × 10²³ molecules H₂ = 1 mol H₂

                Molecules of H₂ = 0.761 × (6.022 × 10²³ /1)

                Molecules of H₂ = 4.6 × 10²³ molecules H₂

The reaction forms 4.6 × 10²³ molecules of H₂.

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3 years ago
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