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Svetradugi [14.3K]
2 years ago
9

I'd: 9872093250, password: qqqqq, join the meeting​

Chemistry
2 answers:
Sladkaya [172]2 years ago
8 0
The password is suspicious
Ierofanga [76]2 years ago
7 0

Answer:

yoooolllllllooooh

Explanation:

scooby Doo bydoooonwhere are you?

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Which letter corresponds with the location of f orbitals on the periodic table?
Lerok [7]

Answer:

The correct would be C i think :)

Explanation:

Stay postivie :)

8 0
2 years ago
Which part of atom is responsible for chemical bonding
yulyashka [42]

Answer:

electrons are responsible for the chemical bonding

6 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
Which one of these is an example of a physical change?
aleksklad [387]
Hey there!

The correct answer to your question is option A.

Sugar dissolving in warm water is an example of a physical change.
This is because sugar dissolving in the water does not create a substance or material. All of the other options are examples of chemical changes, because they eventually become a substance/material.

Hope this helps you.
Have a great day!
6 0
3 years ago
How many potassium atoms are in 250.0 grams of potassium?
kogti [31]
Use the atomic mass of potassium, k, 39.1 g/mol, and the formula n = mass / atomic-mass.

Where n is the number of moles.

n = 250.0 g / 39.1 g/mol = 6.3939 mol.

Now multiply by Avogadro number to find the number of atoms:

6.3939 mol * 6.02*1023 atoms/mol = 38.49 * 10^23 atoms = 3.849 * 10^24.

Answer: 3.85*10^24
7 0
3 years ago
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