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allochka39001 [22]
3 years ago
14

Which electrons participate in chemical bonding?

Chemistry
2 answers:
OverLord2011 [107]3 years ago
7 0
B.) Valence Electrons. The nucleus of the valence electrons attracts and pulls atoms together.
VMariaS [17]3 years ago
4 0

Answer:

B

Explanation:

I took the test

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What type of reaction does the following equation represent?
Ostrovityanka [42]

Answer:

B. double-replacement RXN

Explanation:

more specifically, this is a precipitation rxn.

6 0
2 years ago
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1. Who is the father of atomic theory? 2. Who discovered the electron? 3. Who expressed particles by wave equations? 4. Who rese
krek1111 [17]
1. Who is the father of atomic theory?
Dalton

2. Who discovered the electron?
<span>Thomson
</span>
3. Who expressed particles by wave equations?
<span>Schrödinger
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4. Who researched on radioactivity?
Curie

<span>5. Who discovered the "open spaces" model?
</span><span>Rutherford
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6. Who applied quantum theory to atoms? 
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6 0
3 years ago
Read 2 more answers
How many milliliters of 0.0630 m edta are required to react with 50.0 ml of 0.0110 m cu2 ?
gizmo_the_mogwai [7]
Moles Cu+2 = M * V
                     =  0.05 L * 0.011  m
                     = 0.00055 moles

when the molar ratio of Cu2+: EDTA = 1:1 so moles od EDTa also =0.00055 moles

and when the Molarity of EDTa = 0.0630 M

∴ Volume of EDTA =  moles / Molarity
                 = 0.00055 / 0.0630
                 = 0.0087 L = 8.7 L
4 0
3 years ago
Define the term chemotherapy.​
Ksivusya [100]

I'm not able to understand that what is written in the pic.. As I can't understand this language..

Treatment that uses drugs to stop the growth of cancer cells, either by killing the cells or by stopping them from dividing. Chemotherapy may be given by mouth, injection, or infusion, or on the skin, depending on the type and stage of the cancer being treated.

8 0
2 years ago
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The rate of effusion of an unknown gas was measured and found to be 11.9 mL/min. Under identical conditions, the rate of effusio
iren2701 [21]

Answer : The correct option is, (B) CO_2

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

R\propto \sqrt{\frac{1}{M}}

or,

(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}       ..........(1)

where,

R_1 = rate of effusion of unknown gas = 11.9\text{ mL }min^{-1}

R_2 = rate of effusion of oxygen gas = 14.0\text{ mL }min^{-1}

M_1 = molar mass of unknown gas  = ?

M_2 = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the above formula 1, we get:

(\frac{11.9\text{ mL }min^{-1}}{14.0\text{ mL }min^{-1}})=\sqrt{\frac{32g/mole}{M_1}}

M_1=44.2g/mole

The unknown gas could be carbon dioxide (CO_2) that has approximately 44 g/mole of molar mass.

Thus, the unknown gas could be carbon dioxide (CO_2)

5 0
3 years ago
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