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3 years ago

Which subatomic particle is involved in the covalent and ionic bonding processes? A. protons B. electrons C. Nucleus D. neutrons

Chemistry

2 answers:

topjm [15]3 years ago

8 0

B. Electrons because

A - Protons always remain constant. They decay over a LONG period of time and don't do anything.

B - it is correct because they do all the bonding.

C - It's not a subatomic particle

D - Neutrons are not involved in bonding as well.

Reptile [31]3 years ago

8 0

Hello gurl

the answer is B)electrons

hope it helps

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A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 23.0 mL of

Rus_ich [418]

52.0ml of 0.35M CH3COOH : 0.052 L(0.35M) = .0182 mol of CH3COOH.

31.0ml of 0.40M NaOH : .031 L(0.40M) = .0124 mol of NaOH.

After the reaction, .0124 Mol CH3COO- is generated and .058 mol CH3COOH is left un-reacted. The concentration would be 12.4/V and 5.8/V, respectively. Therefore:


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3 years ago

A) 24 g B) 36 g C) 48 g D) 60 g

Hitman42 [59]

Answer:

I'm not sure if this right because i've only taken one year of Chemistry but the answer I got was

A. 24 g

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2 years ago

A. Based on the activation energies and frequency factors, rank the following reactions from fastest to slowest reaction rate, a

deff fn [24]

Answer:

A) $E_{a} = 350KJ/mol$ , $E_{a} = 50KJ/mol$ , $E_{a} = 50KJ/mol$

A = 1.5× $10^{-7}s^{-1}$ , A = 1.9× $10^{-7} s^{-1}$ , A=1.5× $10^{-7} s^{-1}$

B) 4.469

Explanation:

From Arrhenius equation

$K=Ae^{\frac{E_{a} }{RT} }$

where; K = Rate of constant

A = Pre exponetial factor

$E_{a}$ = Activation Energy

R = Universal constant

T = Temperature in Kelvin

Given parameters:

$E_{a} =165KJ/mol$

$T_{1}=505K$

$T_{2}=525K$

$R=8.314JK^{-1}mol^{-1}$

taking logarithm on both sides of the equation we have;

$InK=InA-\frac{E_{a} }{RT}$

since we have the rate of two different temperature the equation can be derived as:

$In(\frac{K_{2} }{K_{1} } )=\frac{E_{a} }{R}(\frac{1}{T_{1} } -\frac{1}{T_{2} } )$

$In(\frac{K_{2} }{K_{1} } )=\frac{165000J/mol}{8.314JK^{-1}mol^{-1} }.(\frac{1}{505} -\frac{1}{525} )$

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