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3 years ago

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What effects does the concentration of reactants have on the rate of a reaction ?

1 answer:

Harlamova29_29 [7]3 years ago

5 0

Question:

<em>What effects does the concentration of reactants have on the rate of a reaction?</em>

Answer:

<em>Reactant concentration. Increasing the concentration of one or more reactants will often increase the rate of reaction. This occurs because a higher concentration of a reactant will lead to more collisions of that reactant in a specific time period.</em>

<em>Increasing the concentration of reactants generally increases the rate of reaction because more of the reacting molecules or ions are present to form the reaction products. ... When concentrations are already high, a limit is often reached where increasing the concentration has little effect on the rate of reaction.</em>

Hope this helps, have a good day. c;

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CH3CH2PH2 can be synthesized by an SN2 reaction. Draw the structures of the alkyl chloride and nucleophile that will give this c

Maru [420]

Answer:

The PH2 becomes attached to the alkyl halide with inversion of configuration

Explanation:

SN2 reaction involves the simultaneous attachment of the nucleophile as the leaving group departs. This leads to an inversion of configuration because the sterochemistry around the carbon bearing the leaving group is reversed. The reaction proceed via a crowded five coordinate transition state. Details are shown in the image attached.

6 0

3 years ago

Phosgene is formed from carbon monoxide and chlorine. At 611 oC, Kc is approximately 16.5 M-1 or L/mol. Calculate Kp (in atm-1)

defon

Answer:

Kp is 0.228/atm

Explanation:

This is the reaction:

CO + Cl<u>₂</u> → CCl₂O

1 mol of carbon monoxide and 1 mol of chlorine produce 1 mol of phosgene.

Formula for Kp which derivates from Kc is:

Kp = Kc (R.T)ⁿᵇ ⁻ ⁿᵃ

Δп = nb (moles in the products) - nₐ(moles in the reactants)

Δп = 1 - 2 = -1

T is T° in K → T°C + 273 = 611°C +273 = 884K

R → Universal constant gas → 0.082 L.atm/mol.K

We replace the data: Kp = 16.5 L/mol (0.082 . 884K)⁻¹ → 0.228/atm

5 0

4 years ago

Please help me which one

mel-nik [20]

Non of the above because protons and neutrons don’t mix with each one there

3 0

2 years ago

3) The acronym m/s stands for?

algol13

Answer:

B-meters per second

Explanation:

7 0

3 years ago

Read 2 more answers

Arrange the following H atom electron transitions in order of increasing frequency of the photon absorbed or emitted:

Temka [501]

The question is incomplete , complete question is:

Arrange the following H atom electron transitions in order of increasing frequency of the photon absorbed or emitted:

(a) n = 2 to n = 4

(b) n = 2 to n = 1

(c) n = 2 to n = 5

(d) n = 4 to n = 3

Answer:

Hence the order of the transition will be : d < a < c < b

Explanation:

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number

Energy of n = 1 in an hydrogen atom:

$E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6 eV$

Energy of n = 2 in an hydrogen atom:

$E_2=-13.6\times \frac{1^2}{2^2}eV=-3.40eV$

Energy of n = 3 in an hydrogen atom:

$E_3=-13.6\times \frac{1^2}{3^2}eV=-1.51eV$

Energy of n = 4 in an hydrogen atom:

$E_4=-13.6\times \frac{1^2}{4^2}eV=-0.85 eV$

Energy of n = 5 in an hydrogen atom:

$E_5=-13.6\times \frac{1^2}{5^2}eV=-0.544 eV$

a) n = 2 to n = 4 (absorption)

$\Delta E_1= E_4-E_2=-0.85eV-(-3.40eV)=2.55 eV$

b) n = 2 to n = 1 (emission)

$\Delta E_2= E_1-E_2=-13.6 eV-(-3.40eV)=-10.2 eV$

Negative sign indicates that emission will take place.

c) n = 2 to n = 5 (absorption)

$\Delta E_3= E_5-E_2=-0.544 eV-(-3.40eV)=2.856 eV$

d) n = 4 to n = 3 (emission)

$\Delta E_4= E_3-E_4=-1.51 eV-(-0.85 eV)=-0.66 eV$

Negative sign indicates that emission will take place.

According to Planck's equation, higher the frequency of the wave higher will be the energy:

$E=h\nu$

h = Planck's constant

$\nu$ frequency of the wave

So, the increasing order of magnitude of the energy difference :

$E_4$

And so will be the increasing order of the frequency of the of the photon absorbed or emitted. Hence the order of the transition will be :

: d < a < c < b

7 0

4 years ago