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Blababa [14]
3 years ago
12

The second-order rate constant for the following gas-phase reaction is 0.041 1/MLaTeX: \cdotâs. We start with 0.438 mol C2F4 in

a 2.42 liter container, with no C4F8 initially present. C2F4 LaTeX: \longrightarrowâ¶ 1/2 C4F8 What is the half-life (in seconds) of the reaction for the initial C2F4 concentration? Enter to 1 decimal place.
Chemistry
1 answer:
pantera1 [17]3 years ago
3 0

Answer:

134.8 seconds is the half-life (in seconds) of the reaction for the initial C_2F_4 concentration

Explanation:

Half life for second order kinetics is given by:

t_{\frac{1}{2}=\frac{1}{k\times a_0}

Integrated rate law for second order kinetics is given by:

\frac{1}{a}=kt+\frac{1}{a_0}

t_{\frac{1}{2} = half life

k = rate constant

a_0 = initial concentration

a = Final concentration of reactant after time t

We have :

C_2F_4 \longrightarrow \frac{1}{2} C_4F_8

Initial concentration of C_2F_4=[a_o]=\frac{0.438 mol}{2.42 L}=0.1810 mol/L

Rate constant = k = 0.041 M^{-1} s^{-1}

t_{\frac{1}{2}=\frac{1}{k\times a_0}

=\frac{1}{0.041 M^{-1} s^{-1}\times 0.1810 mol/L}

t_{1/2}=134.8 s

134.8 seconds is the half-life (in seconds) of the reaction for the initial C_2F_4 concentration

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Tasya [4]

Answer:

V = 38.48 L

Explanation:

Given that,

No. of moles = 1.5 mol

Pressure, P = 700 torr

Temperature, T = 15°C = 288 K

We need to find the volume of the gas. The ideal gas equation is given by :

PV=nRT\\\\V=\dfrac{nRT}{P}\\\\V=\dfrac{1.5\times 62.36\times 288}{700}\\\\V=38.48\ L, R = L.Torr.K⁻¹.mol⁻¹

So, the required volume is equal to 38.48 L.

7 0
3 years ago
A solution has a pOH of 8.7 so what is the pH of the solution? Is the solution acidic, basic, or neutral?
RSB [31]
PH + pOH = 14

pH = 14 - pOH

pH = 14 - 8.7

pH = 5,3

This solution is <u>acidic</u>.

If pH<7 - acidic
If pH=7 - neutral
If pH>7 - basic
5 0
3 years ago
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PLEASE HELP! I DONT UNDERSTAND WHAT TO DO!
goldenfox [79]

Answer:

Na+Cl- + Ag+no3- ---> Na+No3- + Ag+Cl-

A spectator ion is an ion that exists as a reactant and a product in a chemical equation

Explanation:

When a solution of sodium hydroxide, NaOH, is mixed with hydrochloric acid, HCl, the compounds dissociate into the ions Na+, OH-, H+ and Cl-. The hydrogen and hydroxide ions react to form water, but the sodium and chlorine ions stay in solution unchanged.

5 0
3 years ago
Neutral atoms of argon, atomic number 18, have the same number of electrons as each of the following items except: Cl- S -2 K+ C
irinina [24]

Answer:

Ne.

Explanation:

Neutral argon, atomic number 18, has 18 electrons.

Cl⁻:

Neutral atom of Cl, atomic number 17, has 17 electrons.

When it gains electron and be Cl⁻, then it has 18 electrons a neutral Ar.

S²⁻:

Neutral atom of S, atomic number 16, has 16 electrons.

When it gains 2 electrons and be S²⁻, then it has 18 electrons a neutral Ar.

K⁺:

Neutral atom of K, atomic number 19, has 19 electrons.

When it losses electron and be K⁺, then it has 18 electrons a neutral Ar.

<em>Ca²⁺:</em>

Neutral atom of Ca, atomic number 20, has 20 electrons.

When it losses 2 electrons and be Ca²⁺, then it has 18 electrons a neutral Ar.

Ne:

It is a noble gas that has 10 electrons.

<em>So, the right choice is: Ne.</em>

<em></em>

3 0
3 years ago
Can 1750 mL of water dissolve 4.6 moles of Copper Sulfate CuSO4? _________ Why? / Why not?
Wewaii [24]

Answer:

  • <u>No, you cannot dissolve 4.6 moles of copper sulfate, CuSO₄, in 1750mL of water.</u>

Explanation:

This question is part of a Post-Lab exercise sheet.

Such sheet include the saturation concentrations for several salts.

The saturation concentration of Copper Sulfate, CuSO₄, indicated in the table is 1.380M.

That means that 1.380 moles of copper sulfate is the maximum amount that can be dissolved in one liter of solution.

Find the molar concentration for 4.6 moles of copper sulfate in 1,750 mL of water.

You need to assume that the volume of water (1750mL) is the volume of the solution. This is, that the 4.6 moles of copper sulfate have a negligible volume.

<u>1. Volume in liters:</u>

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<u />

<u>2. Molar concentration, molarity, M:</u>

  • M = number of moles of solute / volume of solution in liters

  • M = 4.6 moles / 1.75 liter = 2.6 M

Since the solution is saturated at 1.380M, you cannot reach the 2.6M concentration, meaning that you cannot dissolve 4.6 moles of copper sulfate, CuSO₄ in 1750mL of water.

8 0
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