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3 years ago

What is green chemistry? How is it important?

Chemistry

1 answer:

Ronch [10]3 years ago

5 0

Called sustainable chemistry

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The reaction C4H8(g)⟶2C2H4(g) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ/mol.262 kJ/mol. At 600.0 K,600.0 K, the rate c

crimeas [40]

Answer : The rate constant at 785.0 K is, $1.45\times 10^{-2}s^{-1}$

Explanation :

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

or,

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = rate constant at $600.0K$ = $6.1\times 10^{-8}s^{-1}$

$K_2$ = rate constant at $785.0K$ = ?

$Ea$ = activation energy for the reaction = 262 kJ/mole = 262000 J/mole

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = $600.0K$

$T_2$ = final temperature = $785.0K$

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

$\log (\frac{K_2}{6.1\times 10^{-8}s^{-1}})=\frac{262000J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{785.0K}]$

$K_2=1.45\times 10^{-2}s^{-1}$

Therefore, the rate constant at 785.0 K is, $1.45\times 10^{-2}s^{-1}$

7 0

4 years ago

Use the periodic table to match each of the following element symbols to its name, atomic mass, or atomic number

uysha [10]

Answer:

Explanation:

From the periodic table, the element designated as Sn is tin. Sn is derieved from a latin name of the metal called Stanum

Selenium Se is a group 6 element. It belongs to the same group with oxygen, sulfur and tellurium. This substance has an atomic weight of 78.96g/mol

Oxygen is a group 6 element with an atomic mass of 16. The atomic mass is the number of protons in the nucleus of this atom.

5 0

4 years ago

Read 2 more answers

What is the overall voltage for the nonspontaneous redox reaction involving

Ann [662]

Answer:

Answer D => E°(Mg°/Cu⁺²) = 0.34 + 2.37 = 2.71v

Explanation:

(Oxidation) => Mg°(s) => Mg⁺²(aq) + 2e⁻ E°(Mg°/Mg⁺²) = -2.37 v

(Reduction) => Cu⁺²(aq) + 2e⁻ => Cu°(s) E°(Cu⁺²/Cu°) = +0.34 v

________________________________________________

Net Rxn => Mg°(s) + Cu⁺²(aq) => Mg⁺²(aq) + Cu°(s)

Std Cell Potential (25°C/1Atm) = E°(Redn) = E°(Oxidn) = +0.34v - (-2.37v)

= 0.34v + 2.37v = 2.72v

6 0

3 years ago

Read 2 more answers

What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb

Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

$(\text{CH}_3)_3\text{N}$ in this question acts as a weak base. As seen in the equation in the question, $(\text{CH}_3)_3\text{N}$ produces $\text{OH}^{-}$ rather than $\text{H}^{+}$ when it dissolves in water. The concentration of $\text{OH}^{-}$ will likely be more useful than that of $\text{H}^{+}$ for the calculations here.

Finding the value of $[\text{OH}^{-}]$ from pH:

Assume that $\text{pK}_w = 14$ ,

$\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}$ .

$[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}$ .

Solve for $[(\text{CH}_3)_3\text{N}]_\text{initial}$ :

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}$

Note that water isn't part of this expression.

The value of Kb is quite small. The change in $(\text{CH}_3)_3\text{N}$ is nearly negligible once it dissolves. In other words,

$[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}$ .

Also, for each mole of $\text{OH}^{-}$ produced, one mole of $(\text{CH}_3)_3\text{NH}^{+}$ was also produced. The solution started with a small amount of either species. As a result,

$[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}$ .

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3}$ ,

$[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b}$ ,

$[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}$ .

8 0

4 years ago

calculate the volume od a CO2 cartridge that has a pressure of 850 PSI at a temperature of 21ºc the cartridge contains 0.273 mol

gladu [14]

Answer is: volume of CO₂ is 0,113 dm³.
Ideal gas law = pV = nRT.
p = 850 PSI = 5860543,6992 Pa.
Psi <span>is the abbreviation of pound per square inch.
T = 21</span>°C = 294,15 K.
n = 0,273 mol.
R = 8,314 J/K·mol.
V = nRT ÷ p
V = 0,273 mol · 8,314 J/K·mol · 294,15 K ÷ 5860543,6992 Pa.
V = 0,00011 m³ = 0,113 dm³.

3 0

3 years ago