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For an endothermic reaction, which of the following conditions must be true for all products

mamaluj [8]

<span>the change in entropy is positive and change in enthalpy is negative.and potential energy is lower.and the reason is materials in nature naturaly want to get stabler and more chaotic.

for the second question there is a big formula.Equilibrium θ=mcθ1(water)+mcθ1(metal) /mc(water)+mc(metal) just put numbers and it is easy to solve.equlibrium temperture is 23.5</span>

8 0

3 years ago

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Mechanism of Allan's with H2SO4, CH3CH2OH

lana66690 [7]

<span> that the reaction is preferably called an elimination, rather than a dehydration, although I personally like the term dehydration because it emphasizes the strong dehydrating power of H2SO4 that is unfotunately forgotten too often by undergraduates. I like to think of it as H2SO4 has such a strong affinity for water that it literally pulls water molecules from organics.</span>

3 0

3 years ago

A 13.0 l sample of helium gas has a pressure of 26.0 atm. what volume would this gas occupy at 8.50 atm? assume ideal behavior.

Elena L [17]

The game crashes after every single episode I play it

5 0

3 years ago

A teacher had two clear liquids, Liquid A and Liquid B. The teacher filled one beaker with Liquid A, and then added Liquid B, dr

mrs_skeptik [129]

Answer:its a precipitate!

Explanation:

<h2>Its a solid that forms as the result of a chemical reaction </h2>

3 0

3 years ago

Find the pH of a 0.010 M HNO2 solution.

lidiya [134]

Data:
Molar Mass of HNO2
H = 1*1 = 1 amu
N = 1*14 = 14 amu
O = 3*16 = 48 amu
------------------------
Molar Mass of HNO2 = 1 + 14 + 48 = 63 g/mol

M (molarity) = 0.010 M (Mol/L)

Now, since the Molarity and ionization constant has been supplied, we will find the degree of ionization, let us see:
M (molarity) = 0.010 M (Mol/L)
Use: Ka (ionization constant) = $5.0*10^{-4}$
$\alpha^2 (degree\:of\:ionization) = ?$

$Ka = M * \alpha^2$
$5.0*10^{-4} = 0.010* \alpha^2$
$0.010\alpha^2 = 5.0*10^{-4}$
$\alpha^2 = \frac{5.0*10^{-4}}{0.010}$
$\alpha^2\approx500*10^{-4}$

$\alpha\approx\sqrt{500*10^{-4}}$
$\alpha \approx 2.23*10^{-3}$

Now, we will calculate the amount of Hydronium [H3O+] in nitrous acid (HNO2), multiply the acid molarity by the degree of ionization, we will have:

$[ H_{3} O^+] = M* \alpha$
$[ H_{3} O^+] = 0.010* 2.23*10^{-3}$
$[ H_{3} O^+] \approx 0.0223*10^{-3}$
$[ H_{3} O^+] \approx 2.23*10^{-5} \:mol/L$

And finally, we will use the data found and put in the logarithmic equation of the PH, thus:

Data:
log10(2.23) ≈ 0.34
pH = ?
$[ H_{3} O^+] = 2.23*10^{-5}$

Formula:
$pH = - log[H_{3} O^+]$

Solving:
$pH = - log[H_{3} O^+]$
$pH = -log2.23*10^{-5}$
$pH = 5 - log2.23$
$pH = 5 - 0.34$
$\boxed{\boxed{pH = 4.66}}\end{array}}\qquad\quad\checkmark$

Note:. The pH <7, then we have an acidic solution.

6 0

4 years ago