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4 years ago

When did Pangaea begin to break apart?

Chemistry

2 answers:

LUCKY_DIMON [66]4 years ago

7 0

About 250 years ago ΩαΨ

sergeinik [125]4 years ago

5 0

Answer:

Explanation:The most famous supercontinent had a good run, though — Pangaea didn’t really start to break up until the Early-Middle Jurassic Period (175 million years ago). The scraps of the once proud island of prehistoric civilization eventually settled into the seven continents we know today around 140 million years after that.

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How many valence electrons are in the electron-dot structures for the elements in group 3a(13)?

frosja888 [35]

Three valence electrons

3 0

3 years ago

when this chemical equation is correctly balanced what is the coefficient of the nal molecule I2+Na2S2O3+Nal+Na2S4O6

Kitty [74]

I₂ + 2Na₂S₂O₃ → 2NaI + Na₂S₄O₆

k(NaI)=2

4 0

3 years ago

Kc for the reaction of hydrogen and iodine to produce hydrogen iodide, H2(g) I2(g) ⇌ 2HI(g) is 54.3 at 430°C. Determine the init

Jlenok [28]

Answer : The initial concentration of HI and concentration of $HI$ at equilibrium is, 0.27 M and 0.386 M respectively.

Solution : Given,

Initial concentration of $H_2$ and $I_2$ = 0.11 M

Concentration of $H_2$ and $I_2$ at equilibrium = 0.052 M

Let the initial concentration of HI be, C

The given equilibrium reaction is,

$H_2(g)+I_2(g)\rightleftharpoons 2HI(g)$

Initially 0.11 0.11 C

At equilibrium (0.11-x) (0.11-x) (C+2x)

As we are given that:

Concentration of $H_2$ and $I_2$ at equilibrium = 0.052 M = (0.11-x)

0.11 - x = 0.052

x = 0.11 - 0.052

x = 0.058 M

The expression of $K_c$ will be,

$K_c=\frac{[HI]^2}{[H_2][I_2]}$

$54.3=\frac{(C+2(0.058))^2}{(0.052)\times (0.052)}$

By solving the terms, we get:

C = 0.27 M

Thus, initial concentration of HI = C = 0.27 M

Thus, the concentration of $HI$ at equilibrium = (C+2x) = 0.27 + 2(0.058) = 0.386 M

3 0

4 years ago

Consider the reaction 3Fe2O3(s) + H2(g)2Fe3O4(s) + H2O(g) Using standard thermodynamic data at 298K, calculate the entropy chang

balandron [24]

Answer:

the entropy change for the surroundings when 1.68 moles of Fe2O3(s) react at standard conditions = 49.73 J/K.

Explanation:

3Fe2O3(s) + H2(g)-----------2Fe3O4(s) + H2O(g)

∆S°rxn = n x sum of ∆S° products - n x sum of ∆S° reactants

∆S°rxn = [2x∆S°Fe3O4(s) + ∆S°H2O(g)] - [3x∆S°Fe2O3(s) + ∆S°H2(g)]

∆S°rxn = [(2x146.44)+(188.72)] - [(3x87.40)+(130.59)] J/K

∆S°rxn = (481.6 - 392.79) J/K =88.81J/K.

For 3 moles of Fe2O3 react, ∆S° =88.81 J/K,

then for 1.68 moles Fe2O3 react, ∆S° = (1.68 mol x 88.81 J/K)/(3 mol) = 49.73 J/K the entropy change for the surroundings when 1.68 moles of Fe2O3(s) react at standard conditions.

5 0

3 years ago

Why is it good to use blue and red litmus for the same solution?

Neko [114]

Under acidic conditions, the solution is red, and under alkaline conditions, the solution is blue. Chemical reactions other than acid-base can also cause a color change to litmus paper. For instance, chlorine gas turns blue litmus paper white – the litmus dye is bleached, because of presence of hypochlorite ions.

6 0

3 years ago