1-magma wich flows through channles to the surface will produce a???

Chemistry

1 answer:

AlladinOne [14]4 years ago

6 0

Answer:

thats one hint. but for 1 it will produce lava

Explanation:

number 2

some of the Earth's grandest mountains are composite volcanoes--sometimes called stratovolcanoes. They are typically steep-sided, symmetrical cones of large dimension built of alternating layers of lava flows, volcanic ash, cinders, blocks, and bombs and may rise as much as 8,000 feet above their bases

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A chemistry graduate student is given of a chlorous acid solution. Chlorous acid is a weak acid with . What mass of should the s

DerKrebs [107]

Answer:

11.31g NaClO₂

Explanation:

<em> Is given 250mL of a 1.60M chlorous acid HClO2 solution. Ka is 1.110x10⁻². What mass of NaClO₂ should the student dissolve in the HClO2 solution to turn it into a buffer with pH =1.45? </em>

It is possible to answer this question using Henderson-Hasselbalch equation:

pH = pKa + log₁₀ [A⁻] / [HA]

<em>Where pKa is -log Ka = 1.9547; [A⁻] is the concentration of the conjugate base (NaClO₂), [HA] the concentration of the weak acid</em>

You can change the concentration of the substance if you write the moles of the substances:

[Moles HClO₂] = 250mL = 0.25L×(1.60mol /L) = <em>0.40 moles HClO₂</em>

Replacing in H-H expression, as the pH you want is 1.45:

1.45 = 1.9547 + log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]

-0.5047 = log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]

<em>0.3128 = </em>[Moles NaClO₂] / [0.40 moles HClO₂]

0.1251 = Moles NaClO₂

As molar mass of NaClO₂ is 90.44g/mol, mass of 0.1251 moles of NaClO₂ is:

0.1251 moles NaClO₂ ₓ (90.44g / mol) =

<h3>11.31g NaClO₂</h3>

5 0

3 years ago

Determine a detailed mechanism for the chlorination of benzene using cl2 and fecl3.

RUDIKE [14]

Step (1):
Generation of electrophile: by the action of Lewis acid FeCl₃ on Cl₂ to serve as a source of Cl⁺ (Electrophile)
Step (2):
Addition of electrophile to form carbocation:
addition of electrophile to form C-Cl bond and form carbocation which is stabilized by resonance.
Step (3):
Loss of proton to re-form the aromatic ring by the action of FeCl₄⁻ which removes proton from carbon containing Cl and forming the aromatic ring again