9. Mangnesium can form compounds by acting with different oxidation numbers (2-, 3+, 4+, 6+, 7+). Argue your answer.

If chemical weathering occurs when minerals inside of the rock changes chemically. Will it change the rate of chemical weatherin

Mariulka [41]

Increase of rainfall will accelerate chemical weathering. Especially if the runoff is slightly acidic.

7 0

3 years ago

Acid-catalyzed dehydration of secondary and tertiary alcohols proceeds through an E1 mechanism. The first step is the protonatio

Over [174]

Answer:

Most substituted alkene is produced as a major product

Explanation:

Dehydration of 3-methyl-2-butanol proceeds through E1 mechanism to form alkenes.
Most substituted alkene is produced as major product because of presence of highest number of hyperconjugative hydrogen atoms corresponding to the produced double bond (Saytzeff product).
Here, a H-shift also occurs in one of the intermediate step during dehydration to produce more stable tertiary carbocation.
Reaction mechanism has been shown below.

4 0

3 years ago

(a.) a 0.7549g sample of the compound burns in o2(g) to produce 1.9061g of co2(g) and 0.3370g of h2o(g).

Natali [406]

The individual mass of C, H and O in given sample are 0.5196 g, 0.0374 g and 0.1979 g respectively.

Moles of CO2 formed can be calculated as

= Mass of CO2 / Molar mass of CO2

= 1.9061 / 44 = 0.0433 moles

<h3>Calculation of no. of moles of carbon</h3>

Now, moles of C which is present in one mole of CO2 = 1 mole

Moles of C in 0.0433 moles of CO2 = 0.0433 moles

As we know that, molar mass of C = 12 g / mol

Mass of C in 0.7549 g of given sample can be calculated as

= 0.0433 × 12 =0.5196 g

Mass of H2O formed = 0.3370 g

Similarly, Molar Mass of H2O = 18 g / mol

Moles of H2O = 0.3370 / 18 = 0.0187 moles

Moles of H present in 1 mole of H2O = 2 moles

Moles of H present in 0.0187 mole of H2O = 2 × 0.0187 = 0.0374 moles

Molar mass of H = 1 g / mol

Mass of H contained in 0.7549 g of sample = 1 × 0.0374= 0.0374 g

Mass of O in 0.7549 g sample can be calculated as

= 0.7549 – [(Mass of C ) + (Mass of H) ]

= 0.7549 – [ (0.5196) + (0.0374) ]

= 0.1979 g

Thus, we calculated that the individual mass of C, H and O in given sample are 0.5196 g, 0.0374 g and 0.1979 g respectively.

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DISCLAIMER: THE above question is incomplete. Complete question is given below:

A 0.7549g sample of the compound burns in o2(g) to produce 1.9061g of co2(g) and 0.3370g of h2o(g). Calculate the individual mass of C, H and O in the given sample.

4 0

2 years ago

The equilibrium constant k for the synthesis of ammonia is 6.8x105 at 298 k. what will k be for the reaction at 375 k?

spayn [35]

Answer is: K be for the reaction at 375 K is 326.
Chemical reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g); ΔH = -92,22 kJ/mol.
T₁ = 298 K
T₂ = 375 K
ΔH = -92,22 kJ/mol = -92220 J/mol.
R = 8,314 J/K·mol.
K₁ = 6,8·10⁵.
K₂ = ?The van’t Hoff equation: ln(K₂/K₁) = -ΔH/R(1/T₂ - 1/T₁).
ln(K₂/6,8·10⁵) = 92220 J/mol / 8,314 J/K·mol (1/375K - 1/298K).
ln(K₂/6,8·10⁵) = 11092,13 · (0,00266 - 0,00335).
ln(K₂/6,8·10⁵) = -7,64.
K₂/680000= 0,00048
K₂ = 326,4.

6 0

3 years ago

Read 2 more answers

"You measure 48.9 mL of a solution of sulfuric acid with an unknown concentration, and carefully titrate this solution using a 1

Blizzard [7]

Answer:

C= 0.532M

Explanation:

The equation of reaction is

H2SO4 + 2KOH = K2SO4+ H2O

nA= 1, nB= 2, CA= ?, VA= 48.9ml, CB= 1.5M, VB= 34.7ml

Applying

CAVA/CBVB = nA/nB

(CA× 48.9)/(1.5×34.7)= 1/2

Simplify

CA= 0.532M

6 0

3 years ago