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Charra [1.4K]
2 years ago
12

What is the pH of an aqueous solution of HCl with a hydrogen ion

Chemistry
1 answer:
yulyashka [42]2 years ago
6 0

Answer:

pH = 5.17

Explanation:

pH = -log[H+] = -log(6.7 x 10^-6) = 5.17

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B) water has a bent shape, this would be correct given that VSEPR theory and molecular geometry classify this as a polar shaped compound
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A sample of the chiral molecule limonene is 62% enantiopure. what percentage of each enantiomer is present? what is the percent
ikadub [295]

The mixture contains 62 % one isomer and 38 % the enantiomer.

Let’s say that the mixture contains 62 % of the (<em>R</em>)-isomer.

Then % (<em>S</em>) = 100 % -62 % = 38 %

ee = %  (<em>R</em>) - % (<em>S</em>) = 62 % -38 % = 24 %


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3 years ago
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PLEASE HELP<br> two ways chemistry affects you in your daily life?
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3 years ago
According to markovnikov's rule of the electrophilic addition to an alkene, the electrophile, usually a proton, is more likely t
stealth61 [152]

According to markovnikov's rule of the electrophilic addition to an alkene, the electrophile, usually a proton, is more likely to add to the less-substituted carbon in a double bond.

With additional substituents present in this configuration, the intermediate carbocation is stabilised by being located on the more-substituted carbon.

The nucleophile will then end up in a double bond on the more-substituted carbon in a reaction that follows Markovnikov's rule.The outcome of some addition reactions is described by Markovnikov's rule or Markownikoff's rule in organic chemistry. Vladimir Markovnikov, a Russian scientist, created the rule in 1870.

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8 0
1 year ago
Given the chemical equation: 2 Pb + O2 → 2 PbO, if 51.8 grams of Pb are formed in this reaction, then 8.00 grams of O2 must have
Nutka1998 [239]

Answer:

If 51.8 of Pb is reacting, it will require 4.00 g of O2

If 51.8 g of PbO is formed, it will require 3.47 g of O2.

Explanation:

Equation of the reaction:

2 Pb + O2 → 2 PbO

From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO

Molar mass of Pb = 207 g

Molar mass of O2 = 32 g

Molar mass of PbO = 207 + 32 = 239 g

Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO

= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO

Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.

If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2

7 0
3 years ago
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