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Rama09 [41]
3 years ago
8

Deuterium is an isotope of hydrogen. It has a mass number of two. Which describes a deuterium atom? A. a nucleus of one proton a

nd one neutron, which is orbited by one electron B. a nucleus of one proton, which is orbited by one neutron and one electron C. a nucleus of one neutron, which is orbited by one electron D. a nucleus of two neutrons which is orbited by two electrons
Chemistry
2 answers:
Stells [14]3 years ago
6 0

also known as heavy hydrogen) is one of two stable isotopes of hydrogen (the other being protium, or hydrogen-1). The nucleus of deuterium, called a deuteron, contains one proton and one neutron, whereas the far more common protium has no neutron in the nucleus.

arsen [322]3 years ago
4 0

Answer:

A. a nucleus of one proton and one neutron, which is orbited by one electron.

Explanation:

Isotopes have the same atomic number and different mass number. The atomic number is the numbers of protons in the nucleus and defines each element while the mass number is the sum of protons and neutrons in the nucleus.

An hydrogen atom with a mass number of two means that its nucleus has one proton and one neutron because the atomic number of hydrogen is 1, otherwise it wouldn’t be hydrogen, so it has 1 proton and to complete the mass number of two it must have 1 neutron.

On the other hand, the atomic model says that the nucleus is orbited by electrons. The number of electrons orbiting the nucleus is given by the atomic number. Then an atomic number of 1 means that 1 electron orbits the nucleus with one proton.

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Glycolic acid, which is a monoprotic acid and a constituent in sugar cane, has a pKa of 3.9. A 25.0 mL solution of glycolic acid
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Answer:

pH of resulting solution = 7.98

Explanation:

The balanced equation  

HA + NaOH - Na+ + A- + H2O

Number of moles of A = Number of moles of HA  = Number of moles of NaOH

= 35.8/1000 * 0.020 = 0.000716 mol

Initial concentration of A = 0.000716/0.0608 = 0.01178 M

pKb = 14 – pKa = 14 -3.9 = 10.1

Kb = 10^{-Kb} = 10^{-10.1} = 7.943 * 10^-11

Kb = [HA][OH-]/[A-]

Kb = a^2/(0.01178 -a) = 7.943 * 10^-11

a^2 + 7.943 * 10^-11 a – 9.357 * 10^-13 = 0

a = 9.673 * 10^-7

OH- = a = 9.673 * 10^-7 M

pOH = -log [OH-] = -log (9.673 * 10^-7) = 6.02

pH = 14-6.02 = 7.98

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Part B Redox titrations are used to determine the amounts of oxidizing and reducing agents in solution. For example, a solution
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Answer:

0,508g of H₂O₂

Explanation:

For the reaction:

2KMnO₄(aq) + H₂O₂(aq) + 3H₂SO₄(aq) → 3O₂(g) + 2MnSO₄(aq) + K₂SO₄(aq) + 4H₂O(l)

2 moles of KMnO₄ react with 1 mol of H₂O₂.

In the titration, moles of KMnO₄ required were:

1,68M×0,0178L = 0,0299 moles of KMnO₄. Moles of H₂O₂ are:

0,0299 moles of KMnO₄×\frac{1molH_{2}O_{2}}{2molKMnO_{4}} = 0,01495 moles of H₂O₂. As molar mass of H₂O₂ is 34,01g/mol, mass of H₂O₂ was dissolved is:

0,01495 moles of H₂O₂×\frac{34,01g}{1molH_{2}O_{2}} = <em>0,508g of H₂O₂</em>

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