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ss7ja [257]
3 years ago
15

La valencia que utiliza el hidrógeno al formar un hidruro metálico es

Chemistry
1 answer:
Serjik [45]3 years ago
4 0

Answer: Esta tendencia es tan regular que el poder de combinación, o valencia, de un elemento se definió una vez como el número de átomos de hidrógeno unidos al elemento en su hidruro. El hidrógeno es el único elemento que forma compuestos en los que los electrones de valencia están en la capa n = 1.

Explanation:

¡Espero que esto ayude!

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List two general properties of molecular compounds
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Low melting points and boiling points. ...Low enthalpies of fusion and vaporization These properties are usually one or two orders of magnitude smaller than they are for ionic compounds.Soft or brittle solid forms. ...Poor electrical and thermal conductivity.
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3 years ago
What will be the new volume if 125 mL of He gas at 100 degree Celsius and .979 atm is cooled to 26 degree Celsius and the pressu
ycow [4]
Use Boyle's Law of Pressure: P1 x V1 = P2 x V2. Givens: P1=0.9 atm V1=                4 P2= 0.9 atm Find: V2 Equation: 0.9 atm x 4 x 4 L = 0.20 atm x V2Solve: 36 atmL= 0.20 atm x V2 18 : = V2 Short answer: The new volume is 104 ml. 
6 0
3 years ago
What is the maximum number of p orbitals in any single energy level in an atom?
nordsb [41]
Answer is 3
s -1
p - 3
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7 0
3 years ago
Give the structure that corresponds to the following molecular formula and 1H NMR spectrum: C4H10O2: δ 1.36 (3H, d, J = 5.5 Hz);
kondor19780726 [428]

Answer:

For the determination of a structure through its NMR it is necessary to know its molecular formula as well as the delta values, its coupling and the shift of each signal.

The separation produced is called coupling constant J and is measured in Hz. If the split is produced by two equal protons (equal J) a triple signal known as triplet is produced and if produced by three equal protons, the signal is quadruple and is known as quadruplet. The magnitude of the coupling is varied, depending on the relative disposition of the coupled protons (elevations that separate them, arrangement, spatial arrangement)

OH CH2 CH2 CH2 CH2 OH

(A)   (B)   (C)

1,4-butanediol

In the case of the molecule to study the signal at 1.36 shows a doublet, which corresponds to the hydrogen (C), is split in two for each different proton on the same carbon or on neighboring carbons.

At 3.32 ocurrs a singlet, wich belong to hidrogen (B). The last signal is a quintet, at 4.63 belonging to the hydrogen (C)

Explanation

Nuclear magnetic resonance NMR is a physical phenomenon based on the mechanical-quantum properties of atomic nuclei. NMR also refers to the family of scientific methods that explore this phenomenon to study molecules, macromolecules, as well as tissues and whole organisms.

NMR takes advantage of the fact that atomic nuclei resonate at a frequency directly proportional to the force of a magnetic field exerted, in accordance with the Larmor precession frequency equation, to subsequently disturb this alignment with the use of an alternating magnetic field, of orthogonal orientation.

The behavior of the nuclei in the magnetic field can be influenced in multiple ways, to give different types of information, but the basic information obtained is:

  • Frequency at which each particular nuclei comes out, displacement.
  • Number of nucleis of each type, integral.
  • Number and arrangement of nearby nuclei, multiplicity.
6 0
2 years ago
Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb(NO3)2(aq)+2
Setler79 [48]

Answer:

a) volume of ammonium iodide required =349 mL

b) the moles of lead iodide formed = 0.0436 mol

Explanation:

The reaction is:

Pb(NO_{3})_{2}+2NH_{4}I -->PbI_{2}+2NH_{4}NO_{3}

It shows that one mole of lead nitrate will react with two moles of ammonium iodide to give one mole of lead iodide.

Let us calculate the moles of lead nitrate taken in the solution.

Moles=molarityX volume (L)

Moles of lead nitrate = 0.360 X 0.121 =0.0436 mol

the moles of ammonium iodide required = 2 X0.0436 = 0.0872 mol

The volume of ammonium iodide required will be:

volume=\frac{moles}{molarity}=\frac{0.0872}{0.250}=0.349L=349mL

the moles of lead iodide formed = moles of lead nitrate taken = 0.0436 mol

7 0
3 years ago
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