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3 years ago

Which element can form an oxide with the general formula m2o or mo where m is a metal

Chemistry

1 answer:

boyakko [2]3 years ago

7 0

Answer:

Because the compound has a formula of M2O, the number of valence electrons of M should be 1. Therefore, (1) Group 1 is the correct answer because elements in group 1 have 1 valence electron.

Explanation:

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Which of the following molecules is diamagnetic according to molecular orbital theory? A) N₂⁺ B) N₂⁻ C) CN D) CN⁻

Gnom [1K]

Answer:

D) CN⁻

Explanation:

Hund's Rule of Maximum Multiplicity state that electrons go into degenerate orbitals of sub-levels (p,d, and f ) singly before pairing commences. Hund's rule is useful in determining the number of unpaired electrons in an atom. As such, it explains some magnetic properties of elements.

An element whose atoms or molecules contain unpaired electrons is paramagnetic. i.e., weakly attracted to substances in a magnetic field.

On the other hand, the element whose atoms or molecules are filled up with paired electrons is known as diamagnetic, i.e., not attracted by magnetic substances.

According to the molecular orbital theory, the diamagnetic molecule is CN⁻ because of the absence of unpaired electrons.

6 0

3 years ago

In NMR if a chemical shift(δ) is 211.5 ppm from the tetramethylsilane (TMS) standard and the spectrometer frequency is 556 MHz,

Vika [28.1K]

Answer:

The answer is: 11759 Hz

Explanation:

Given: Chemical shift: δ = 211.5 ppm, Spectrometer frequency = 556 MHz = 556 × 10⁶ Hz

In NMR spectroscopy, the chemical shift (δ), expressed in ppm, of a given nucleus is given by the equation:

$\delta (ppm) = \frac{Observed\,frequency (Hz)}{Frequency\,\, of\,\,the\,Spectrometer (MHz)} \times 10^{6}$

$\therefore Observed\,frequency (Hz)= \frac{\delta (ppm)\times Frequency\,\, of\,\,the\,Spectrometer (MHz)}{10^{6}}$

$Observed\,frequency= \frac{211.5 ppm \times 556 \times 10^{6} Hz}{10^{6}} = 11759 Hz$

<u>Therefore, the signal is at 11759 Hz from the TMS.</u>

6 0

3 years ago

_____ and _______ are necessary for rusting.

Anastaziya [24]

Answer :-

<h3>Water and Air are necessary for rusting .</h3>

3 0

2 years ago

Read 2 more answers

A bond exists between A and B in the compound AB. A and B are sharing electrons. The type of bond is

Jlenok [28]

Answer: Covalent Bond.

6 0

3 years ago

aseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 5.5 g of methane is

maksim [4K]

Answer:

There is 9.6 grams of CO2 produced

Explanation:

Step 1: Data given

Mass of methane = 5.50 grams

Molar mass of methane = 16.04 g/mol

Mass of oxygen = 13.9 grams

Molar mass of oxygen = 32.0 g/mol

Step 2: The reaction

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

Step 3: Calculate number of moles

Moles = mass / molar mass

Moles methane = 5.50 grams / 16.04 g/mol

Moles methane = 0.343 moles

Moles oxygen = 13.9 grams / 32.0 g/mol

Moles oxygen = 0.434 moles

For 1 mol CH4 we need 2 moles O2 to produce 1 mol CO2 and 2 moles H2O

O2 is the limiting reactant. It will completely react (0.434 moles).

There will react 0.434/2 = 0.217 moles CH4

There will remain 0.343-0.217 = 0.126 moles CH4

There will be produced 0.434 moles of H2O and

0.434/2 =0.217 moles of CO2

Step 4: Calculate mass of products

Mass = moles * molar mass

Mass CO2 = 0.217 moles ¨44.01 g/mol

Mass CO2 = 9.6 grams

Mass H2O = 0.434 moles * 18.02

Mass H2O = 7.8 grams

4 0

2 years ago