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2 years ago

1. Except for special cases such as peroxides, oxygen has an oxidation number of _______ in compounds. A. +2 B. –1 C. +1 D. –2

Chemistry

1 answer:

egoroff_w [7]2 years ago

7 0

Answer:

The Typical Oxidation number of Oxygen is –2.

And as the question said... its Oxidation changes to –1 when it is in Peroxides and –½ when its in SuperOxides.

Correct Answer : Option D.

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Covalent bonds take place between..

umka2103 [35]

Answer:

Covalent Bonds are formed when two non-metals share electrons

Hope this helps

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3 years ago

The difference in an area with high concentration and an area with low concentration is called.

Varvara68 [4.7K]

The difference in an area with high concentration and an area with low concentration is called the concentration gradient.

<h3>What is Concentration Gradient ?</h3>

A concentration gradient occurs when the concentration of particles is higher in one area than another.

In passive transport, particles will diffuse down a concentration gradient, from areas of higher concentration to areas of lower concentration, until they are evenly spaced.

This difference in an area with high concentration and an area with low concentration is called the concentration gradient.

Learn more about diffusion here ;

brainly.com/question/24746577

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5 0

2 years ago

_______ is the classic structural formula for all the addictive narcotics

oee [108]

The answer to this statement is codein. Codein, or 3-Methylmorphine by its IUPAC name, is an opiate used as pain reliever and suppressor of coughs. Its structural formula is shown in the picture. The patient's dosage of these narcotics should be strictly prescribed by the doctor. When patients take this, they feel euphoria, hence, they tend to crave for that feeling once it's gone. Too much dosage of this drug would lead to addiction. Examples of drugs with this structural formula are Cotabflu, Nalex AC, T-Koff and Pediatuss.

7 0

3 years ago

What is the energy of an electromagnetic wave with a frequency of 8 x 10^12

omeli [17]

Answer:

Energy, $E=5.3\times 10^{-21}\ J$

Explanation:

It is required to find the energy of an electromagnetic wave with a frequency of $8\times 10^{12}\ Hz$ . The energy of a wave in terms of its frequency is given by :

$E=h\nu$

$\nu$ = frequency of em wave

$E=6.63\times 10^{-34}\times 8\times 10^{12}\\\\E=5.3\times 10^{-21}\ J$

So, the energy of an electromagnetic wave is $5.3\times 10^{-21}\ J$

4 0

3 years ago

N₂O(g) + 3 H₂(g) N₂H4(1) + H₂O(1) AH = -317 kJ/mol

docker41 [41]

Answer:

Explanation:

Recall that ΔH is the sum of the heats of formation of the products minus the heat of formation of the reactants multiplied by their respective coefficients. That is:

$\displaystyle \Delta H^\circ_{rxn} = \sum \Delta H^\circ_{f} \left(\text{Products}\right) - \sum \Delta H^\circ_{f} \left(\text{Reactants}\right)$

Therefore, from the chemical equation, we have that:

$\displaystyle \begin{aligned} (-317\text{ kJ/mol}) = \left[\Delta H^\circ_f \text{ N$_2$H$_4$} + \Delta H^\circ_f \text{ H$_2$O} \right] -\left[3 \Delta H^\circ_f \text{ H$_2$}+\Delta H^\circ_f \text{ N$_2$O}\right] \end{aligned}$

Remember that the heat of formation of pure elements (e.g. H₂) are zero. Substitute in known values and solve for hydrazine:

$\displaystyle \begin{aligned} (-317\text{ kJ/mol}) & = \left[ \Delta H^\circ _f \text{ N$_2$H$_4$} + (-285.8\text{ kJ/mol})\right] -\left[ 3(0) + (82.1\text{ kJ/mol})\right] \\ \\ \Delta H^\circ _f \text{ N$_2$H$_4$} & = (-317 + 285.8 + 82.1)\text{ kJ/mol} \\ \\ & = 50.9\text{ kJ/mol} \end{aligned}$

In conclusion, our answer is A.

5 0

2 years ago