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3 years ago

4. Nitrogen and oxygen gases react to form dinitrogen monoxide gas (N2O). What volume of O, is

Chemistry

1 answer:

solmaris [256]3 years ago

8 0

Answer:

no be t ha t s had ice

Explanation:

uah an

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givi [52]

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3 years ago

6th grade science . :((

Nadusha1986 [10]

Answer:

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Explanation:

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3 years ago

Identify the Lewis acid and Lewis base from among the reactants in each of the following equations.

Alex

Answer:

Part A

Ag+ is the Lewis acid and NH3 is the Lewis base.

Part B

AlBr3 is the Lewis acid and NH3 is the Lewis base.

Part C

AlCl3 is the Lewis acid and Cl− is the Lewis base.

Explanation:

A Lewis acid is any specie that accepts a lone pair of electrons. Ag^+, AlBr3 and AlCl3 all accepted lone pairs of electrons according to the three chemical reaction equations shown. Hence, they are Lewis acids.

A Lewis base donates a lone pair of electrons. They include neutral molecules having lone pair of electrons such as NH3 or negative ions such as Cl- .

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3 years ago

Describe how nucleic acid basis pair up

castortr0y [4]

The rules of base pairing (or nucleotide pairing) are: A with T: the purine adenine (A) always pairs with the pyrimidine thymine (T) C with G: the pyrimidine cytosine (C) always pairs with the purine guanine (G)

The nucleotides in a base pair are complementary which means their shape allows them to bond together with hydrogen bonds. The A-T pair forms two hydrogen bonds. The C-G pair forms three. The hydrogen bonding between complementary bases holds the two strands of DNA together.

7 0

3 years ago

The solubility of o2 at 20c is 1.38 x10^-3. the partial presure of o2 in the air at sea level is 0.27 atm. using henery;s law, c

netineya [11]

<u>Answer:</u> The solubility of oxygen at 682 torr is $4.58\times 10^{-3}M$

<u>Explanation:</u>

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{A}=K_H\times p_{A}$

Or,

$\frac{C_{1}}{C_{2}}=\frac{p_{1}}{p_2}$

where,

$C_1\text{ and }p_1$ are the initial concentration and partial pressure of oxygen gas

$C_2\text{ and }p_2$ are the final concentration and partial pressure of oxygen gas

We are given:

Conversion factor used: 1 atm = 760 torr

$C_1=1.38\times 10^{-3}M\\p_1=0.27atm\\C_2=?\\p_2=682torr=0.897atm$

Putting values in above equation, we get:

$\frac{1.38\times 10^{-3}}{C_2}=\frac{0.27atm}{0.897atm}\\\\C_2=\frac{1.38\times 10^{-3}\times 0.897atm}{0.27atm}=4.58\times 10^{-3}M$

Hence, the solubility of oxygen gas at 628 torr is $4.58\times 10^{-3}M$

4 0

3 years ago