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2 years ago

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What are the common properties of the following element groups:

1 answer:

son4ous [18]2 years ago

8 0

Answer:

Alkali and alkaline-earth cations are energetically stable with an empty valence shell. Atoms of alkali and alkaline-earth elements achieve octets after losing one (alkali) or two (alkaline-earth) electrons such that they minimize the potential energy of the system.

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Binary Covalent (molecular) Compounds

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The hydrochloride form of cocaine has a solubility of 1.00 g in 0.400 mL water. Calculate the molarity of a saturated solution o

Tanya [424]

Answer:

The molarity of the solution is 7.4 mol/L

Explanation:

From the question above

0.400 ml of water contains 1.00 g of hydrochloride form of cocaine

Therefore 1000 ml of water will contain x g of hydrochloride form of cocaine

x = 1000 / 0.400

x = 2500 g

2500g of hydrochloride form of cocaine is present in 1000 ml of water.

Mole of hydrochloride form of cocaine = mass /molar mass of hydrochloride

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3 years ago

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The true absorbance for a 1.0 x 10 −5 M solution is 0.7526. If the percentage stray light for a spectrophotometer is 0.56%, calc

Korvikt [17]

Answer:

The percentage deviation is $\Delta M = 1.87$ %

Explanation:

From the question we are told that

The concentration is of the solution is $C = 1.0*10^{-5} M$

The true absorbance A = 0.7526

The percentage of transmittance due to stray light $z = 0.56$ % $=\frac{0.56}{100} = 0.0056$

Generally Absorbance is mathematically represented as

$A = -log T$

Where T is the percentage of true transmittance

Substituting value

$0.7526 = - log T$

$T = 10^{-0.7526}$

$= 0.177$

$= 17.7$ %

The Apparent absorbance is mathematically represented

$A_p = -log (T +z)$

Substituting values

$A_p = -log(0.177 + 0.0056)$

$= -log(0.1826)$

= 0.7385

The percentage by which apparent absorbance deviates from known absorbance is mathematically evaluated as

$\Delta A = \frac{A -A_p}{A} * \frac{100}{1}$

$= \frac{0.7526 - 0.7385}{0.7526} * \frac{100}{1}$

$\Delta A = 1.87$ %

Since Absorbance varies directly with concentration the percentage deviation of the apparent concentration from know concentration is

$\Delta M = 1.87$ %

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3 years ago