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lana [24]
4 years ago
6

What are doldrums? Give an example.

Chemistry
2 answers:
qaws [65]4 years ago
4 0

Answer:

<em> Doldrums- a state or period of inactivity, stagnation, or depression.</em>

Example:

<em>Doldrums</em><em> is defined as a gloomy feeling, low spirits or a time of inactivity. An example of doldrums is being stuck in the house during a week long snow storm.</em>

dusya [7]4 years ago
4 0

Answer:

doldrums-a state or period of inactivity, stagnation, or depression.

Ex. being stuck in the house during a week long snow storm

Explanation:

xd

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The fuel used in many disposable lighters is liquid butane, C4H10. Butane has a molecular weight of 58.1 grams in one mole. How
viktelen [127]
The exact molecular mass for butane (C4H10) is 
12.0096*4+1.0079*10=58.1174  which is 58.1 to 3 significant figures.

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48.0384 : 58.1174

The mass of carbon in 2.50 grams of butane can be obtained by proportion, namely
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What is the empirical formula of a compound that contains 49.4% K, 20.3% S, and 30.3% by mass? A) KSO3 B) K2SO3 C) KSO2 D) KSO E
yulyashka [42]

Answer: K_2SO_3.

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of K = 49.4 g

Mass of S = 20.3 g

Mass of O = 30.3 g

Step 1 : convert given masses into moles.

Moles of K=\frac{\text{ given mass of K}}{\text{ molar mass of K}}= \frac{49.4g}{40g/mole}=1.23moles

Moles of S= \frac{\text{ given mass of S}}{\text{ molar mass of S}}= \frac{20.3g}{32g/mole}=0.63moles[/tex]

Moles of O = \frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{30.3g}{16g/mole}=1.89moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For K = \frac{1.23}{0.63}=2

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