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tangare [24]
3 years ago
8

The voltage in a flashlight bulb is 6.0 volts and has a resistance of 24 ohms. Calculate the current

Chemistry
2 answers:
taurus [48]3 years ago
5 0

Answer:

Thanks for the points

Explanation:

Genrish500 [490]3 years ago
3 0
The answer will be .25 because R=v/I so it would be 24=0.6/I you divide it and you’ll get .25 or 0.25
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In the absence of other information, you can predict that an organic molecule with the empirical formula of c12h24o12 is a_____.
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The empirical formula of c12h24o12 is a carbohydrate.

<h3>Carbohydrate</h3>

A carbohydrate is a biomolecule made up of carbon (C), hydrogen (H), and oxygen (O) atoms, often with a hydrogen-oxygen atom ratio of 2:1 (as in water), and so having the empirical formula Cm(H2O)n (where m may or may not be different from n). All molecules that meet this exact stoichiometric criterion are not, however, automatically categorized as being carbohydrates.

The term is most frequently used in biochemistry, where it is used as a synonym for saccharide, a class of compounds that includes sugars, starches, and cellulose. The four chemical categories of saccharides are monosaccharides, disaccharides, oligosaccharides, and polysaccharides. The smallest carbohydrates, monosaccharides and disaccharides, are sometimes referred to as sugars.

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1 year ago
name two gases other than nitrogen, oxygen water, argon, and carbon dioxide that are found in the atmosphere​
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2 years ago
Be sure to answer all parts. A mixture of CO2 and Kr weighs 47.9 g and exerts a pressure of 0.751 atm in its container. Since Kr
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Answer:a)  Mass of CO_2 in mixture = 26.3 grams

b) Mass of Kr in mixture that can be recovered= 21.5 grams

Explanation:

Given :

Total Pressure = pressure of krypton + pressure of carbon dioxide = 0.751 atm

pressure of krypton = 0.231 atm

Thus pressure of carbon dioxide = 0.751 - 0.231 =0.52 atm

As we know:

p=x\times P

where,

p = partial pressure

P = total pressure = 0.751 atm

x = mole fraction

For CO_2

x_{CO_2}=\frac{p_{CO_2}}{P}=\frac{0.52}{0.751}=0.70

{\text {Mass of} CO_2}=moles\times {\text {Molar mass}}=0.70\times 44=30.8g

For Kr

x_{Kr}=\frac{p_{Kr}}{P}=\frac{0.231}{0.751}=0.30

{\text {Mass of} Kr}=moles\times {\text {Molar mass}}=0.30\times 84=25.2g

Total mass = Mass of CO_2 + Mass of krypton = 30.8 + 25.2 = 56 g

Percentage of CO_2=\frac{30.8}{56}\times 100=55\%

a) Thus Mass of CO_2 in mixture =\frac{55}{100}\times 47.9=26.3g

Percentage of Kr=\frac{25.2}{56}\times 100=45\%

b) Thus Mass of Kr in mixture that can be recovered=\frac{45}{100}\times 47.9=21.5g

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