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2 years ago

The diagram is a model of one way that materials move into a cell.

Chemistry

2 answers:

Vlad [161]2 years ago

8 0

Answer:

B. The cell membrane surrounds particles outside the cell.

Explanation:

The last step is when the cell membrane completely surrounds particles outside the cell.

This process is often known as endocytosis.

endocytosis is the process whereby a cell ingests materials by engulfing them using the cell membrane.
In this process, the cell membrane completely covers the food.

alex41 [277]2 years ago

7 0

Answer:

The answer above is incorrect. The correct answer is A:

A vacuole carries particles into the cell

Explanation:

Ape-x

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Ultrasound is used to break up kidney stones. How do sound waves break up kidney stones?

dolphi86 [110]

Answer:

eswl uses shock waves to break kidney stones into small pieces that can more easily travel through the urinary tract.

Explanation:

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4 0

3 years ago

What is the lower concentration limit (vol%) at which a mixture of ethanol in air can explode?

liubo4ka [24]

Answer:

Lower explosive limit (LEL) of ethanol = 3.3%

Explanation:

In the case of alcohol, ethanol presents certain fire hazards. Its momentary flash point is 55ºF (12.9ºC), while the momentary flash point of gasoline is -45ºF (-42.8ºC), and the E85 mixture ranges between -20ºF and -4ºF (between -28 , 9ºC and -20ºC), and has a wider range of flammability limits than gasoline. For emergency response teams, this implies that during a release of the typical ethanol / gasoline mixture, the fuel can be expected to behave like gasoline: It is heavier than air - as we mentioned earlier - and can produce vapors and form flammable mixtures in the air, under most environmental conditions.

General properties and comparison with other inflambles products:

Flash point momentary Gasoline = -45 ° F

Ethanol = 55 ° F

E 85 = between -20º and -4º F

Flammability limits

Lower explosive limit (LEL) of ethanol = 3.3%

Upper Explosive Limit (UEL) = 19%

Lower explosive limit (LEL) of the mixture E 85 = 1.4%

Upper Explosive Limit (UEL) 85 = 19%

Lower explosive limit (LEL) of gasoline = 1.4%

Upper Explosive Limit (UEL) = 7.6%

They have a wider range than gasoline

4 0

2 years ago

If there are 3.10 moles of O, how many moles of each of the compounds are present?

Burka [1]

Explanation:

The question pretty much requires us to find the amount of moles of each compounds based on the number of moles of O given.

H2SO4

1 mol of H2SO4 contains 4 mol of O

x mol of H2SO4 would contain 3.10 mol of O

x = 3.10 * 1 / 4 = 0.775 mol of H2SO4

C2H4O2

1 mol of C2H4O2 contains 2 mol of O

x mol of C2H4O2 would contain 3.10 mol of O

x = 3.10 * 1 / 2 = 1.55 mol of C2H4O2

NaOH

1 mol of NaOH contains 1 mol of O

x mol of NaOH would contain 3.10 mol of O

x = 3.10 * 1 / 1 = 3.10 mol of NaOH

3 0

2 years ago

Use this oxidation-reduction reaction to answer questions about half-reactions:

Advocard [28]

Answer:

N, O, R

Explanation:

Thanks for answering bro, you saved me

5 0

2 years ago

Calculate the mass of CO2 that can be produced if the reaction of 54.0 g of propane and sufficient oxygen has a 64.0% yield.

Oduvanchick [21]

Answer:

103.9 g

Explanation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

First we convert 54.0 g of propane (C₃H₈) into moles, using its molar mass:

54.0 g ÷ 44 g/mol = 1.23 mol C₃H₈

Then we convert 1.23 moles of C₃H₈ into moles of CO₂, using the stoichiometric coefficients:

1.23 mol C₃H₈ * $\frac{3molCO_2}{1molC_3H_8}$ = 3.69 mol CO₂

We convert 3.69 moles of CO₂ into grams, using its molar mass:

3.69 mol CO₂ * 44 g/mol = 162.36 g

And apply the given yield:

162.36 g * 64.0/100 = 103.9 g

7 0

2 years ago