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3 years ago

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Which part of the cell is a gel-like material that contains chemicals used to break down large molecules so that they can be use

d by other organelles?
Cytoplasm
Mitochondrion
Nucleus
Ribosome
Thanks

2 answers:

weqwewe [10]3 years ago

8 0

The answer you seek is Cytoplasm

jeyben [28]3 years ago

8 0

Cytoplasm is your answer

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A) Calculate the average density of the following object (assume it is a perfect sphere). SHOW ALL YOUR WORK (formulas used, num

-BARSIC- [3]

Answers:

A) 2040 kg/m³; B) 58 600 km

Explanation:

A) Density

$V = \frac{ 4}{3 }\pi r^{3} = \frac{ 4}{3 }\pi\times (\text{1150 km})^{3} = 6.37 \times 10^{9} \text{ km}^{3}$

$\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{1.3\times 10^{22} \text{ kg} }{6.37 \times 10^{9} \text{ km}^{3}}\times (\frac{\text{1 km}}{\text{1000 m}})^{3} = \text{2040 kg/m}^{3}$

<em>B) Radius</em>

$\text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{5.68\times 10^{26} \text{ kg} }{687 \text{ kg/m}^{3} }= 8.268 \times 10^{23} \text{ m}^{3}$

$V = \frac{ 4}{3 }\pi r^{3}$

$r^{3} = \frac{3V }{4 \pi }\$

$r= \sqrt [3]{ \frac{3V }{4 \pi } }$

$r= \sqrt [3]{ \frac{3\times 8.268 \times 10^{23} \text{ m}^{3}}{4 \pi } }= \sqrt [3]{ 1.974 \times 10^{23} \text{ m}^{3}}= 5.82 \times 10^{7} \text{ m}=\text{58 200 km}$

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3 years ago

What are the effects of viagra?

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3 years ago

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Is MgCO3 organic or inorganic

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Its inorganic as MgCO3 is contains no carbon more hydrogen which is a crutial component of all organic compounds

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3 years ago

Calculate the volume of the gas when the pressure of the gas is 1.60 atm at a temperature of 298 K. There are 160. mol of gas in

LekaFEV [45]

Answer:

2445 L

Explanation:

Given:

Pressure = 1.60 atm

Temperature = 298 K

Volume = ?

n = 160 mol

Using ideal gas equation as:

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 08206 L.atm/K.mol

Applying the equation as:

1.60 atm × V = 160 mol × 0.08206 L.atm/K.mol × 298 K

<u>⇒V = 2445.39 L</u>

Answer to four significant digits, Volume = 2445 L

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3 years ago

How many grams of aluminum oxide (Al2O3) will be formed from 10.0 grams of aluminum (Al)? 4 AL + 3 02 --> 2 AL203

d1i1m1o1n [39]

Answer:

Mass = 18.9 g

Explanation:

Given data:

Mass of Al₂O₃ formed = ?

Mass of Al = 10.0 g

Solution:

Chemical equation:

4Al + 3O₂ → 2Al₂O₃

Number of moles of Al:

Number of moles = mass/molar mass

Number of moles = 10.0 g/ 27 g/mol

Number of moles = 0.37 mol

Now we will compare the moles of Al and Al₂O₃.

Al : Al₂O₃

4 : 2

0.37 : 2/4×0.37 = 0.185 mol

Mass of Al₂O₃:

Mass = number of moles × molar mass

Mass = 0.185 mol × 101.9 g/mol

Mass = 18.9 g

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3 years ago