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2 years ago

Can someone help me with my science it’s super hard :(

Chemistry

1 answer:

Lunna [17]2 years ago

5 0

Answer: here is a description of what they look like since I cannot see the diagram.

Mitochondria- wavy line

Ribosomes- little dots

Chromosome- X

Nucleus- Circle with bite taken out of it

Endoplasm- Bubble

Nuclear membrane- Outer layer/bubble

Golgi-wavy circle thing

Vacuole- Bigger circle (but not nucleus)

Cytoplasm- its the jelly on the inside

Cell wall- only in plant cells, looks like a box

Cell membrane- The outmost part of the animal cell, the barrier.

I sure hope this helps!!!

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In the polypeptide Phe-Tyr-Glu-Asp-Ser-Ile-Leu-Ser what is the N-terminal amino acid?

avanturin [10]

Answer:

N-terminal Phe, C-terminal Ser

Explanation:

Amino acids connect like

NH2 -CH(R1) -CO -NH-CH(R2)-CO-.....-NH-CH(Rn)COOH

So, 1st amino acid is N -terminal , and it is Phe.

Last amino acid is C- terminal, and it is Ser.

7 0

3 years ago

In Rutherford's Gold foil experiment, were the alpha particles directed to different areas of the gold foil or only the same spo

MrRissso [65]

Answer:

See explanation

Explanation:

In the Rutherford experiment, alpha particles were directed at the same spot on a thin gold foil.

As the alpha particles hit the foil, most of the alpha particles went through the foil. In Rutherford's interpretation, most of the particles went through because the atom consisted largely of empty space.

However, some of the alpha particles were deflected through large angles, in Rutherford's interpretation, the deflected alpha particles had hit the dense positive core of the atom which he called the nucleus.

This accounted for their scattering through large angles throughout the foil in all directions.

8 0

3 years ago

A solution of barium nitrate has 61.2g of barium nitrate in 1 liter of solution. How many mg of barium are there in 7.5 quarts

Nuetrik [128]

Answer:

220.44g Ba²⁺ ions in solution

Explanation:

Given parameters:

Mass of barium nitrate = 61.2g

Volume of solution = 1 liter

Unkown:

Mass of barium in 7.5quarts of solution?

Solution

We must first convert quarts to its liter equivalence:

1 quarts = 0.95 liter

7.5 quarts = 0.95 x 7.5; 7.125liter

Now, let us find the mass of barium nitrate in a solution of 7.125liter:

Given:

1 liter of solution contains 61.2g of barium nitrate:

7.125 liter will contain 7.125 x 61.2 = 436.05g of barium nitrate.

The formula of the compound is Ba(NO₃)₂:

In solution we have Ba²⁺ + NO₃⁻

Ba(NO₃)₂ → Ba²⁺ + 2NO₃⁻

Number of moles of Ba(NO₃)₂ = $\frac{mass of Ba(NO₃)₂}{Molar mass of Ba(NO₃)₂}$

Molar mass of Ba(NO₃)₂ = 137 + 2[14 + 3(16)] = 271g/mol

Number of moles of Ba(NO₃)₂ = $\frac{436.05}{271}$ = 1.609mole

1 mole of Ba(NO₃)₂ will produce 1 mole of Ba²⁺ ions in solution

therefore, 1.609mole of Ba(NO₃)₂ will also yield 1.609mole of Ba²⁺ ions in solution

Mass of Ba²⁺ ions = Number of moles of Ba²⁺ ions x molar mass of Ba²⁺ ions

Mass of Ba²⁺ ions = 1.609 x 137 = 220.44g

3 0

3 years ago

In an electrically heated boiler, water is boiled at 140°C by a 90 cm long, 8 mm diameter horizontal heating element immersed in

RideAnS [48]

Explanation:

The given data is as follows.

Volume of water = 0.25 $m^{3}$

Density of water = 1000 $kg/m^{3}$

Therefore, mass of water = Density × Volume

= $1000 kg/m^{3} \times 0.25 m^{3}$

= 250 kg

Initial Temperature of water ( $T_{1}$ ) = $20^{o}C$

Final temperature of water = $140^{o}C$

Heat of vaporization of water ( $dH_{v}$ ) at $140^{o}C$ is 2133 kJ/kg

Specific heat capacity of water = 4.184 kJ/kg/K

As 25% of water got evaporated at its boiling point ( $140^{o}C$ ) in 60 min.

Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg

Heat required to evaporate = Amount of water evapotaed × Heat of vaporization

= 62.5 (kg) × 2133 (kJ/kg)

= $133.3 \times 10^{3}$ kJ

All this heat was supplied in 60 min = 60(min) × 60(sec/min) = 3600 sec

Therefore, heat supplied per unit time = Heat required/time = $\frac{133.3 \times 10^{3}kJ}{3600 s}$ = 37 kJ/s or kW

The power rating of electric heating element is 37 kW.

Hence, heat required to raise the temperature from $20^{o}C$ to $140^{o}C$ of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)

= 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)

= 125520 kJ

Time required = Heat required / Power rating

= $\frac{125520}{37}$

= 3392 sec

Time required to raise the temperature from $20^{o}C$ to $140^{o}C$ of 0.25 $m^{3}$ water is calculated as follows.

$\frac{3392 sec}{60 sec/min}$

= 56 min

Thus, we can conclude that the time required to raise the temperature is 56 min.

4 0

3 years ago

What allows electrical charges to move freely

Zolol [24]

Explanation:

Certain materials, called conductors, allow electric charge to move pretty freely through them. ... Other materials, like plastic and rubber, are called insulators because they don't allow electric charges to move through them.

8 0

3 years ago