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3 years ago

Molecules that are too large to flow across the membrane need help from what structures?

2 answers:

6 0

Some molecules, such as carbon dioxide and oxygen, can diffuse across the plasma membrane directly, but others need help to cross its hydrophobic core. In facilitated diffusion, molecules diffuse across the plasma membrane with assistance from membrane proteins, such as channels and carriers.

Troyanec [42]3 years ago

5 0

Answer:

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The water cycle imagine you are a rain drop

Soloha48 [4]

Evaporation,condensation,precipitation,sublimation,transpirtation,runoff and infiltration

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3 years ago

Pls help me What is the mass in grams of 2.64 mol of sulfur dioxide, SO2?

notsponge [240]

Answer: The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles SO2, or 64.0638 grams.

Hope it helps.

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3 years ago

a cube of iron (cp = 0.450 j/g•°c) with a mass of 55.8 g is heated from 25.0°c to 49.0°c. how much heat is required for this pro

krek1111 [17]

Q = ?

Cp = 0.450 j/g°C

Δt = 49.0ºC - 25ºC => 24ºC

m = 55.8 g

Q = m x Cp x Δt

Q = 55.8 x 0.450 x 24

Q = 602.64 J

hope this helps!

7 0

3 years ago

Read 2 more answers

The amount of energy it takes to remove an electron from an atom is

lara31 [8.8K]

the first ionization energy of an element is the energy needed toremove the outermost, or highest energy, electron from a neutral atom in the gas phase.

3 0

3 years ago

A 6.165 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 10.27 grams of CO2 and 3

e-lub [12.9K]

Answer:

Explanation:

mass of carbon in 10.27 g of CO₂ = 12 x 10.27 / 44 = 2.80 g

mass of hydrogen ( H ) in 3.363 g of H₂O = 2 x 3.363 / 18

= .373 g

These masses would have come from the sample of 6.165 g .

Rest of 6.165 g of sample is oxygen .

So oxygen in the sample = 6.165 - ( 2.8 + .373 ) = 2.992 g

Ratio of C , H , O in the sample

2.8 : .373 : 2.992

C: H : O : : 2.8 : .373 : 2.992

Ratio of moles

C: H : O : : 2.8/12 : .373/1 : 2.992 / 16

C: H : O : : .2333 : .373 : .187

C: H : O : : .2333/.187 : .373/.187 : .187/.187

C: H : O : : 1.247 : 1.99 : 1

C: H : O : : 5 : 8 : 4 ( after multiplying by 4 )

Hence empirical formula

C₅H₈O₄

Molecular formula ( C₅H₈O₄ )n

n ( 5 x 12 + 8 x 1 + 4 x 16 ) = 132

n x ( 60 + 8 + 64 ) = 132

n = 1

Molecular formula = C₅H₈O₄.

3 0

3 years ago