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3 years ago

What does a cell use to eliminate a substance that is too large to leave by diffusion

1 answer:

6 0

Endocytosis where the substance will be surrounded by a cell membrane and then broken down for food or just discarded as waste.

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A 0.465 g sample of an unknown compound occupies 245 ml at 298 k and 1.22 atm. what is the molar mass of the unknown compound?

nekit [7.7K]

<span>we can find the number of moles of gas using the ideal gas law equation

PV = nRT

where P - pressure - 1.22 atm

V - volume - 0.245 L

n - number of moles

R - gas constant - 0.08206 L.atm/mol.K

T - temperature - 298 K

substituting the values in the equation

1.22 atm x 0.245 L = n x 0.08206 L.atm/mol.K x 298 K

n = 0.0122 mol

molar mass of compound = mass present / number of moles therefore molar mass = 0.465 g / 0.0122 mol = 38.1 g/mol
the answer is d) 38.0 g/mol </span>

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3 years ago

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What is the length of one mole of the typical cell phone that is approximately 15 cm in length?

Salsk061 [2.6K]

i dont know how to remove my answer so ignore this

4 0

3 years ago

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Where is the Oort cloud located

Karolina [17]

To be honest I don’t even know

5 0

3 years ago

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Carbon dioxide dissolves in water to form carbonic acid, which is primarily dissolved CO2. Dissolved CO2 satisfies the equilibri

lorasvet [3.4K]

Explanation:

The reaction equation will be as follows.

$CO_{2}(aq) + H_{2}O \rightleftharpoons H^{+}(aq) + HCO^{-}_{3}(aq)$

Calculate the amount of $CO_{2}$ dissolved as follows.

$CO_{2}(aq) = K_{CO_{2}} \times P_{CO_{2}}$

It is given that $K_{CO_{2}}$ = 0.032 M/atm and $P_{CO_{2}}$ = $1.9 \times 10^{-4}$ atm.

Hence, $[CO_{2}]$ will be calculated as follows.

$[CO_{2}]$ = $K_{CO_{2}} \times P_{CO_{2}}$

= $0.032 M/atm \times 1.9 \times 10^{-4}atm$

= $0.0608 \times 10^{-4}$

or, = $0.608 \times 10^{-5}$

It is given that $K_{a} = 4.46 \times 10^{-7}$

As, $K_{a} = \frac{[H^{+}]^{2}}{[CO_{2}]}$

$4.46 \times 10^{-7} = \frac{[H^{+}]^{2}}{0.608 \times 10^{-5}}$

$[H^{+}]^{2}$ = $2.71 \times 10^{-12}$

$[H^{+}]$ = $1.64 \times 10^{-6}$

Since, we know that pH = $-log [H^{+}]$

So, pH = $-log (1.64 \times 10^{-6})$

= 5.7

Therefore, we can conclude that pH of water in equilibrium with the atmosphere is 5.7.

3 0

2 years ago

Name the following aromatic hydrocarbon:

Tresset [83]

3-Methylpentane is the IUPAC name for the substance.

whether in a continuous chain or a ring, the longest chain of carbons joined by a single bond serves as the basis for IUPAC nomenclature. According to a precise set of priorities, all deviations—whether they involve numerous bonds or atoms other than carbon and hydrogen—are denoted by prefixes or suffixes.

+3-Methylpentane is the IUPAC name for the substance in question. It has a lengthy chain of 5 carbon atoms, which gives it the prefix pent-, and a single bond is what gives it the postfix -ane (alkane). Given that the methyl group is present at the third carbon, it is 3-methylpentane.

Learn more about IUPAC Nomenclature here-

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8 0

1 year ago