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zheka24 [161]
3 years ago
9

What cell part do all substances have to pass through to get in or out of all cells​

Chemistry
1 answer:
DIA [1.3K]3 years ago
7 0

I think it’s the cell membrane

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When a particle is moving fast it has a lot of kinetic energy in it. ?
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Random particle motion in liquids and gases is a difficult concept for in temperature, the particles move faster as they gain kinetic energy.

Explanation:

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What is the symbol for copper
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The symbol for copper is CU, which means copper cuprum, the Latin word for copper.
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What is the net ionic equation of sodium sulfate and silver nitrate
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AgNO3 ( aq ) + NaI ( aq ) = NaNO3 ( aq ) + AgI ( s ) .

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3 years ago
er solutions can be produced by mixing a weak acid with its conjugate base or by mixing a weak base with its conjugate acid. The
liubo4ka [24]

Answer:

You need to add 19,5 mmol of acetates

Explanation:

Using the Henderson-Hasselbalch equation:

pH = pKa + log₁₀ [base]/[acid]

For the buffer of acetates:

pH = pKa + log₁₀ [CH₃COO⁻]/[CH₃COOH]

As pH you want is 5,03, pka is 4,74 and milimoles of acetic acid are 10:

5,03 = 4,74 + log₁₀ [CH₃COO⁻]/[10]

1,95 = [CH₃COO⁻]/[10]

<em>[CH₃COO⁻] = 19,5 milimoles</em>

Thus, to produce an acetate buffer of 5,03 having 10 mmol of acetic acid, you need to add 19,5 mmol of acetates.

I hope it helps!

7 0
3 years ago
2C2H6 + 7O2 ------&gt; 4CO2 + 6H2O
Romashka-Z-Leto [24]

We convert the masses of our reactants to moles and use the stoichiometric coefficients to determine which one of our reactants will be limiting.

Dividing the mass of each reactant by its molar mass:

(10 g C2H6)(30.069 g/mol) = 0.3326 mol C2H6

(10 g O2)(31.999 g/mol) = 0.3125 mol O2.

Every 2 moles of C2H6 react with 7 moles of O2. So the number of moles of O2 needed to react completely with 0.3326 mol C2H6 would be (0.3326)(7/2) = 1.164 mol O2. That is far more than the number of moles of O2 that we are given: 0.3125 moles. Thus, O2 is our limiting reactant.

Since O2 is the limiting reactant, its quantity will determine how much of each product is formed. We are asked to find the number of grams (the mass) of H2O produced. The molar ratio between H2O and O2 per the balanced equation is 6:7. That is, for every 6 moles of H2O that is produced, 7 moles of O2 is used up (intuitively, then, the number of moles of H2O produced should be less than the number of moles of O2 consumed).

So, the number of moles of H2O produced would be (0.3125 mol O2)(6 mol H2O/7 mol O2) = 0.2679 mol H2O. We multiply by the molar mass of H2O to convert moles to mass: (0.2679 mol H2O)(18.0153 g/mol) = 4.826 g H2O.

Given 10 grams of C2H6 and 10 grams of O2, 4.826 g of H2O are produced.

8 0
2 years ago
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