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3 years ago

What will happen to the pH of a buffer if acid/base ratio is increased by 100?

Chemistry

1 answer:

sp2606 [1]3 years ago

4 0

Answer:

It will decrease by 2 units.

Explanation:

The Henderson-Hasselbalch equation for a buffer is

pH = pKa + log(base/acid)

Let's assume your acid has pKa = 5.

(a) If the base: acid ratio is 1:1,

pH(1) = 5 + log(1/1) = 5 + log(1) = 5 + 0 = 5

(b) If the base: acid ratio is 1:100,

pH(2) = 5 + log(1/100) = 5 + log(0.01) = 5 - 2 = 3

(c) Difference

ΔpH = pH(2) - pH(1) = 5 - 3 = -2

If you increase the acid:base ratio to 100:1, the pH will decrease by two units.

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250 ml of seawater and we inked each molecule with pink color, then we mixed this 250 ml in the ocean. After mixing you took 250

Marysya12 [62]

Answer:

9.77 × 10⁹ molecules

Explanation:

Since the density of water is 1 g/cm³ = 1000 g/L

So, we find the mass of sea water in 250 mL = 0.250 L

We know density = mass/volume

mass = density × volume = 1000 g/L × 0.250 L = 250 g

Now we calculate the number of moles of sea water in 250 g, 250 mL of sea water.

number of moles n = mass of sea water,m/molar mass of water, M

molar mass of water, M = 18 g/moL

n = m/M = 250 g/18 g/mol = 13.89 mol

We now calculate the number of molecules present in the 250 mL of sea water.

n = N/N' where N = number of molecules, N' = avogadro's number = 6.022 × 10²³/mol

So,N =nN' = 13.89 mol × 6.022 × 10²³/mol = 8.36 × 10²⁴ molecules

Now, the volume of the ocean is 1.337 × 10¹⁸ m³ = 1.337 × 10¹⁵ L

Since the 250 mL sea water is mixed with the ocean, the number of molecules per liter is 8.36 × 10²⁴ molecules/1.337 × 10¹⁵ L = 6.25 × 10⁹ molecules/L

If we now take 250 mL out of the ocean, the number of molecules in this 250 mL will be 6.25 × 10⁹ molecules/L × 250 mL = 6.25 × 10⁹ molecules/L × 0.250 L = 9.77 × 10⁹ molecules.

So, we have 9.77 × 10⁹ molecules of pink molecules in the 250 mL after mixing in the ocean.

6 0

3 years ago

How does the law of conservation of mass apply to this reaction: C2H4+O2=H2O+Co2?

pav-90 [236]

Consider the formation of water molecule. Hydrogen combines with oxygen to form a water molecule.

In this case, the total mass of the reactants = total mass of the products. Also, the number of atoms of hydrogen and oxygen in the reactants side and the products side are equal.

that's all I know

3 0

3 years ago

Read 2 more answers

How many joules of heat are needed to raise the temperature of 47.5 g of aluminum from 21°C to 94°C , if the specific heat of al

Montano1993 [528]

Answer:

3120.75J

Explanation:

So, we have the formula $q = mc\Delta t$ . For this example, q is the heat energy in Joules, m is the mass in grams, c is the specific heat capacity in $\frac{J}{{g\°C}}$ , and $\Delta$ t is the change in temperature. In this case, m = 47.5g, c = 0.9 $\frac{J}{{g\°C}}$ , and $\Delta$ t = 94-21 = 73°C. Plugging in the values, we get the joules of heat required to raise 47.5g of Al from 21°C to 94°C which is stated above. You can double check my answer but that should be it. An important thing to be aware of are the units. Sometimes, the heat capacity may not be $\frac{J}{{g\°C}}$ . I may be in Kelvin or something. Anyways, hope this helps.

5 0

4 years ago

A student has a mixture of salt (NaCl) and sugar (C12H22O11). To determine the percentage, the student measures out 5.84 grams o

julsineya [31]

Answer:

The volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml

Explanation:

Here we have the reaction of AgNO₃ and NaCl as follows;

AgNO₃(aq) + NaCl(aq)→ AgCl(s) + NaNO₃(aq)

Therefore, one mole of silver nitrate, AgNO₃, reacts with one mole of sodium chloride, NaCl, to produce one mole of silver choride, AgCl, and one mole of sodium nitrate, NaNO₃,

Therefore, since 5.84 grams of NaCl which is 58.44 g/mol, contains

$Number \, of \, moles, n = \frac{Mass}{Molar \ mass} = \frac{5.84}{58.44} = 0.09993 \ moles \ of \ NaCl$

0.09993 moles of NaCl will react with 0.09993 moles of AgNO₃

Also, as 1.0 M solution of AgNO₃ contains 1 mole per 1 liter or 1000 mL, therefore, the volume of AgNO₃ that will contain 0.09993 moles is given as follows;

0.09993 × 1 Liter/mole= 0.09993 L = 99.93 mL

Therefore, the volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml.

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4 years ago

Why cant humans live off sunlight

Liula [17]

They will have no air and no energy or food. No food for animals, so animals die, and humans die with the lack of food, meat and plants

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3 years ago