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2 years ago

What is the ph of a solution of 0.400 m k2hpo4, potassium hydrogen phosphate?

1 answer:

3 0

When we can get Pka for K2HPO4 =6.86 so we can determine the Ka :

when Pka = - ㏒ Ka

6.86 = -㏒ Ka

∴Ka = 1.38 x 10^-7

by using ICE table:

H2PO4- → H+ + HPO4
initial 0.4 m 0 0

change -X +X +X

Equ (0.4-X) X X

when Ka = [H+][HPO4] / [H2PO4-]

by substitution:

1.38 X 10^-7 = X^2 / (0.4-X) by solving for X

∴X = 2.3x 10^-4

∴[H+] = X = 2.3 x 10^-4

∴PH = -㏒[H+]

= -㏒ (2.3 x 10^-4)
∴PH = 3.6

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When 70. milliliter of 3.0-molar Na2CO3 is added to 30. milliliters of 1.0-molar NaHCO3 the resulting concentration of Na+ is 2

tiny-mole [99]

Answer : The resulting concentration of $Na^+$ ion is, 4.5 M

Explanation : Given,

Concentration of $Na_2CO_3$ = $M_1$ = 3.0 M = 3.0 mol/L

Volume of $Na_2CO_3$ = $V_1$ = 70 mL = 0.07 L

Concentration of $NaHCO_3$ = $M_2$ = 1.0 M = 1.0 mol/L

Volume of $NaHCO_3$ = $V_2$ = 30 mL = 0.03 L

First we have to calculate the moles of $Na_2CO_3$ and $NaHCO_3$

$\text{Moles of }Na_2CO_3=\text{Concentration of }Na_2CO_3\times \text{Volume of }Na_2CO_3=3.0mol/L\times 0.07L=0.21mol$

and,

$\text{Moles of }NaHCO_3=\text{Concentration of }NaHCO_3\times \text{Volume of }NaHCO_3=1.0mol/L\times 0.03L=0.03mol$

Now we have to calculate the moles of $Na^+$ ions.

As, 1 mole of $Na_2CO_3$ will give 2 moles of $Na^+$ ions

So, 0.21 moles of $Na_2CO_3$ will give $2\times 0.21=0.42$ moles of $Na^+$ ions

and,

As, 1 mole of $NaHCO_3$ will give 1 mole of $Na^+$ ions

So, 0.03 moles of $NaHCO_3$ will give 0.03 moles of $Na^+$ ions

So,

Total number of moles of $Na^+$ ions = 0.42 + 0.03 =0.45 mole

Total volume of both solution = 70 mL + 30 mL = 100 mL = 0.1 L

Now we have to calculate the concentration of $Na^+$ ions.

$\text{Concentration of }Na^+=\frac{\text{Moles of }Na^+}{\text{Volume of solution}}=\frac{0.45mol}{0.1L}=4.5mol/L=4.5M$

Therefore, the resulting concentration of $Na^+$ ion is, 4.5 M

6 0

2 years ago

Which nuclear emission moving through an electric field would be deflected toward the positive electrode?

Tom [10]

As a rule, opposite charges attract. So,you would expect that the particle that would deflect towards the positive electrode has a negative charge. Now, an alpha particle is basically a neutral Helium atom. Beta particle is an electron. Gamma particle has no charge (neutral). <em>Thus, the answer is B.</em>

6 0

3 years ago

Gamma rays are often used to kill microorganisms in food, in an attempt to make the food safer. Some people contend that this ir

nikdorinn [45]

Answer:

b . Irradiated food is shown to not be radioactive.

Explanation:

If it can be proven that irradiated food is not radioactive, then it will effective dispute the idea that irradiated food are less safe to eat.

An irradiated food is one in which ionizing radiations have been employed to improve food quality.
Thus, bacteria and other food spoilers can be exterminated from the food.
Most irradiated food do not contain radiation and are fit for consumption.

If it can be proven, that this is true, then it will challenge the idea that irradiated foods are not safe.

4 0

1 year ago

29. A certain nut crunch cereal contains 11.0 grams of sugar (sucrose, C12H22O11) per serving size of 60.0 grams. How many servi

adoni [48]

Answer:

The answer is: 51.8 g (86% of serving size)

Explanation:

In order to solve the problem, we have to first determine the number of moles there are in 11.0 g of sucrose. Sucrose has a molecular weight of 342 g (we calculate this from the molar mass of the elements : 12 x 12 g/mol C + 22 x 1 g/mol H + 11 x 16 g/mol O). So, we divide the mass (11.0 g) into the molecular weight of sucrose:

11.0 g sucrose x 1 mol/342 g sucrose= 0.032 mol

We have 0.032 mol of sucrose in a serving of 60 g. But we need less moles (0.0278 mol):

0.032 mol ------------ 60 g serving

0.0278 mol------------ x= 0.0278 mol x 60 g serving/0.032 mol

x= 51.8 g

So, lesser than 1 serving of 60 g must be eaten to consume 0.0278 mol os sucrose. Exactly, 51.8 g (which stands for a 86% of the serving size).

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2 years ago

The chemical formula for artificial sweetener is C7H5NO3S. Match the elements with the numbers found in one molecule of artifici

klio [65]

its is 1.hydrogen 2. oxygen 3. argon 4. carbon

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2 years ago