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svp [43]
3 years ago
7

A student titrates an unknown amount of potassium hydrogen phthalate (KHC8H4O4, abbreviated as KHP) with 21.30 mL of a 0.1161 M

NaOH solution. KHP (molar mass
Chemistry
1 answer:
schepotkina [342]3 years ago
7 0

Answer:

0.5050g of KHP were titrated

Explanation:

<em>... (molar mass = 204.22 g/mol) has one acidic hydrogen. What mass of KHP was titrated (reacted completely) by the sodium hydroxide solution?</em>

<em />

The KHP is a salt used as standard to determine concentrations of bases as NaOH solutions. The reaction that occurs is:

KHP + NaOH → KP⁻Na⁺ + H₂O

<em>Only 1 hydrogen is acidic. Only 1 hydrogen reacts with NaOH</em>

In the reaction you can see that 1 mole of NaOH reacts per mole of KHP

To find mass of KHP we need to determine its moles finding moles of NaOH (The moles of KHP required to reach endpoint = Moles of NaOH added):

<em>Moles NaOH:</em>

21.30mL = 0.02130L * (0.1161mol / L) = 0.002473 moles of NaOH = Moles KPH

With mass of KHP and its molar mass we can solve the mass of KHP:

0.002473 moles of KHP * (204.22g/mol) =

<h3>0.5050g of KHP were titrated</h3>
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The normal freezing point of a certain liquid
slavikrds [6]

Answer : The molal freezing point depression constant of liquid X is, 4.12^oC/m

Explanation :  Given,

Mass of urea (solute) = 5.90 g

Mass of liquid X (solvent) = 450 g  = 0.450 kg

Molar mass of urea = 60 g/mole

Formula used :  

\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of urea}}{\text{Molar mass of urea}\times \text{Mass of liquid X Kg}}

where,

\Delta T_f = change in freezing point

\Delta T_s = freezing point of solution = -0.5^oC

\Delta T^o = freezing point of liquid X = 0.4^oC

i = Van't Hoff factor = 1 (for non-electrolyte)

K_f = Molal-freezing-point-depression constant = ?

m = molality

Now put all the given values in this formula, we get

0.4^oC-(-0.5^oC)=1\times K_f\times \frac{5.90g}{60g/mol\times 0.450kg}

K_f=4.12^oC/m

Therefore, the molal freezing point depression constant of liquid X is, 4.12^oC/m

3 0
3 years ago
If a reaction mixture initially contains 0.168 MSO2Cl2, what is the equilibrium concentration of Cl2 at 227 ∘C?
Semenov [28]

Answer:

see explanation below

Explanation:

The question is incomplete. Here's the complete question:

<em>Consider the following reaction: </em>

<em>SO2Cl2 -----> SO2(g) + Cl2(g) </em>

<em>Kc= 2.99 x 10^-7 at 227 degrees celcius </em>

<em>If a reaction mixture initially contains 0.168 MSO2Cl2, what is the equilibrium concentration of Cl2 at 227 ∘C?</em>

This is a problem of equilibrium, therefore, we need to solve this using the expression of equilibrium constant. To do that, we need to wirte an ICE chart and solve from there:

       SOCl2 ---------> SO2 + Cl2     Kc = 2.99x10⁻⁷

i)       0.168                  0        0

c)         -x                    +x       +x

e)     0.168-x                x         x

Writting the Kc expression:

Kc = [SO2] [Cl2] / [SOCl2]

Replacing the values from the chart:

2.99x10⁻⁷ = x² / 0.168 - x

However, Kc is a very very small value, therefore, we can assume that the value of "x" would be very small too, and we can neglect the 0.168-x and just round it to 0.168:

2.99x10⁻⁷ = x²/0.168

2.99x10⁻⁷ * 0.168 = x²

√5.02x10⁻⁸ = x

x = 2.24x10⁻⁴ M

This means then, that the concentration of Cl2 in equilibrium would be:

<em>[Cl₂] = 2.24x10⁻⁴ M</em>

5 0
3 years ago
HELP ITS OVER DUE AND GRADE BOOK ENDS MARCH 5TH
Ymorist [56]
Mutualism- both organisms benefit
Commensalism- one organism benefits while the other neither is harmed or helped
Parasitism- one organism is benefited while the other is harmed

4 P (flea benefits but dog is harmed)
5 M (both get to eat the honey)
6 C (bird gets a place to live without harming or helping the tree)
7 P (lice get a place to live but humans are harmed)
8 M (both are helped)
9 C (flower isn’t harmed or helped but bee is helped)
10 P (the tree is harmed but the mistletoe is benefitted)
4 0
3 years ago
What do we call the type of chemical reaction that occurs in this step? Explain how the name of this type of reaction relates to
icang [17]
Figure it out yourself
6 0
3 years ago
A 50.0 mg sample of iodine-131 was placed in a container 32.4 days ago. if its half-life is 8.1 days, how many milligrams of iod
sertanlavr [38]

3.124mg of I-131 is present after 32.4 days.

The 131 I isotope emits radiation and particles and has an 8-day half-life. Orally administered, it concentrates in the thyroid, where the thyroid gland is destroyed by the particles.

What is Half life?

The time required for half of something to undergo a process: such as. a : the time required for half of the atoms of a radioactive substance to become disintegrated.

Half of the iodine-131 will still be present after 8.1 days.

The amount of iodine-131 will again be halved after 8.1 additional days, for a total of 8.1+8.1=16.2 days, reaching (1/2)(1/2)=1/4 of the initial amount.

The quantity of iodine-131 will again be halved after 8.1 more days, for a total of 16.2+8.1+8.1=32.4 days, to (1/4)(1/2)(1/2)=1/16 of the initial quantity.

If the original dose of iodine-131 was 50mg, the residual dose will be (50mg)*(1/16)=3.124mg after 32.4 days.

Learn more about the Half life of radioactie element with the help of the given link:

brainly.com/question/27891343

#SPJ4

7 0
1 year ago
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