What are the significant figures in 250?

In an acid-base titration experiment, 50 mL of a 0.05 M solution of acetic acid (Ka= 1.75 x 10-5 ) was titrated with a 0.05M sol

Viktor [21]

Answer:

(3) 5.36

Explanation:

Since this is a titration of a weak acid before reaching equivalence point, we will have effectively a buffer solution. Then we can use the Henderson-Hasselbalch equation to answer this question.

The reaction is:

HAc + NaOH ⇒ NaAc + H₂O

V NaOH = 40 mL x 1 L/1000 mL = 0.040 L

mol NaOH reacted with HAc = 0.040 L x 0.05 mol/L = 0.002 mol

mol HAC originally present = 0.050 L x 0.05 mol/L = 0.0025 mol

mol HAc left after reaction = 0.0025 - 0.002 = 0.0005

Now that we have calculated the quantities of the weak acid and its conjugate base in the buffer, we just plug the values into the equation

pH = pKa + log ((Ac⁻)/(HAc))

(Notice we do not have to calculate the molarities of Ac⁻ and HAc because the volumes cancel in the quotient)

pH = -log (1.75 x 10⁻⁵) + log (0.002/0.0005) = 5.36

THe answer is 5.36

5 0

3 years ago

Read 2 more answers

Suppose an enzyme and its substrate obey the lock and key model of enzyme catalysis. Which of the following would be true of the

Maurinko [17]

Answer:

only one substrate could be converted to product by the enzyme

Explanation:

If an enzyme obeys the lock and key model, it then follows that the design of the enzyme is unique to a particular substrate.

Just as only the correct key can effectively fit into a lock, only the appropriate substrate whose design matches that of the enzyme can fit into the enzyme.

Owing to the lock and key model of enzyme activity, only one enzyme could interact with a substrate to yield particular product.

4 0

3 years ago

Is gold a compound or an element

timurjin [86]

Gold is an element (Au) you can identify it on a periodic table

3 0

4 years ago

Question 23 (5 points)

iren [92.7K]

Answer:

λ = 6.25 × 10⁻¹⁰ m

x-ray

General Formulas and Concepts:

Speed of Light = Wavelength times Frequency

c = λν

Explanation:

Step 1: Define

C = 3.0 × 10⁸ m/s

ν = 4.80 × 10¹⁷ Hz (s⁻¹)

λ = unknown

Step 2: Solve for wavelength

3.0 × 10⁸ m/s = λ(4.80 × 10¹⁷ Hz)

λ = 6.25 × 10⁻¹⁰ m

Step 3: Check

We are given 3 sig figs. Follow sig fig rules.

6.25 has 3 sig figs. No changes necessary.

Step 4: Identify

6.25 × 10⁻¹⁰ m is in the x-ray spectrum according the the Electromagnetic Spectrum

5 0

3 years ago

What are the names of the following three compounds (see attached)

balu736 [363]

Answer:

1. 4-ethyl-1-heptene

2. 6-ethyl-2-octene

3. 1-butyne

Explanation:

The compounds are named according to IUPAC rules.

Compound 1:

Identify the longest carbon chain. This chain is called the parent chain.
Identify all of the substituents (groups appending from the parent chain).
The parent chain is numbered so that the multiple bonds have the lowest numbers (double has the priority over alkyl substituents).
The longest chain contains 7 carbon atoms, so taken the name hept.
The double bond between C1 and C2, so take no. 1 and add the suffix ene to hept "1-heptene".
The ethyl group is the alkyl substituent on position 4.
So the name is 4-ethyl-1-heptene.

Compound 2:

Identify the longest carbon chain. This chain is called the parent chain.
Identify all of the substituents (groups appending from the parent chain).
The parent chain is numbered so that the multiple bonds have the lowest numbers (double has the priority over alkyl substituents).
The longest chain contains 8 carbon atoms, so taken the name oct.
The double bond between C2 and C3, so take no. 2 and add the suffix ene to oct "2-octene".
The ethyl group is the alkyl substituent on position 6.

So the name is 6-ethyl-2-octene.

Compound 3:

Identify the longest carbon chain. This chain is called the parent chain.
Identify all of the substituents (groups appending from the parent chain), there is no substituents.
The parent chain is numbered so that the multiple bonds have the lowest numbers (Triple bond here take the lowest number).
The longest chain contains 4 carbon atoms, so taken the name but.
The triple bond between C1 and C2, so take no. 1 and add the suffix yne to but "1-butyne".

So the name is 1-butyne.

7 0

3 years ago