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Lera25 [3.4K]
3 years ago
11

A substance containing atoms of different elements that are bonded together is called a(n):

Chemistry
1 answer:
ser-zykov [4K]3 years ago
3 0

Answer:Molecules

Explanation:

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Which of these compounds is a proton acceptor ?
Vikki [24]

Answer:

Sodium Hydroxide

Please mark me as brainlest because I have help you pleaseeeee.

Explanation:

Sodium hydroxide is a common proton acceptor also known as lye. It is the active ingredient in oven cleaners and drain cleaners because it can react with fats to produce compounds that dissolve in water. It is also used in many school laboratories to demonstrate acid-base reactions.

5 0
2 years ago
Calculate the pH after the addition of 10.0 mL of 0.240 M sodium hydroxide to 50.0 mL of 0.120 M acetic acid.
Goryan [66]

Answer:

pH = 4.58

Explanation:

The reaction of NaOH with acetic acid, CH₃COOH occurs as follows:

NaOH + CH₃COOH → CH₃COO⁻Na⁺ + H₂O

<em>Moles that react:</em>

NaOH = 10mL = 0.010L * (0,240mol / L) = 0.0024 moles NaOH

CH₃COOH = 50.0mL = 0.050L * (0.120mol / L) = 0.0060 moles CH₃COOH

That means after the reaction you will have:

CH₃COOH: 0.0060 mol - 0.0024 mol = 0.0036 moles

CH₃COO⁻Na⁺: 0.0024 moles

in solution, you will have the mixture of a weak acid (Acetic acid), with its conjugate base (sodium acetate, CH₃COO⁻Na⁺). And pH of this buffer can be determined using H-H equation:

pH = pKa + log [A⁻] / [HA]

For Acetic buffer pKa = 4.76:

pH = 4.76 + log [CH₃COO⁻Na⁺] / [CH₃COOH]

<em>Where [] is molarity of each species or moles</em>

<em />

Replacing:

pH = 4.76 + log [0.0024 moles] / [0.0036 moles]

<h3>pH = 4.58</h3>

<em />

6 0
3 years ago
Could someone please help me with this?<br><br>A) 2.1 cm<br>B) 1.5 cm<br>C) 1.8 cm<br>D) 2.4 cm
katrin2010 [14]

Answer: It is C I just did this question in math class

Explanation:

7 0
1 year ago
Name the bonds found between amino acids in a polypeptide chain
Anon25 [30]

Answer:

Explanation:

dunno

3 0
3 years ago
A voltaic cell made of a Cr electrode in a solution of 1.0 M in Cr3+ and a gold electrode in a solution that is 1.0 M in Au3+.
Vikki [24]

Answer: a) Anode: Cr\rightarrow Cr^{3+}+3e^-

Cathode: Au{3+}+3e^-\rightarrow Au

b) Anode : Cr

Cathode : Au

c) Au^{3+}+Cr\rightarrow Au+Cr^{3+}

d) E_{cell}=2.14V

Explanation: - 

a) The element Cr with negative reduction potential will lose electrons undergo oxidation and thus act as anode.The element Au with positive reduction potential will gain electrons undergo reduction and thus acts as cathode.

At cathode: Au{3+}+3e^-\rightarrow Au

At anode: Cr\rightarrow Cr^{3+}+3e^-

b) At cathode which is a positive terminal, reduction occurs which is gain of electrons.

At anode which is a negative terminal, oxidation occurs which is loss of electrons.

Gold acts as cathode ad Chromium acts as anode.

c) Overall balanced equation:

At cathode: Au{3+}+3e^-\rightarrow Au     (1)

At anode: Cr\rightarrow Cr^{3+}+3e^-        (2)

Adding (1) and (2)

Au^{3+}+Cr\rightarrow Au+Cr^{3+}

d)E^0_(Cr^{3+}/Cr)= -0.74 V

E^0_(Au^{3+}/Au)= 1.40 V  

E^0{cell}=E^0{cathode}-E^0{anode}=1.40-(-0.74)=2.14V

Using Nernst equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Au^{3+}]}{[Cr^{3+}]^}

where,

n = number of electrons in oxidation-reduction reaction = 3

E^o_{cell} = standard electrode potential = 2.14 V

E_{cell}=2.14-\frac{0.0592}{3}\log \frac{[1.0}{[1.0]}

E_{cell}=2.14

Thus the standard potential for an electrochemical cell with the cell reaction is 2.14 V.

6 0
3 years ago
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