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MrRa [10]
3 years ago
12

Which of the following represents the greatest number of moles? please help! :)

Chemistry
2 answers:
babymother [125]3 years ago
8 0
The answer is A helium I’m pretty sure
zloy xaker [14]3 years ago
5 0

Answer:

Helium

Explanation:

He- 3.99739969150064 (4)

S-0.498986433806336 (0.5)

Li-3.02550064832148 (3)

B-2.959948200906496 (3)

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What happens to elements when they combine together chemically ?
lara [203]

Answer:

When two distinct elements are chemically combined—i.e., chemical bonds form between their atoms—the result is called a chemical compound. Most elements on Earth bond with other elements to form chemical compounds, such as sodium (Na) and Chloride (Cl), which combine to form table salt (NaCl).

7 0
2 years ago
What is the empirical formula of a molecule containing 65.5% carbon, 5.5% hydrogen, and 29.0% oxygen
natta225 [31]
Carbon(C):
number of moles= mass/molar mass(Mr)
=65.5/12
=5.5 moles

Hydrogen(H):
number of moles=mass/molar mass (Mr)
=5.5/1
=5.5 moles

Oxygen (O):
number of moles = mass/molar mass (Mr)
=29.0/16
=1.8 moles

EF= lowest number of moles over each of the elements

So,
C= 5.5/1.8 = 3
H= 5.5/1.8 = 3
O= 1.8/1.8 = 1

Therefore Emperical formula= C3H3O
6 0
2 years ago
What are the constituents of baking powder?
timurjin [86]
Baking powder is a mixture of baking soda and a mild edible acid such as tartaric acid
5 0
3 years ago
Use the following half-reactions to write three spontaneous reactions, calculate E°cell for each reaction, and rank the oxidizin
Ainat [17]

Answer:

See explaination

Explanation:

1)

we know that

half cell with higher reduction potential is cathode

so

cathode :

N20 + 2H+ + 2e- ---> N2 + H20

anode :

Cr(s) ---> Cr+3 + 3e-

so

overall reaction is

3 N20 + 6H+ + 2 Cr ---> 3N2 + 3H20 + 2Cr+3

now

Eo cell = Eo cathode - Eo anode

so

EO cell = 1.77 + 0.74

Eo cell = 2.51 V

now

in this case

oxidizing agents are N20 and Cr+3

reducing agents are Cr and N2

higher the reduction potential , stronger the oxidizing agent

lower the reduction potential , stronger the reducing agent

so

oxidzing agents

N20 > Cr+3

reducing agents

Cr > N2

2)

cathode :

Au+ + e- --> Au

anode :

Cr ---> Cr+3 + 3e-

overall reaction

3Au+ + Cr ---> 3Au + Cr+3

Eo cell = 1.69 + 0.74

Eo cell = 2.43

now

oxidizing agents :

Au+ > Cr+3

reducing agents :

Cr > Au

3)

cathode :

N20 + 2H+ + 2e- ---> N2 + H20

andoe :

Au ---> Au+ + e-

overall

2 Au + N20 + 2H+ --> 2 Au+ + N2 + H20

Eo cell = 1.77 - 1.69

Eo cell = 0.08

oxidizing agents

N20 > Au+

reducing agents

Au > N2

8 0
3 years ago
Calculate the molality of an ethylene glycol (C2H6O2) soltuion, that has a molarity of 2.07 M. The density of the solution is 1.
vlada-n [284]

Answer:

2.03

Explanation:

Let's <u>assume we have 1 L of the solution</u>:

  • There would be 2.07 ethylene glycol moles.
  • The solution would weigh (1000 mL * 1.02 g/mL) = 1020 g.

With that information we can <u>calculate the molality</u>:

  • molality = moles of solute / kg of solvent
  • molality = 2.07 moles / (1020 ÷ 1000) = 2.03 m

Keep in mind that this is only an estimate, as we used the kg of the solution and not of the solvent.

3 0
2 years ago
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