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Vinvika [58]
3 years ago
15

How do I do this and what are the answers?

Chemistry
1 answer:
vodka [1.7K]3 years ago
6 0
I cannot see your question to help you... sorry
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Negative ions have ______ (More or Less) protons than electrons.
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Answer:

Less

Explanation: I took the test.

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3 years ago
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The chemical formula for sodium citrate is Na3C6H5O7. Which statement is true? Sodium citrate is a compound with a total of 21 a
Scrat [10]

Answer:

Sodium citrate is a compound with a total of 21 atoms

6 0
3 years ago
How many grams of Kr are in a 8.55 L cylinder at 12.3 ∘C and 8.33 atm?
NISA [10]

Answer:

I think its 0.301

Explanation:

8 0
2 years ago
The concentration of ozone in ground-level air can be determined by allowing the gas to react with an aqueous solution of potass
Natali5045456 [20]

Answer:

a) 2KI + O₃ + H₂O ----> I₂ + O₂ + 2KOH

b) Ozone concentration = 0.246 ppb

Explanation:

a) The balanced equation for the reaction is

2KI + O₃ + H₂O ----> I₂ + O₂ + 2KOH

b) We first convert 17 μg of KI to number of moles

Number of moles = (mass)/(molar mass)

Molar mass of KI = 166 g/mol

Mass of KI that reacted = 17 μg = (17 × 10⁻⁶) g

Number of moles = (17 × 10⁻⁶)/166

Number of moles of KI that reacted = (1.0241 × 10⁻⁷) moles

From the stoichiometric balance of the reaction,

2 moles of KI reacts with 1 mole of O₃

Then, (1.0241 × 10⁻⁷) moles of KI will react with (1.0241 × 10⁻⁷ × 1/2) moles of O₃

Number of moles of O₃ that reacted = (5.12 × 10⁻⁸) moles.

To express the amount of O₃ in 10.0 L of air in ppb, we need to convert the amount of O₃ that reacted.

Mass = (number of moles) × (molar mass)

Molar mass of O₃ = 48 g/mol

Mass of O₃ that reacted = (5.12 × 10⁻⁸) × 48 = 0.0000024578 g = (2.46 × 10⁻⁶) g

Concentration in ppb = (Mass of solute in μg)/(volume of solution in L)

Mass of solute = Mass of O₃ = (2.46 × 10⁻⁶) g = 2.46 μg

Volume of solution = 10.0 L

Concentration of O₃ in air in ppb = 2.46/10 = 0.246 ppb

4 0
3 years ago
The bond between sodium and chlorine in the compound sodium chloride (NaCl) is a/an:
Deffense [45]
The classic case of ionic bonding, the sodium chloride molecule forms by the ionization of sodium and chlorine atoms and the attraction of the resulting ions. An atom of sodium has one 3s electron outside a closed shell, and it takes only 5.14 electron volts of energy to remove that electron.
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