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Vesna [10]
3 years ago
14

Water can be an atom. True False

Chemistry
2 answers:
Tresset [83]3 years ago
8 0

Answer:

false

Explanation:

yep

ad-work [718]3 years ago
3 0

true because I asked my sister

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A nonvolatile solute is dissolved in benzene so that it has a mole fraction of 0.139. What is the vapor pressure of the solution
lapo4ka [179]

Answer:

The vapor pressure of the solution is 3.69 torr

Explanation:

Step 1: Data given

Mole fraction of benzene in the solution = 0.139

P° of benzene is 26.5 torr

Step 2: Calculate the vapor pressure of the solution

Psolution = Xbenzene * P°benzene

⇒with Psolution = the vapor pressure of the solution

⇒with Xbenzene = the mole fraction of benzene = 0.139

⇒with P°benzene = the vapor pressure of pure benzene = 26.5 torr

Psolution = 0.139 * 26.5 torr

Psolution = 3.69 torr

The vapor pressure of the solution is 3.69 torr

7 0
3 years ago
Light waves travel much__________________________________ than sound waves.
LUCKY_DIMON [66]

Yeah, a lot faster than sound.

3 0
3 years ago
If an atom has three protons three neutrons and two electrons what is the electrical charge of the atom
Olenka [21]

Answer:

let assume the atom be x so the charge will be x^+

8 0
3 years ago
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Difference between NaOH and HCI ​
morpeh [17]
Compare HCl, NaOH, and NaCl: HCl is a stronger acid than water. NaCl is a weaker base than NaOH. Strong acids react with strong bases to form weaker acids and bases. ... Compare NaOH, NH3, and H2O, and NH4Cl: NaOH is a stronger base than NH3.
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Glycolic acid, which is a monoprotic acid and a constituent in sugar cane, has a pKa of 3.9. A 25.0 mL solution of glycolic acid
Phoenix [80]

Answer:

pH = 8.0

Explanation:

First, we have to calculate the moles of NaOH.

35.8 \times 10^{-3}L.\frac{0.020mol}{L} =7.2\times 10^{-4}mol

Let's consider the balanced equation.

C₂H₄O₃ + NaOH ⇒ C₂H₃O₃Na + H₂O

The molar ratio C₂H₄O₃: NaOH: C₂H₃O₃Na is 1: 1: 1. So, when 7.2 × 10⁻⁴ moles of NaOH react completely with 7.2 × 10⁻⁴ moles of C₂H₄O₃ they form 7.2 × 10⁻⁴ moles of C₂H₃O₃Na.

The concentration of C₂H₃O₃Na is:

\frac{7.2\times 10^{-4}mol}{60.8 \times 10^{-3}L} =0.012M

C₂H₃O₃Na dissociates according to the following equation:

C₂H₃O₃Na(aq) ⇒ C₂H₃O₃⁻(aq) + Na⁺(aq)

C₂H₃O₃⁻ comes from a weak acid so it undergoes basic hydrolisis.

C₂H₃O₃⁻ + H₂O ⇄ C₂H₄O₃ + OH⁻

If we know that pKa for C₂H₄O₃ is 3.9, we can calculate pKb for C₂H₃O₃⁻ using the following expression:

pKa + pKb = 14

pKb = 14 -3.9 = 10.1

10.1 = -log Kb

Kb = 7.9 × 10⁻¹¹

We can calculate [OH⁻] using the following expression:

[OH⁻] = √(Kb.Cb)               <em>where Cb is the initial concentration of the base</em>

[OH⁻] = √(7.9 × 10⁻¹¹ × 0.012M) = 9.7 × 10⁻⁷ M

Now, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log (9.7 × 10⁻⁷) = 6.0

pH + pOH = 14

pH = 14 - pOH = 14 - 6.0 = 8.0

7 0
3 years ago
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