Since Na has a 1+ charge and O has a -2 charge, by reversing the charges and placing them as subscripts for the other atoms the formula is Na2O1 or simply Na2O.
Answer:
They went from atoms to oxygen molecule.
Explanation:
I'm not sure if this is the answer your teacher is looking for, but in simple terms you had 2 oxygen atoms. Together they make an oxygen molecule, which is the stuff we breathe.
Answer:
158 L.
Explanation:
What is given?
Pressure (P) = 1 atm.
Temperature (T) = 112 °C + 273 = 385 K.
Mass of methane CH4 (g) = 80.0 g.
Molar mass of methane CH4 = 16 g/mol.
R constant = 0.0821 L*atm/mol*K.
What do we need? Volume (V).
Step-by-step solution:
To solve this problem, we have to use ideal gas law: the ideal gas law is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas. The formula is:

Where P is pressure, V is volume, n is the number of moles, R is the constant and T is temperature.
So, let's find the number of moles that are in 80.0 g of methane using its molar mass. This conversion is:

So, in this case, n=5.
Now, let's solve for 'V' and replace the given values in the ideal gas law equation:

The volume would be 158 L.
Answer:
- Last choice: <em><u>- 3.72°C</u></em>
Explanation:
The freezing point depression in a solvent is a colligative property: it depends on the number of solute particles.
The equation to predict the freezing point depression in a solvent is:
Where,
- ΔTf is the freezing point depression of the solvent,
- Kf is the cryoscopic molal constant of the solvent, and i is the Van'f Hoff factor, which is the number of ions produced by each unit formula of the ionic compound.
The calcualtions are in the attached pdf file. Please, open it by clicking on the image of the file.
Answer:
12 g of choloracetic acid
Explanation:
The buffer equilibrium is:
HCH₂ClCO₂ ⇄ CH₂ClCO₂⁻ + H⁺
pka= -log ka =
Ka: 1,3x10⁻³ = [CH₂ClCO₂⁻] [H⁺] / [HCH₂ClCO₂]
By Henderson-Hasselbalch equation:
pH = pka + log₁₀ [A⁻] / [HA]
3,01 = 2,89 + log₁₀ [A⁻] / [HA]
1,318 = [A⁻] / [HA]
As molar concentration of chloroacetic acid (HA) is 0,20M
[A⁻] = 0,26 mol/L
The volume is 500 mL ≡ 0,5 L
0,26mol/L × 0,5 L = 0,13 moles of chloroacetic acid. In grams:
0,13 mol × (94,5g / 1mol) = <em>12 g of choloracetic acid</em>
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I hope it helps!