Answer:
Explanation:
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In this case, taking into account that HCl has one molecule of hydrogen per mole of compound which weights 36.45 g/mol, we compute the number of molecules of hydrogen in hydrochloric acid by considering the given mass and the Avogadro's number:
Now, from the 180 g of water, we see two hydrogen molecules per molecule of water, thus, by also using the Avogadro's number we compute the molecules of hydrogen in water:
Thus, the total number of molecules turns out:
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Answer:
2.9 M
Explanation:
Step 1: Given data
Moles of barium chloride (solute): 4.4 moles
Volume of solution: 1.5 liters
Step 2: Calculate the molarity of barium chloride in the solution
The molarity is a way to quantitatively express the concentration of a solute in a solution. The molarity is equal to the moles of solute divided by the volume, in liters, of solution.
The molecular formula of the compound that has a percent composition of 38.7% carbon, 9.76% hydrogen, 51.5% oxygen is C2H6O2.
<h3>How to calculate molecular formula?</h3>
The molecular formula can be calculated from the empirical formula. The empirical formula of the compound is calculated as follows:
- C = 38.7% = 38.7g
- H = 9.76% = 9.76g
- O = 51.5% = 51.5g
Next, we convert the mass to moles by dividing by their atomic mass:
- C = 38.7 ÷ 12 = 3.23mol
- H = 9.76 ÷ 1 = 9.76mol
- O = 51.5÷ 16 = 3.22mol
Next, we divide by the smallest (3.22)
Hence, the empirical formula of the compound is CH3O
If the molar mass of the compound is 62g/mol;
(CH3O)n = 62
31n = 62
n = 2
(CH3O)2 = C2H6O2
Therefore, the molecular formula of the compound that has a percent composition of 38.7% carbon, 9.76% hydrogen, 51.5% oxygen is C2H6O2.
Learn more about molecular formula at: brainly.com/question/14425592
Answer:
a. 100.0 mL of 0.10 M NH₃ with 100.0 mL of 0.15 M NH₄Cl.
c. 50.0 mL of 0.15 M HF with 20.0 mL of 0.15 M NaOH.
Explanation:
A buffer system is formed in 1 of 2 ways:
- A weak acid and its conjugate base.
- A weak base and its conjugate acid.
Determine whether mixing each pair of the following results in a buffer.
a. 100.0 mL of 0.10 M NH₃ with 100.0 mL of 0.15 M NH₄Cl.
YES. NH₃ is a weak base and NH₄⁺ (from NH₄Cl ) is its conjugate base.
b. 50.0 mL of 0.10 M HCl with 35.0 mL of 0.150 M NaOH.
NO. HCl is a strong acid and NaOH is a strong base.
c. 50.0 mL of 0.15 M HF with 20.0 mL of 0.15 M NaOH.
YES. HF is a weak acid and it reacts with NaOH to form NaF, which contains F⁻ (its conjugate base).
d. 175.0 mL of 0.10 M NH₃ with 150.0 mL of 0.12 M NaOH.
NO. Both are bases.
