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pishuonlain [190]
3 years ago
10

PLS HELP I WILL MARK BRAINLIEST NO LINKS OR PDFS AND PLSEXPLAIN YOUR ANSWER CLEARLY THANKYOU

Chemistry
2 answers:
Butoxors [25]3 years ago
8 0

Answer:If each side of the equation has the same number of atoms of a given element, that element is balanced. If all elements are balanced, the equation is balanced. - online resource

Explanation: if not im sorry

galina1969 [7]3 years ago
6 0
There are the same amount of each type of atoms on the left side as there are on the right side.
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How many carbons would the following molecules contain?
Aleks04 [339]
A is 14
B is 9    is it right
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3 years ago
If a buffer contains 1.05M B and 0.750M BH+ has the pH of 9.5. What would be the pH after 0.005mol of HCL is added to 0.5L of so
Leviafan [203]

Answer:

Final pH: 9.49.

Round to two decimal places as in the question: 9.5.

Explanation:

The conjugate of B is a cation that contains one more proton than B. The conjugate of B is an acid. As a result, B is a weak base.

What's the pKb of base B?

Consider the Henderson-Hasselbalch equation for buffers of a weak base and its conjugate acid ion.

\displaystyle \text{pOH} = \text{pK}_b + \log{\frac{[\text{Salt}]}{[\text{Base}]}}.

\text{pOH} = \text{pK}_w - \text{pH}.

\text{pK}_w = 14.

\text{pOH} = 14 - 9.5 = 4.5

\displaystyle \text{pK}_b = \text{pOH} -\log{\frac{[\text{Salt}]}{[\text{Base}]}}\\\phantom{\text{pK}_b} = 4.5 - \log{\frac{0.750}{1.05}} \\\phantom{\text{pK}_b} =4.64613.

What's the new salt-to-base ratio?

The 0.005 mol of HCl will convert 0.005 mol of base B to its conjugate acid ion BH⁺.

Initial:

  • n(\text{B}) = c\cdot V = 1.05 \times 0.5 = 0.525\;\text{mol};
  • n(\text{BH}^{+}) = c\cdot V = 0.750 \times 0.5 = 0.375\;\text{mol}.

After adding the HCl:

  • n(\text{B}) = 0.525 - 0.005 = 0.520\;\text{mol};
  • n(\text{BH}^{+}) = 0.375+ 0.005 = 0.380\;\text{mol}.

Assume that the volume is still 0.5 L:

  • \displaystyle [\text{B}] = \frac{n}{V} = \frac{0.520}{0.5} = 1.04\;\text{mol}\cdot\text{dm}^{-3}.
  • \displaystyle [\text{BH}^{+}] = \frac{n}{V} = \frac{0.380}{0.5} = 0.760\;\text{mol}\cdot\text{dm}^{-3}.

What's will be the pH of the solution?

Apply the Henderson-Hasselbalch equation again:

\displaystyle \text{pOH} = \text{pK}_b + \log{\frac{[\text{Salt}]}{[\text{Base}]}} = 4.64613 + \log{\frac{0.760}{1.04}} = 4.50991

\text{pH} = \text{pK}_w - \text{pOH}= 14 - 4.50991 = 9.49.

The final pH is slightly smaller than the initial pH. That's expected due to the hydrochloric acid. However, the change is small due to the nature of buffer solutions: adding a small amount of acid or base won't significantly impact the pH of the solution.

3 0
3 years ago
How does wavelength affect the pitch of a sound?
boyakko [2]

Answer:

D

Explanation:

because the higher the wavelength the shorter the pitch and the shorter the wavelength the higher the pitch

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2 years ago
5. What would you expect to happen if the cup of olive oil in question 4a is poured into a container of
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shi i dont know

Explanation:

8 0
3 years ago
How many liters of oxygen gas, at standard
Karo-lina-s [1.5K]

Answer:

Explanation:

  • For the balanced reaction:

<em>4Fe(s) + 3O₂(g) → 2Fe₂O₃(s)​.</em>

It is clear that 4 mol of Fe react with 3 mol of O₂ to produce 2 mol of Fe₂O₃.

  • Firstly, we need to calculate the no. of moles of 35.8 grams of Fe metal:

no. of moles of Fe = mass/molar mass = (35.8 g)/(55.845 g/mol) = 0.64 mol.

  • Now, we can find the no. of moles of O₂ is needed to react with the proposed amount of Fe:

<em><u>Using cross multiplication:</u></em>

4 mol of Fe is needed to react with → 3 mol of O₂, from stichiometry.

0.64 mol of Fe is needed to react with → ??? mol of O₂.

∴ The no. of moles of O₂ needed = (3 mol)(0.64 mol)/(4 mol) = 0.48 mol.

  • Finally, we can get the volume of oxygen using the information:

<em>It is known that 1 mole of any gas occupies 22.4 L at standard P and T (STP).</em>

<em></em>

<em><u>Using cross multiplication:</u></em>

1 mol of O₂ occupies → 22.4 L, at STP conditions.

0.48 mol of O₂ occupies → ??? L.

∴ The no. of liters of O₂ = (0.48 mol)(22.4 L)/(1 mol) = 10.752 L.

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3 years ago
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