The molar concentration of the KI_3 solution is 0.0833 mol/L.
<em>Step 1</em>. Calculate the <em>moles of S_2O_3^(2-)</em>
Moles of S_2O_3^(2-) = 25.00 mL S_2O_3^(2-) ×[0.200 mmol S_2O_3^(2-)/(1 mL S_2O_3^(2-)] = 5.000 mmol S_2O_3^(2-)
<em>Step 2</em>. Calculate the <em>moles of I_3^(-)
</em>
Moles of I_3^(-) = 5.000 mmol S_2O_3^(2-)))) × [1 mmol I_3^(-)/(2 mmol S_2O_3^(2-)] = 2.500 mmol I_3^(-)
<em>Step 3</em>. Calculate the <em>molar concentration of the I_3^(-)</em>
<em>c</em> = "moles"/"litres" = 2.500 mmol/30.00 mL = 0.083 33 mol/L
Answer:
omega.
Explanation:
The omega numbers simply reference how many carbons away from the methyl end of the fatty acid chain that the first carbon-carbon double bond appears. If the double bond is three carbons away, it's called an omega-3 fatty acid.
You can just write omega.
The distance covered will be:
The correct answer is
D).
The element symbol for oxygen<span> is O and its atomic number is 8. The mass numbers for oxygen must be 8 + 8 = 16; 8 + 9 = 17; 8 + 10 = 18</span>
Answer:
4.07
Explanation:
There is some info missing. I think this is the original question.
<em>A solution is prepared at 25 °C that is initially 0.057 M in nitrous acid (HNO₂), a weak acid with Ka = 4.5 × 10⁻⁴, and 0.30 M in sodium nitrite (NaNO₂). Calculate the pH of the solution. Round your answer to 2 decimal places.</em>
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Nitrous acid is a weak acid and nitrite (coming from sodium nitrite) is its conjugate base. Together, the form a buffer system. We can calculate its pH using the Henderson-Hasselbach equation.
pH = pKa + log [base]/[acid]
pH = -log 4.5 × 10⁻⁴ + log 0.30/0.057
pH = 4.07