The answer is B because <span>It would be useful to memorize that sentence. Once you know that, you can figure out whatever else happens at the anode, the cathode, in the solution, and in the external circuit.</span>
Here are the possible answers for the following questions above:
1. H-CC-H (name) - C<span>. ethyne
</span>2. cyclic compound with both saturated and unsaturated characteristics - G<span>. benzene
</span>3. CnH2n - E<span>. general formula for alkenes
</span>4. reaction typical of unsaturated hydrocarbons - A<span>. addition
</span>5. CnH2n-2 - F<span>. general formula for alkynes
</span>6. series name of hydrocarbons with triple bond - D<span>. alkyne
</span>7. CnH2n+2 - B<span>. general formula of alkanes</span>
Answer:
561 g P₂O₃
Explanation:
To find the mass of P₂O₃, you need to (1) convert moles H₃PO₃ to moles P₂O₃ (via mole-to-mole ratio from equation coefficients) and then (2) convert moles P₂O₃ to grams P₂O₃ (via molar mass). It is important to arrange the ratios/conversions in a way that allows for the cancellation of units. The final answer should have 3 sig figs to match the amount of sig figs in the given value.
Atomic Mass (P): 30.974 g/mol
Atomic Mass (O): 15.998 g/mol
Molar Mass (P₂O₃): 2(30.974 g/mol) + 3(15.998 g/mol)
Molar Mass (P₂O₃): 109.942 g/mol
1 P₂O₃ + 3 H₂O -----> 2 H₃PO₃
10.2 moles H₃PO₃ 1 mole P₂O₃ 109.942 g
---------------------------- x -------------------------- x ------------------- = 561 g P₂O₃
2 moles H₃PO₃ 1 mole
Answer:
0.263M of CH₃COOH is the concentration of the solution.
Explanation:
The reaction of acetic acid (CH₃COOH) with NaOH is:
CH₃COOH + NaOH → CH₃COO⁻Na⁺ + H₂O
<em>1 mole of acetic acid reacts per mole of NaOH to produce sodium acetate and water.</em>
<em />
In the equivalence point, moles of acetic acid are equal to moles of NaOH and moles of NaOH are:
0.0375L × (0.175 moles / L) = 6.56x10⁻³ moles of NaOH = moles of CH₃COOH.
As the sample of acetic acid had a volume of 25.0mL = 0.025L:
6.56x10⁻³ moles of CH₃COOH / 0.0250L =
<em>0.263M of CH₃COOH is the concentration of the solution</em>
