Is this a real question? It’s B
a. 34 mL; b. 110 mL
a. A tablet containing 150 Mg(OH)₂
Mg(OH)₂ + 2HCl ⟶ MgCl₂ + 2H₂O
<em>Moles of Mg(OH)₂</em> = 150 mg Mg(OH)₂ × [1 mmol Mg(OH)₂/58.32 mg Mg(OH)₂
= 2.572 mmol Mg(OH)₂
<em>Moles of HCl</em> = 2.572 mmol Mg(OH)₂ × [2 mmol HCl/1 mmol Mg(OH)₂]
= 5.144 mmol HCl
Volume of HCl = 5.144 mmol HCl × (1 mmol HCl/0.15 mmol HCl) = 34 mL HCl
b. A tablet containing 850 mg CaCO₃
CaCO₃ + 2HCl ⟶ CaCl₂ + CO₂ + H₂O
<em>Moles of CaCO₃</em> = 850 mg CaCO₃ × [1 mmol CaCO₃/100.09 mg CaCO₃
= 8.492 mmol CaCO₃
<em>Moles of HCl</em> = 8.492 mmol CaCO₃ × [2 mmol HCl/1 mmol CaCO₃]
= 16.98 mmol HCl
Volume of HCl = 16.98 mmol HCl × (1 mL HCl/0.15 mmol HCl) = 110 mL HCl
Answer:
A compound is a unique substance that forms when two or more elements combine chemically. Compounds form as a result of chemical reactions. The elements in compounds are held together by chemical bonds. A chemical bond is a force of attraction between atoms or ions that share or transfer valence electrons
Answer: One carrier interested in taxonomy is a veterinarian
Explanation:
Veterinarians as said, are interested in taxonomy. They are interested with animal organisms of all types with all species of animals. Hope this helps. Let me know if you need anything else.
Answer:
Molarity of the packet is 0.5M
Explanation:
In the reaction of acetic acid with NaOH:
CH₃COOH + NaOH → CH₃COO⁻ + H₂O + Na⁺
<em>1 mole of acetic acid reacts with 1 mole of NaOH.</em>
<em />
When you are titrating the acid with NaOH, you reach equivalence point when moles of acid = moles of NaOH.
Moles of NaOH are:
3.0mL = 3.0x10⁻³L ₓ (0.1 mol / L) =<em> 3.0x10⁻⁴ moles</em> of NaOH = moles of CH₃COOH.
Now, you find the moles of acetic acid in the hot sauce packet. But molarity is the ratio between moles of the acid and liters of solution.
As you don't know the volume of your packet, <em>you can assume its density as 1g/mL. </em>Thus, volume of 0.6g of hot sauce is 0.6mL = 6x10⁻⁴L.
And molarity of the packet is:
3.0x10⁻⁴ moles acetic acid / 6x10⁻⁴L =
<h3>0.5M</h3>