I think it’s B not rlly sure
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:There is colder and drier air coming
Explanation:
Answer:
24.309
Explanation:
mass×percentage+.. = average amu
23.985×78.7/100+24.986×10.13/100+25.983×11.17/100
to three decimal places
=24.309
Answer:
Equilibrium concentration of 
is 12.5 M
Explanation:
Given reaction: 
Here, ![K_{c}=\frac{[C_{2}H_{5}OH]}{[C_{2}H_{4}][H_{2}O]}](https://tex.z-dn.net/?f=K_%7Bc%7D%3D%5Cfrac%7B%5BC_%7B2%7DH_%7B5%7DOH%5D%7D%7B%5BC_%7B2%7DH_%7B4%7D%5D%5BH_%7B2%7DO%5D%7D)
where 
represents equilibrium constant in terms of concentration and species inside third bracket represent equilibrium concentrations
Here, ![[C_{2}H_{4}]=0.015M](https://tex.z-dn.net/?f=%5BC_%7B2%7DH_%7B4%7D%5D%3D0.015M)
, ![[C_{2}H_{5}OH]=1.69M](https://tex.z-dn.net/?f=%5BC_%7B2%7DH_%7B5%7DOH%5D%3D1.69M)
and 
So, ![[H_{2}O]=\frac{[C_{2}H_{5}OH]}{[C_{2}H_{4}]\times K_{c}}=\frac{1.69}{0.015\times 9.0}=12.5M](https://tex.z-dn.net/?f=%5BH_%7B2%7DO%5D%3D%5Cfrac%7B%5BC_%7B2%7DH_%7B5%7DOH%5D%7D%7B%5BC_%7B2%7DH_%7B4%7D%5D%5Ctimes%20K_%7Bc%7D%7D%3D%5Cfrac%7B1.69%7D%7B0.015%5Ctimes%209.0%7D%3D12.5M)
Hence equilibrium concentration of is 12.5 M
