Condensation because condensation is a process of freezing
        
                    
             
        
        
        
Answer:
0.74 grams of methane 
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) /  (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) /  (32.00 g/mol) = 0.2812 mol. 
- it is clear that O₂ is the limiting reactant and methane will leftover.
 
using cross multiplication
1 mol of  CH₄ needs → 2 mol of O₂
???  mol of  CH₄  needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol 
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol 
now we convert moles into grams 
mass of CH₄ left over = no. of mol of CH₄ left over *  molar mass 
                                     = 0.0469 mol * 16 g/mol = 0.7504 g 
 
So, the right choice is 0.74 grams of methane 
 
        
             
        
        
        
Answer:
Half-life = 3 minutes
Explanation:
Using the radioactive decay equation we can solve for reaction constant, k. And by using:
K = ln2 / Half-life
We can find half-life of polonium-218
Radioactive decay:
Ln[A] = -kt + ln [A]₀
Where:
[A] could be taken as mass of polonium after t time: 1.0mg
k is Reaction constant, our incognite
t are 12 min
[A]₀ initial amount of polonium-218: 16mg
Ln[A] = -kt + ln [A]₀
Ln[1.0mg] = -k*12min + ln [16mg]
-2.7726 = - k*12min
k = 0.231min⁻¹
Half-life = ln 2 / 0.231min⁻¹
<h3>Half-life = 3 minutes</h3>
 
        
             
        
        
        
Answer:
11.31g NaClO₂
Explanation:
<em> Is given 250mL of a 1.60M chlorous acid HClO2 solution. Ka is 1.110x10⁻². What mass of NaClO₂ should the student dissolve in the HClO2 solution to turn it into a buffer with pH =1.45? </em>
It is possible to answer this question using Henderson-Hasselbalch equation:
pH = pKa + log₁₀ [A⁻] / [HA]
<em>Where pKa is -log Ka = 1.9547; [A⁻] is the concentration of the conjugate base (NaClO₂), [HA] the concentration of the weak acid</em>
You can change the concentration of the substance if you write the moles of the substances:
[Moles HClO₂] = 250mL = 0.25L×(1.60mol /L) = <em>0.40 moles HClO₂</em>
Replacing in H-H expression, as the pH you want is 1.45:
1.45 = 1.9547 + log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]
-0.5047 = log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]
<em>0.3128 = </em>[Moles NaClO₂] / [0.40 moles HClO₂]
0.1251 = Moles NaClO₂
As molar mass of NaClO₂ is 90.44g/mol, mass of 0.1251 moles of NaClO₂ is:
0.1251 moles NaClO₂ ₓ (90.44g / mol) = 
<h3>11.31g NaClO₂</h3>