Balance Chemical Equation for this reaction is,
2 CH₄ + O₂ → 2CH₃OH
According to this eq, 22.4 L (1 moles) of Oxygen requires 44.8 L (2 mole) CH₄ for complete reaction.
So, the volume of CH₄ required to consume 0.66 L of O₂ is calculated as,
22.4 L O₂ required to consume = 44.8 L CH₄
0.660 L O₂ will require = X L of CH₄
Solving for X,
X = (44.8 L × 0.660 L) ÷ 22.4 L
X = 1.320 L of CH₄
Result:
1.320 L of CH₄ <span>gas is needed to react completely with 0.660 L of O</span>₂<span> gas to form methanol (CH</span>₃OH<span>).</span>
Nitrogen I believe . I need 20 characters.
The enthalpy<span> of </span>solution<span>, </span>enthalpy<span> of dissolution, or heat of </span>solution<span> is the</span>enthalpy<span> change associated with the dissolution of a substance in a solvent at constant pressure resulting in infinite dilution. The </span>enthalpy<span> of </span>solution<span> is most often expressed in kJ/mol at constant temperature. </span>
Answer:
4121 years
Explanation:
From;
0.693/t1/2 = 2.303/t log No/N
t1/2= half life of the carbon-14
No= count rate of the living tissue
N= count rate of the sample
t = age of the sample
0.693/5730 =2.303/t log (13.5/8.2)
1.21 * 10^-4 = 2.303/t * 0.2165
1.21 * 10^-4 = 0.4986/t
t = 0.4986/1.21 * 10^-4
t = 4121 years
The volume of the 0.15 M LiOH solution required to react with 50 mL of 0.4 M HCOOH to the equivalence point is 133.3 mL
<h3>Balanced equation </h3>
HCOOH + LiOH —> HCOOLi + H₂O
From the balanced equation above,
The mole ratio of the acid, HCOOH (nA) = 1
The mole ratio of the base, LiOH (nB) = 1
<h3>How to determine the volume of LiOH </h3>
- Molarity of acid, HCOOH (Ma) = 0.4 M
- Volume of acid, HCOOH (Va) = 50 mL
- Molarity of base, LiOH (Mb) = 0.15 M
- Volume of base, LiOH (Vb) =?
MaVa / MbVb = nA / nB
(0.4 × 50) / (0.15 × Vb) = 1
20 / (0.15 × Vb) = 1
Cross multiply
0.15 × Vb = 20
Divide both side by 0.15
Vb = 20 / 0.15
Vb = 133.3 mL
Thus, the volume of the LiOH solution needed is 133.3 mL
Learn more about titration:
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