The IUPAC name of the compound <span>ch3–ch2–c ≡ c–ch3 is
PEN-2-YNE or
2-PENTYNE.
Attached below is a diagram that fully explains how the name was given and derived.</span>
The number of mole of NaOH that was used, given that 3.85 mL of 10.0 M NaOH is used in the reaction is 0.0385 mole
<h3>How do I determine the number of mole of NaOH?</h3>
We know that molarity is related to number of mole and volume according to the following formula:
Molarity = number of mole / Volume
With the above formula, we can determine the number of mole of NaOH. Details below
The following data were obtained from the question:
- Volume = 3.85 mL = 3.85 / 1000 = 0.00385 L
- Molarity = 10.0 M
- Number of mole of NaOH =?
Molarity = number of mole / Volume
Cross multiply
Number of mole = molarity × volume
Number of mole of NaOH = 10 × 0.00385
Number of mole of NaOH = 0.0385 mole
Thus, the number of mole of NaOH is 0.0385 mole
Learn more about number of mole:
brainly.com/question/20522500
#SPJ1
I believe the answer is B. Definitely not D or A, and C doesn't makes sense.
H₂SO₄ + 2NH₄OH → (NH₄)₂SO₄ + 2H₂O
Explanation:
Unbalanced reaction equation:
H₂SO₄ + NH₄OH → (NH₄)₂SO₄ + H₂O
We can use a simple mathematical approach to balance this equation by solving simple algebraic equations.
aH₂SO₄ + bNH₄OH → c(NH₄)₂SO₄ + dH₂O
a,b,c and d are simple coefficients that will balance the equation:
Conservation of H: 2a + 5b = 8c + 2d
S: a = c
O: 4a + b = 4c + d
N: b = 2c
lets assume that a = 1
c= 1
b = 2
Solving for d from 4a + b = 4c + d
d = 4a + b -4c = 4(1) + 2 - 4(1) = 2
H₂SO₄ + 2NH₄OH → (NH₄)₂SO₄ + 2H₂O
Learn more:
Balancing of equations brainly.com/question/6078553
#learnwithBrainly
