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Nady [450]
3 years ago
11

What is the same for all of the drilling sites we examined?

Chemistry
2 answers:
deff fn [24]3 years ago
7 0
A picture of the sites?
yarga [219]3 years ago
4 0

Answer:

the same is what is this question like what did u exame

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30 point. What is formed when a hydroxyl group is substituted for a hydrogen?
Andru [333]
Haloalkanes






Yeah it’s that
6 0
2 years ago
A beaker contains 0.125 L of a 3.00 M solution. If the volume goes up to 0.325 L, what is the new molarity?
igor_vitrenko [27]

Answer:

1.15 M

Explanation:

Step 1: Given data

  • Initial volume (V₁): 0.125 L
  • Initial concentration (C₁): 3.00 M
  • Final volume (V₂): 0.325 L
  • Final concentration (C₂): ?

Step 2: Calculate the final concentration of the solution

We want to prepare a dilute solution from a concentrated one by adding water. We can calculate the concentration of the dilute solution using the dilution rule.

C₁ × V₁ = C₂ × V₂

C₂ = C₁ × V₁/V₂

C₂ = 3.00 M × 0.125 L/0.325 L = 1.15 M

8 0
3 years ago
Suppose you are provided with a 30.86 g sample of potassium chlorate to perform this experiment. What is the mass of oxygen you
oee [108]

Answer:

The mass of oxygen is 12.10 g.

Explanation:

The decomposition reaction of potassium chlorate is the following:

2KClO₃(s) → 2KCl(s) + 3O₂(g)

We need to find the number of moles of KClO₃:

\eta_{KClO_{3}} = \frac{m}{M}

Where:

m: is the mass = 30.86 g

M: is the molar mass = 122.55 g/mol

\eta_{KClO_{3}} = \frac{30.86 g}{122.55 g/mol} = 0.252 moles                                      

Now, we can find the number of moles of O₂ knowing that the ratio between KClO₃ and O₂ is 2:3

\eta_{O_{2}} = \frac{3}{2}*0.252 moles = 0.378 moles

Finally, the mass of O₂ is:

m = 0.378 moles*32 g/mol = 12.10 g

Therefore, the mass of oxygen is 12.10 g.

I hope it helps you!

6 0
3 years ago
Muscle physiologists study the accumulation of lactic acid [ch3ch(oh)cooh] during exercise. food chemists study its occurrence i
Drupady [299]
The provided information are:
volume of 0.85 M lactic acid = 225 ml
volume of 0.68 M sodium lactate = 435 ml
Ka of the lactate buffer = 1.38 x 10⁻⁴
 The equation for dissociation of lactic acid is:
CH₃CH(OH)COOH(aq) + H₂O ⇄ CH₃CH(OH)COO⁻(aq) + H₃O⁺(aq)
The pH of buffer is calculated from Henderson-Hasselbalch equation, which is:
pH = pKa + log \frac{[conjugated base]}{[Acid]}
pKa = - log Ka = - log (1.38 x 10⁻⁴) = 3.86 
The number of moles of lactic acid and lactate are as follows:
n (Lactic acid) = 225 ml x 0.85 mmol/ml = 191.25 mmol
n (Lactate) = 435 ml x 0.68 mmol/ml = 295.8 mmol
The number of moles of lactic acid and lactate in total volume of the solution:
[CH₃CH(OH)COOH] = n (lactic acid) / 660 ml = 191.25 mmol / 660 ml = 0.29 M
[CH₃CH(OH)COO⁻] = n (lactate) / 660 ml = 0.45 M
pH = 3.86 + log \frac{0.45 M}{0.29 M} = 3.86 + 0.191 = 4.05
So the pH of given solution is 4.05
7 0
3 years ago
The reaction of an acid and a base will always produce what kind of salt?
Illusion [34]

Answer:

the answer of the question is c

7 0
2 years ago
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