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uranmaximum [27]
2 years ago
11

How many mL of 0.013 M potassium hydroxide are required to reach the equivalence point in the titration of 75 mL 0.166 M hydrocy

anic acid?
Chemistry
1 answer:
umka21 [38]2 years ago
6 0

Answer:

957.7mL

Explanation:

Using the formula below;

CaVa = CbVb

Where;

Ca = concentration of acid (M)

Va = volume of acid (mL)

Cb = concentration of base (M)

Vb = volume of base (mL)

According to the information provided in this question:

Ca = 0.166 M

Cb = 0.013 M

Va = 75mL

Vb = ?

Using CaVa = CbVb

0.166 × 75 = 0.013 × Vb

12.45 = 0.013Vb

Vb =12.45/0.013

Vb = 957.7mL

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True

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All the food chains in an ecosystem that are connected together form a ?
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All food chain in an ecosystem that are connected to each other forms a food web....

Therefore the answer is food web

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Which type of chemical reaction occurs in 2KCIO3 → AKCI + 302?
Oksanka [162]

Answer:

Decomposition reaction

Explanation:

Chemical equation:

2KClO₃ → 2KCl + 3O₂

The given reaction is decomposition reaction.

Decomposition reaction:

It is the reaction in which one reactant is break down into two or more product.

AB → A + B

Synthesis reaction:

It is the reaction in which two or more simple substance react to give one or more complex product.

Single replacement:

It is the reaction in which one elements replace the other element in compound.

AB + C → AC + B

Double replacement:

It is the reaction in which two compound exchange their ions and form new compounds.

AB + CD  → AC +BD

8 0
3 years ago
Suppose 215 g of NO3- flows into a swamp each day. What volume of CO2 would be produced each day at 17.0°C and 1.00 atm?
charle [14.2K]

Answer:

The answer is "41.23 \ L\  N_2"

Explanation:

2 NO_3^{-} + 10 e^{-} + 12 H^{+} \longrightarrow N_2 + 6 H_2O\\\\= \frac{( 215 \ g \ NO_3^{-})}{(62.0049  \frac{\ g NO_3^{-}}{mol})} \times  \frac{(1 \ mol \ N_2}{ 2 \ mol \ NO_3^{-})}\\\\

=3.46746789 \times 0.5\\\\= 1.733 \ mol \ N_2 \\\\\to V = \frac{nRT}{P} \\\\= (1.733 \ mol) \times (0.08205746 \frac{L\ atm}{Kmol}) \times \frac{ (17 + 273) K}{(1.00 atm)}\\\\= 41.23

8 0
3 years ago
Two isotopes of lithium are found in nature Li6 has a mass of 6. 02u and Li7 has a mass of 7.02u . Use the atomic weight of lith
alexandr402 [8]

The isotope that is more abundant, given the data is isotope Li7

<h3>Assumption</h3>
  • Let Li6 be isotope A
  • Let Li7 be isotope B

<h3>How to determine whiche isotope is more abundant</h3>
  • Molar mass of isotope A (Li6) = 6.02 u
  • Molar mass of isotope B (Li7) = 7.02 u
  • Atomic mass of lithium = 6.94 u
  • Abundance of A = A%
  • Abundance of B = (100 - A)%

Atomic mass = [(mass of A × A%) / 100] + [(mass of B × B%) / 100]

6.94 = [(6.02 × A%) / 100] + [(7.02 × (100 - A)) / 100]

6.94 = [6.02A% / 100] + [702 - 7.02A% / 100]

6.94 = [6.02A% + 702 - 7.02A%] / 100

Cross multiply

6.02A% + 702 - 7.02A% = 6.94 × 100

6.02A% + 702 - 7.02A% = 694

Collect like terms

6.02A% - 7.02A% = 694 - 702

-A% = -8

A% = 8%

Thus,

Abundance of B = (100 - A)%

Abundance of B = (100 - 8)%

Abundance of B = 92%

SUMMARY

  • Abundance of A (Li6) = 8%
  • Abundance of B (Li7) = 92%

From the above, isotope Li7 is more abundant.

Learn more about isotope:

brainly.com/question/24311846

#SPJ1

4 0
2 years ago
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