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Xelga [282]
3 years ago
14

What is the difference between the endpoint and the equivalence point?

Chemistry
1 answer:
True [87]3 years ago
5 0

Endpoint . this is when the color of the indicator changes  while  equivalence is when added titrant is chemically equivalent completely to the analyte sample  

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Pls help me i dont know this
r-ruslan [8.4K]

Answer:

1. 12.6 moles

2. 8.95 moles

3.  2A + 5B → 3C

4. 48 moles

Explanation:

1. 2Fe + 3Cl₂ → 2FeCl₃

We assume the chlorine in excess. Ratio is 2:2

2 moles of Fe, can produce 2 moles of chloride

12.6 moles of Fe will produce 12.6 moles of chloride.

2. 2Fe + 3Cl₂ → 2FeCl₃

For the same reaction, first of all we need to convert the mass to moles:

500 g . 1mol / 55.85 g = 8.95 mol

As ratio is 2:2, the moles we have are the same, that the produced

4. The reaction for the combustion is:

2C₂H₆ (g) + 7O₂ (g) → 4CO₂ (g) + 6H₂O (l)

We assume the oxygen in excess.

Ratio is 2:6, so 2 mol of ethane produce 6 moles of water

Therefore 16 moles of ethane may produce (16 .6) / 2 = 48 moles

3 0
3 years ago
Based on the activity series, which one of the reactions below will occur (Spontaneous)? A. Mn (s) + NiCl2 (aq) → MnCl2 (aq) + N
kap26 [50]

Answer:

Mn (s) + NiCl2 (aq) → MnCl2 (aq) + Ni

Explanation:

The order of displacement of metals from aqueous solution by another metal is defined by the activity series of metals.

The activity series arranges metals in order of reactivity and increasing electrode potentials. The less negative the electrode potential of a metal is, the less reactive it is and the lower it is found in the activity series.

Nickel has a less negative electrode potential than manganese hence it is displaced from an aqueous solution of its salt by manganese spontaneously.

8 0
3 years ago
Sodium borohydride, NaBH4, and boron trifluoride, BF3, are compounds of boron. What are the shapes around boron in the borohydri
STatiana [176]
Compound 1: Sodium borohydride
In sodium borohydride (NaBH4), B is a central metal.
Electronic configuration of B is 1s2 2s2 2p1.
B undergoes sp3 hybridization in NaBH4, to generate 4 hybrid orbitals. These hybrid orbitals, forms sigma bond with 4 'H' atoms. Due to this, the structure of sodium borohydride in tetrahedral.
........................................................................................................................

Compound 2: B<span>oron trifluoride
</span>In boron trifluoride (BF3), B is a central metal.
Electronic configuration of B is 1s2 2s2 2p1.
B undergoes sp2 hybridization in NaBH4, to generate 3 hybrid orbitals. These hybrid orbitals, forms sigma bond with 3 'H' atoms. Due to this, the structure of <span>boron trifluoride</span> is <span>triangular planner</span>.
7 0
3 years ago
Read 2 more answers
Brainliest for correct answer please show all work
Korvikt [17]

Answer:

1) Na₃PO₄ + 3 KOH ➙ 3 NaOH + K₃PO₄

2) MgF₂ + Li₂CO₃➙ MgCO₃ + 2 LiF

3) P₄ + 3 O₂➙ 2 P₂O₃

Explanation:

To balance an equation, ensure that the number of atoms of each element is equal on both sides.

Reactants would be those on the left of the arrow while products are on the right of the arrow.

Balance O and H atoms last.

<u>Question 1:</u>

__Na₃PO₄ + __KOH ➙ __NaOH + __K₃PO₄

Reactants: 3Na, 1P, 1K, 5O, 1H

Products: 1Na, 1P, 3K, 5O, 1H

<em>Balance the number of Na:</em>

__Na₃PO₄ + __KOH ➙ 3 NaOH + __K₃PO₄

Reactants: 3Na, 1P, 1K, 5O, 1H

Products: 3Na, 1P, 3K, 7O, 3H

<em>Balance the number of K:</em>

__Na₃PO₄ + 3 KOH ➙ 3 NaOH + __K₃PO₄

Reactants: 3Na, 1P, 3K, 7O, 3H

Products: 3Na, 1P, 3K, 7O, 3H

<em>The equation is now balanced.</em>

<u>Question 2:</u>

__MgF₂ + __Li₂CO₃➙ __MgCO₃ + __LiF

Reactants: 1Mg, 2F, 2Li, 1C, 3O

Products: 1Mg, 1F, 1Li, 1C, 3O

<em>Balance</em><em> </em><em>n</em><em>u</em><em>m</em><em>b</em><em>e</em><em>r</em><em> </em><em>of</em><em> </em><em>L</em><em>i</em><em> </em><em>and</em><em> </em><em>F</em><em> </em><em>atoms</em><em>:</em>

__MgF₂ + __Li₂CO₃➙ __MgCO₃ + 2 LiF

Reactants: 1Mg, 2F, 2Li, 1C, 3O

Products: 1Mg, 2F, 2Li, 1C, 3O

<em>The</em><em> </em><em>equation</em><em> </em><em>is</em><em> </em><em>now</em><em> </em><em>balanced</em><em>.</em>

<u>Question 3:</u>

__P₄ + __O₂➙ __P₂O₃

Reactants: 4P, 2O

Products: 2P, 3O

<em>Balance</em><em> </em><em>the</em><em> </em><em>number</em><em> </em><em>of</em><em> </em><em>P</em><em> </em><em>atoms</em><em>:</em>

__P₄ + __O₂➙ 2 P₂O₃

Reactants: 4P, 2O

Products: 4P, 6O

<em>Balance</em><em> </em><em>the</em><em> </em><em>number</em><em> </em><em>of</em><em> </em><em>O</em><em> </em><em>atoms</em><em>:</em>

__P₄ + 3 O₂➙ 2 P₂O₃

Reactants: 4P, 6O

Products: 4P, 6O

<em>The</em><em> </em><em>equation</em><em> </em><em>is</em><em> </em><em>now</em><em> </em><em>balanced</em><em>.</em>

4 0
2 years ago
Can someone answer this
Yuliya22 [10]
The correct answer is B 25
8 0
3 years ago
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