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MAVERICK [17]
3 years ago
12

Which of the following have the greasiest ionization energy Bi, As, K or Ga?

Chemistry
1 answer:
lara [203]3 years ago
6 0
The answer to this is As
You might be interested in
A sample of mass 6.814 grams is added to another sample weighing 0.08753 grams.
Feliz [49]

Answer:

17.5609g

Explanation:

According to the question, a sample of mass 6.814 grams is added to another sample weighing 0.08753 grams. That is weight of sample 1 + weight of sample 2;

6.814 + 0.08753 = 6.90153grams

Next, the subsequent mixture is then divided into exactly 3 equal parts i.e. 6.90153grams divided by 3

= 6.90153/3

= 2.30051grams.

One of the equal parts is 2.30051grams, which is then multiplied by 7.6335 times I.e. 2.30051 × 7.6335 = 17.5609grams

Therefore, the final mass is 17.5609grams

3 0
3 years ago
The total volume required to reach the endpoint of a titration required more than the 50 mL total volume of the buret. An initia
Margaret [11]

Answer:

The endpoint volume is 50.52 ±  0.14 mL

Explanation:

In a titration always is necessary to subtract the blank volume to the titrant volume to obtain the real volume of the titrant. Thus in this case, the total endpoint volume is the sum of the initial volume delivered and the second volume delivered, minus the blank volume:

V = (49.16±0.06 mL) + (1.69±0.04 mL) - (0.33±0.04 mL)

V = (49.16 + 1.69 - 0.33) ± (0.06+0.04+0.04) mL

V = 50.52 ±  0.14 mL

It is necessary to consider the sum of the errors too.

7 0
3 years ago
How many moles are in 25 g KMnO4
Ronch [10]

Answer:

0.158 moles KMnO4

Explanation:

According to the Periodic Table,

K = 39.10 g/mol

Mn = 54.94 g/mol

O = 16.00 g/mol

KMnO4 = 39.10 g/mol + 54.94 g/mol + 4(16.00 g/mol) = 158.04 g/mol

25.0 grams KMnO4              1 mole

-----------------------------  x --------------------------  = 0.158 moles KMnO4

                                          158.04 grams

8 0
2 years ago
What are three similarities between biotic and abiotic factors?
Viktor [21]
1. Both part of the ecosystem
2. There are biotic objects on abiotic objects ( caterpillars on trees ) and abiotic objects on biotic things ( pollen on bees )
3. Both are made of atoms
6 0
3 years ago
A cell was prepared by dipping a Cu wire and a saturated calomel electrode into 0.10 M CuSO4 solution. The Cu wire was attached
lara [203]

Answer:

a)  cu2+ + 1Hg (l) 1Cl- equilibrium cu (s) + Hg2Cl2 (s)

b)  0.068 V.

Explanation:

A) Cu2+ + 2e- euilibrium cu (s)

 Hg2Cl2 + 2e- equilibrium 2Hg (l) + 1cl-

Cell Reaction: cu2+ + 1Hg (l) 1Cl- equilibrium cu (s) + Hg2Cl2 (s)

B) To calculate the cell voltage

E = E_o Cu2+/Cu - (0.05916 V / 2) log 1/Cu2+

putting values we get

 = 0.339V + (90.05916V/2)log(0.100) = 0.309V

 E_cell = E Cu2+/Cu - E SCE = 0.309 V - 0.241 V = 0.068V.

6 0
2 years ago
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