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ololo11 [35]
4 years ago
15

To name the compound written as cobr2, you would write:

Chemistry
2 answers:
inna [77]4 years ago
8 0

cobalt(II) bromide. (apex)

Tcecarenko [31]4 years ago
7 0
Answer is: cobalt(II) bromide.

Bromine has oxidation number -1 (because it is nonmetal from goup XVII and gain one electron to have electron configuration as closestgas noble) , so cobalt must have oxidation number +2, because compound has neutral charge: 
x + 2·(-1) = 0.
x = +2.
Cobalt(II) bromide<span> red, solid</span><span> inorganic compound, soluble in water.</span>

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NaCl + H2SO4 ---&gt; Na2SO4 + HCl<br> Balance the double replacement chemical reaction.
VikaD [51]

Answer:

2NaCl+H2SO4-->Na2SO4+2HCl

Explanation:

There are two Na on the right, so put a 2 in front of NaCl on the left. This makes 3 Cl also, so put a 2 in front of HCl on the right. There are already 2 H on the left, so the equation is balanced.

7 0
4 years ago
Nuclear energy is associated with
maria [59]
Protons and neutrons
6 0
3 years ago
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Is tap water a compound, a homogenous mixture, or a heterogeneous mixture?
goldenfox [79]
It is a homogenous mixture. It is clear and does not seperate
4 0
4 years ago
What is the concentration of ammonia in a solution if 23.4 mL of a 0.117 M solution of HCl are needed to titrate a 100.0 mL samp
Georgia [21]

Answer: 0.0274 M

Explanation:-

The balanced chemical solution is:

NH_4OH(aq)+HCl(aq)\rightarrow NH_4Cl(aq)+H_2O(l)

According to the neutralization law,

n_1M_1V_1=n_2M_2V_2

where,

M_1 = molarity of HCl solution = 0.117 M

V_1 = volume of HCl solution = 23.4 ml

M_2 = molarity of NH_4OH solution = ?

V_2 = volume of NH_4OH solution = 100.0 ml

n_1 = valency of HCl = 1

n_2 = valency of NH_4OH = 1

1\times 0.117M\times 23.4=1\times M_2\times 100.0

M_2=0.0274

Therefore, the concentration of ammonia in a solution will be 0.0274 M

8 0
3 years ago
In Universe L, recently discovered by an intrepid team of chemists who also happen to have studied interdimensional travel, quan
Advocard [28]

Answer:

Manganese, Fifth transition element

[X] 3d⁶ 4s¹

Iron, Sixth transition element

[X] 3d⁶ 4s²

Explanation:

Complete Question

In Universe L, recently discovered by an intrepid team of chemists who also happen to have studied interdimensional travel, quantum mechanics works as it does in our universe, except that there are six d orbitals instead of the usual number we observe here. Use these facts to write the ground-state electron configurations of the sixth and seventh elements in the first transition series in Universe L. Note; you may use [X] to stand for the electron configuration of the noble gas at the end of the row before the first transition series.

Solution

In our universe, there are 5 d orbitals.

And according to Aufbau's principles that electrons fill the lower energy orbitals before they fill higher energy orbitals and Hund's Rule that states that electrons are fed singly to all the orbitals of a subshell before pairing occurs.

The fifth and sixth transition elements in our universe is then Manganese and Iron respectively.

Manganese - [Ar] 3d⁵ 4s²

Iron - [Ar] 3d⁶ 4s²

So, in the new universe L, where there are six d orbitals, for manganese, the fifth transition metal, because half filled orbitals are more stable than partially filled orbitals (that woukd have been rhe case if we leave 5 electrons on the 3d orbital), the 4s orbital is filled to half of its capacity and the one electron removed from the 4s is used to fill the six 3d orbital to half of its capacity too.

For the sixth transition element, the new extra electron just fills the lower energy 4s orbital, leaving the six 3d orbitals all half-filled.

Hence, they both have ground state configurations of

- Manganese, Fifth transition element

[X] 3d⁶ 4s¹

- Iron, Sixth transition element

[X] 3d⁶ 4s²

Hope this Helps!!!

7 0
3 years ago
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