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Gnom [1K]
2 years ago
5

(a) Give systematic names or the chemical formulas for the following. Remember, names must

Chemistry
1 answer:
lions [1.4K]2 years ago
7 0

Answer:

[Ni(CO)2C2O4]- Dicarbonyloxalatonickel(0)

K3[Fe(NCO)3(OCN)3-potassiumtricyanatotriisocyanatoferrate II

diammineaquahydroxoplatinum(II) ion- [Pt(NH3)2(H2O)(OH)]^+

Al[Co(CO3)2(CN)2- Aluminiumdicarbonatodicyanocobalt III

ethylenediaminedinitritozinc(II)- Zn(en)(NO2)2

Explanation:

The international Union of Pure and applied chemistry published a set rules for the nomenclature of inorganic compounds in 1985. This set of rules have been consistently revised ever since.

The rules stipulate the order for naming positive ions, negative ions and neutral ligands and the allowed order of preference for naming. It also stipulates the proper endings for ligands which are positive ions, negative ions or neutral ligands as the case may be.

The compounds named above were named in accordance with the revised IUPAC nomenclature for inorganic compounds.

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Choose the answer that best completes the following statement: When an aluminum atom reacts so as to attain a noble gas electron
boyakko [2]

The options

Select one:

a. a 3- ion forms.

b. the noble gas configuration of argon is achieved.

c. the result is a configuration of 1s2 2s2 2p6.

d. the atom gains five electrons.

Answer:

c. the result is a configuration of 1s2 2s2 2p6.

Explanation:

Aluminium atom has atomic number of 13 , hence the number of electron is 13 for a neutral atom of aluminium. When aluminium atom reacts with other elements it usually gives out three electron to attain the octet configuration.

The cation representation of aluminium is Al3+ because it has loss three electron to attain the octet rule. Aluminium will be left with 10 electrons after losing 3 of it electrons. The electronic configuration will be represented as follows after losing three electrons;

1S² 2S² 2P∧6 .

At this stage the octet rule has been achieved as it will be represented as

2  8.  The first energy shell now contains two electron and the second energy shell contains 8 electrons.

The configuration of  Neon has been formed in the process.

4 0
3 years ago
A 50.0-g sample of liquid water at 25.0 °c is mixed with 23.0 g of water at 79.0 °c. the final temperature of the water is ___
I am Lyosha [343]

<span><span>m1</span>Δ<span>T1</span>+<span>m2</span>Δ<span>T2</span>=0</span>

<span><span>m1</span><span>(<span>Tf</span>l–l<span>T<span>∘1</span></span>)</span>+<span>m2</span><span>(<span>Tf</span>l–l<span>T<span>∘2</span></span>)</span>=0</span>

<span>50.0g×<span>(<span>Tf</span>l–l25.0 °C)</span>+23.0g×<span>(<span>Tf</span>l–l57.0 °C)</span>=0</span>

<span>50.0<span>Tf</span>−1250 °C+23.0<span>Tf</span> – 1311 °C=0</span>

<span>73.0<span>Tf</span>=2561 °C</span>

<span><span>Tf</span>=<span>2561 °C73.0</span>=<span>35.1 °C</span></span>

8 0
3 years ago
Example of nobel gas​
yanalaym [24]

Answer:

helium , krypton,xenon,radon, argon are noble gasses

5 0
3 years ago
Read 2 more answers
Any help with these questions would help i'm lost
SSSSS [86.1K]

Answer:

1.  31.25 mL

2.  1.98 g/L

3.  0.45 g/mL

Explanation:

For each of the problems, you need to perform unit conversions.  You need to use the information given to you to convert to a specific unit.

1.  You need volume (mL).  You have density (g/mL) and mass (g).  Divide mass by density.  You will cancel out mL and be left with g.

(50.0 g)/(1.60 g/mL) = 31.25 mL

2.  You are given grams and liters.  You need to find density with units g/L.  This means that you have to divide grams by liters.

(0.891 g)/(0.450 L) = 1.98 g/L

3.  You have to find density again but this time with units g/mL.  Divide the given mass by the volume.

(10.0 g)/(22.0 mL) = 0.45 g/mL

5 0
3 years ago
C4H10 is an example of a (n) a. Molecular formula b. Structural formula c. Atomic formula d. Geometric formula​
Oxana [17]

Answer:

molecular formula

Explanation:

hope it helps

8 0
3 years ago
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