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Brrunno [24]
3 years ago
13

4. Given the chemical formulas of the following compounds, name each compound and state the rules you used to determine each nam

e.
• RbF
• CuO
• (NH4)2C2O4
(Note: C2O4– is called oxalate.)
Answer:
Chemistry
2 answers:
earnstyle [38]3 years ago
3 0
I honestly dont even kno
marissa [1.9K]3 years ago
3 0

I am taking the same test! ;)

1. RbF --> Rubidium Fluoride

Change fluorine to fluoride

2. CuO --> Copper (II) Oxide

Change oxygen to oxide. Oxide has a charge of -2. Since no subscripts are written, it means they have the same opposite charge. So, we use Copper (II).

3. (NH₄)₂C₂O₄ --> Ammonium Oxalate

NH₄ is ammonia, but we change it to ammonium for poly atomic ions. 

I hope this helps!

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Oil is more dense than water. How will this affect the weight of objects that will float on it?
horrorfan [7]
Oil is more dense than alcohol, but less dense than water. The molecules that make up the oil are larger than those that that make up water, so they cannot pack as tightly together as the water molecules can. They take up more space per unit area and are less dense.
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A crime-scene is roped off with yellow tape. The rectangular area is 5.6 meters long 10.75 meters wide. What is the area of the
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7 0
3 years ago
What is the affect of increasing the water's mass?how does it reflect it's temperature?
Nikolay [14]
I assume what you're asking about is, how does the temperature changes when we increase water's mass, according the formula for heat ? 
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8 0
3 years ago
2) Which liberates the most energy?
evablogger [386]

Answer: F(g)+e^-\rightarrow F^-(g)

Explanation:

Electron gain enthalpy is defined as energy released on addition of electron to an isolated gaseous atom.  

The amount of energy released will be maximum when the tendency to attract electrons is maximum. As flourine has atomic number of 9 and has electronic configuration of 2,7. It can readily gain 1 electron to attain stable noble gas configuration and hence liberates maximum energy.

F(g)+e^-\rightarrow F^-(g)

4 0
3 years ago
Hydrogen sulfide burns form sulfur dioxide:
Helga [31]

Answer: 404.04 kJ.

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

moles of H_2S

\text{Number of moles}=\frac{26.7g}{34.1g/mol}=0.78moles

2H_2S(g)+3O_2(g)\rightarrow 2SO_2(g)+2H_2O(g)    \Delta H=-1036kJ

According to stoichiometry :

2 moles of H_2S on burning produces = 1036 kJ

Thus 0.78 moles of H_2S on burning produces =\frac{1036kJ}{2}\times 0.78=404.04

Thus the enthalpy change when burning 26.7 g of hydrogen sulfide is 404.04 kJ.

7 0
3 years ago
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