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I am Lyosha [343]
3 years ago
11

Cuales son los nombres de los siguientes iones NH4+1 Y NO3-1

Chemistry
1 answer:
saw5 [17]3 years ago
5 0
NH4+1 ammonium
NO3-1 nitrate

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Explain how matter changes from one state to another such as liquid
beks73 [17]

Answer:

When heated or cooled, matter can transform from one state to another. When you heat ice (a solid), it turns into water (a liquid). MELTING is the term for this transformation. When water is heated, it becomes steam (a gas).

Explanation:

i hope thats the answer you want

8 0
2 years ago
When lithium reacts with bromine to form the compound LiBr, each lithium atom:
yKpoI14uk [10]
(3) loses one electron and becomes positively charged 
Lithium has one valence electron and Bromine has seven. Therefore Lithium will give up its one to Bromine for both to have an octet 
5 0
3 years ago
Read 2 more answers
An object has a mass of 3.50 grams and a density of 5.61 g/mL. What is the volume of this object? *
Elina [12.6K]

Answer:

<h2>0.62 mL</h2>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

volume =  \frac{mass}{density} \\

From the question we have

volume =  \frac{3.50}{5.61}  \\  = 0.6238859...

We have the final answer as

<h3>0.62 mL</h3>

Hope this helps you

5 0
2 years ago
An unknown piece of metal weighing 95.0 g is heated to 98.0°C. It is dropped into 250.0 g of water at 23.0°C. When equilibrium i
lutik1710 [3]

Answer:

C_{metal}=126.6\frac{J}{g\°C}

Explanation:

Hello!

In this case, when two substances at different temperature are put in contact and an equilibrium temperature is attained, we can evidence that the heat lost by the hot substance (metal) is gained by the cold substance (water) and we can write:

Q_{metal}=-Q_{water}

Which can be also written as:

m_{metal}C_{metal}(T_{EQ}-T_{metal})=-m_{water}C_{water}(T_{EQ}-T_{water})

Thus, since we need the specific heat of the metal, we solve for it as shown below:

C_{metal}=\frac{m_{water}C_{water}(T_{EQ}-T_{water})}{-m_{metal}(T_{EQ}-T_{metal})} \\\\C_{metal}=\frac{250.0g*4.184\frac{J}{g\°C}(29.0\°C-98.0\°C)}{95.0g(29.0\°C-23.0\°C)} \\\\C_{metal}=126.6\frac{J}{g\°C}

Best regards.

7 0
2 years ago
Justify air is a mixture not a compound
Andreas93 [3]
A compound has to be chemically bonded, however, air is not chemically bonded.
This can be proven by freezing air. By freezing air, it yields different liquids at different temperature. Liquid nitrogen has a different boiling point than liquid oxygen.
If air was a compound, they would all have a single boiling point and a single freezing point.


Hope this helps :)
4 0
3 years ago
Read 2 more answers
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