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zubka84 [21]
3 years ago
13

When balancing equations does po4 counts as one element or on p and 4 o?

Chemistry
2 answers:
Butoxors [25]3 years ago
4 0
PO4 is not one element, It is one ION having one P and four O
Radda [10]3 years ago
3 0
As one element .because po4 is phosphate for e.g.: NaPO4
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The kinetic molecular theory states that all particles of an ideal gas are
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THE KINETIC MOLECULAR THEORY STATES THAT ALL PARTICLES OF AN IDEAL GAS ARE IN CONSTANT MOTION AND EXHIBITS PERFECT ELASTIC COLLISIONS.

Explanation:

An ideal gas is an imaginary gas whose behavior perfectly fits all the assumptions of the kinetic-molecular theory. In reality, gases are not ideal, but are very close to being so under most everyday conditions.

The kinetic-molecular theory as it applies to gases has five basic assumptions.

  • Gases consist of very large numbers of tiny spherical particles that are far apart from one another compared to their size.
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  • Collisions between gas particles and between particles and the container walls are elastic collisions.
  • The average kinetic energy of gas particles is dependent upon the temperature of the gas.
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3 years ago
If you add a neutron to an atom how does it change ?
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Answer:

Its atomic mass increases by 1

An isotope of that element is fotmed with mass differences by 1

Explanation:

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3 years ago
If a piece of cadmium with a mass of 37.60 g and a temperature of 100.0 oC is dropped into 25.00 cc of water at 23.0 oC, what wi
zlopas [31]

Answer:

T_{eq}=28.9\°C

Explanation:

Hello!

In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:

Q_{Cd}+Q_{W}=0

Thus, we insert mass, specific heat and temperatures to obtain:

m_{Cd}C_{Cd}(T_{eq}-T_{Cd})+m_{W}C_{W}(T_{eq}-T_{W})=0

In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:

T_{eq}=\frac{m_{Cd}C_{Cd}T_{Cd}+m_{W}C_{W}T_{W}}{m_{Cd}C_{Cd}+m_{W}C_{W}}

Now, we plug in to obtain:

T_{eq}=\frac{37.60g*0.232\frac{J}{g\°C}*100.00\°C+25.00g*4.184\frac{J}{g\°C}*23.0\°C}{37.60g*0.232\frac{J}{g\°C}+25.00g*4.184\frac{J}{g\°C}}\\\\T_{eq}=28.9\°C

NOTE: since the density of water is 1g/cc, we infer that 25.00 cc equals 25.00 g.

Best regards!

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3 years ago
What is the best definition of solar radiation?(Giving brainliest)
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Answer:

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