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3 years ago

What are the variables of Boyle's law?

Chemistry

2 answers:

Llana [10]3 years ago

6 0

Boyles law states P1V1=P2V2
P being pressure and V being volume
So volume and pressure

marusya05 [52]3 years ago

6 0

Answer:

Pressure and volume

Explanation:

Boyle's law was discovered by Robert Boyle in the seventeenth century, defined the basis for explaining the relationship between pressure and volume that exists in a gas.

Through a series of experiments he discovered that a gas under a constant temperature decreases the volume when we increase the pressure, and its volume increases when we reduce the pressure.

Boyle's law: P1 x V1 = P2 x V2

The pressure (P) of a gas is inversely proportional to the volume (V) of a gaseous mass at a constant temperature.

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g, Assuming the precipitate is totally insoluble in water, which aqueous ions will be present in the solution (collected in the

Allushta [10]

Answer:

Cl⁻, Na⁺, OH⁻

Explanation:

The titration is:

CuCl₂(aq) + 2 NaOH(aq) → Cu(OH)₂(s) + 2 NaCl(aq)

In solution, before the reaction, the ions are Cu²⁺ and Cl⁻. The addition of NaOH (Na⁺ + OH⁻) produce the precipitation of Cu²⁺ forming Cu(OH)₂(s). When you reach the equivalence point, there is no Cu²⁺ because precipitates completely. All OH⁻ ions reacts when are added but when Cu²⁺ is finished, excess OH⁻ ions still in solution helping to detect the equivalence point.

Thus, ions present after the equivalence point are:<em> Cl⁻, Na⁺</em> (Don't react, spectator ions), and <em>OH⁻</em>.

3 0

3 years ago

Calculate the density of each object

poizon [28]

Object one is 5.2 g/cm3
object two is 3.46g/ml

7 0

3 years ago

Sodium phosphate dodecahydrate, na3po4(12h2o), is crystalline salt that is water soluble. describe what is present in an aqueous

Elza [17]

Hey there!:

Sodium cations and phosphate anions .

hope this helps!

7 0

3 years ago

Need ASAP will give brainlist

Leokris [45]

Answer:

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2 years ago

Write the balanced standard combustion reaction for the c5h7on3s2

AlexFokin [52]

Answer:

4C5H7ON3S2 + 25O2 —> 20CO2 + 14H2O + 6N2 + 4S2

Explanation:

When C5H7ON3S2 under go Combustion, the following are obtained as illustrated below:

C5H7ON3S2 + O2 —> CO2 + H2O + N2 + S2

Now, let us balance the equation. This is illustrated below:

C5H7ON3S2 + O2 —> CO2 + H2O + N2 + S2

There are 3 atoms of N on the left side and 2 atoms on the right side. It can be balance by putting 4 in front of C5H7ON3S2 and 6 in front of N2 as shown below:

4C5H7ON3S2 + O2 —> CO2 + H2O + 6N2 + S2

There are 20 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 20 in front of CO2 as shown below:

4C5H7ON3S2 + O2 —> 20CO2 + H2O + 6N2 + S2

There are 28 atoms on H on the left side and 2 atoms on the right side. It can be balance by putting 14 in front of H2O as shown below:

4C5H7ON3S2 + O2 —> 20CO2 + 14H2O + 6N2 + S2

There are 8 atoms of S on the left side and 2 atoms on the right side. It can be balance by putting 4 in front of S2 as shown below:

4C5H7ON3S2 + O2 —> 20CO2 + 14H2O + 6N2 + 4S2

Now, there are a total of 54 atoms of O on the right side and 6 atoms on the left side

It can be balance by putting 25 in front of O2 as shown below:

4C5H7ON3S2 + 25O2 —> 20CO2 + 14H2O + 6N2 + 4S2

Now, we can see that the equation is balanced.

4 0

3 years ago