The question is incomplete, the complete question is;
Suna passes an electric current through a sample of clear, colorless, and odorless liquid. As the experiment continues, bubbles form, and the volume of liquid decreases. Suna collects samples of two colorless, odorless gases that bubble out of the liquid. One of the gases burns. Neither the original liquid nor the other gas burns. Which is the best explanation of her results? The electric current changed some of the sample to gas even though the sample was not breaking down. Therefore, the original liquid is a compound. The electric current released a gas that was odorless and colorless, like the original sample. Therefore, the original liquid is an element. The sample was broken down by the electric current and formed a new substance that could burn. Therefore, the original liquid is a compound. The sample lost some of its volume, but the gas still had the same chemical makeup as the original sample. Therefore, the original liquid is an element.
Answer:
The sample was broken down by the electric current and formed a new substance that could burn. Therefore, the original liquid is a compound.
Explanation:
When electric current is passed through a compound, the compound may become broken down to release its constituents. We refer to this phenomenon as electrolysis. We can now say that the substance has been 'decomposed' electrolytically.
Since the original sample was decomposed to yield a gas that could burn and one that couldn't burn even though the original sample couldn't burn, then the original sample is a compound.
.0002345 I believe this is correct
I'm going to go with Physical.
Answer:
Balancing Nuclear Equations
To balance a nuclear equation, the mass number and atomic numbers of all particles on either side of the arrow must be equal.
Explanation:
follows:
6
3
Li
+
2
1
H
→
4
2
He
+
?
To balance the equation above for mass, charge, and mass number, the second nucleus on the right side must have atomic number 2 and mass number 4; it is therefore also helium-4. The complete equation therefore reads:
6
3
Li
+
2
1
H
→
4
2
He
+
4
2
He
Or, more simply:
6
3
Li
+
2
1
H
→
2
4
2
He
image
Lithium-6 plus deuterium gives two helium-4s.: The visual representation of the equation we used as an example.
Compact
