Your Answer Will Be Intensive Property
Answer:
1) Salts X and Y
2) The solubility of the salts
3) a) The solvent
b) The solvent temperature
Explanation:
1) The independent variable is the variable that is suspected to be the cause of the subject of the investigation
The given investigation is meant to investigate the solubility of different salts
Therefore, the solubility is expected to be dependent on the type of salt, and the independent variable is the type of salt, X or Y
2) The dependent variable is the effect meant to be observed in the investigation, which is the solubility of the salt in water at room temperature
3) The control variables are the variables which are held constant during the investigation, including;
a) The solvent used if the investigation; water
b) The temperature of the solvent; Room temperature
Molecular mass of H2O is: 16+2*1=18
so moles of 40g of H2O is: 40/18=2.22
Answer:
See solution.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to set up the formula for the calculation of the by-mass percentage of the metal:

Thus, we solve for the molar mass of the metal to obtain:

For the subsequent problems, we proceed as follows:
a.

b.

c.

Regards!
Answer:
2H⁺ + NO₃⁻ + 1e⁻ → NO₂ + H₂O
Explanation:
NO₃⁻ → NO₂
In left side, Nitrogen acts with +5 by oxidation number
In right side, the oxidation number is +4
This is a reduction reaction, because the oxidation number has decreased. So the N has gained electrons.
NO₃⁻ + 1e⁻ → NO₂
In acidic medium, we have to add water, where there are less oxygens to ballance the amount. We have 2 O in left side, and 3 O in right side, so we have to add 1 H₂O on left side.
NO₃⁻ + 1e⁻ → NO₂ + H₂O
Now that oxygens are ballanced, we have to ballance the hydrogens by adding protons in the opposite side
2H⁺ + NO₃⁻ + 1e⁻ → NO₂ + H₂O