Oxygen gas produced : 0.7 g
<h3>Further explanation</h3>
Given
10.0 grams HgO
9.3 grams Hg
Required
Oxygen gas produced
Solution
Reaction⇒Decomposition
2HgO(s)⇒2Hg(l)+O₂(g)
Conservation of mass applies to a closed system, where the masses before and after the reaction are the same
mass of reactants = mass of products
mass HgO = mass Hg + mass O₂
10 g = 9.3 g + mass O₂
mass O₂ = 0.7 g
Answer:
The current temperature is -15° (15° below zero)
Explanation:
The temperature drops 10°:
T-10°
It will reach 25° below zero:
T - 10° = -25°
We add 10° in both members of the equation:
T - 10° +10° = -25° +10°
The equation is simplified as follows:
T = -25° +10°
T = -15°
The integer -15° can be expressed as 15° below zero.
Answer:
Yes, Pb3(PO4)2.
Explanation:
Hello there!
In this case, according to the given balanced chemical reaction, it is possible to use the attached solubility series, it is possible to see that NaNO3 is soluble for the Na^+ and NO3^- ions intercept but insoluble for the Pb^3+ and PO4^2- when intercepting these two. In such a way, we infer that such reaction forms a precipitate of Pb3(PO4)2, lead (II) phosphate.
Regards!
Answer:
This question appears incomplete
Explanation:
There is no such element known as "Ballardium (Bu)" in the periodic table. However, there are elements with a bit of similarity in spellings and pronunciation such as Beryllium (Be) which is found in group 2 (meaning it is an alkali earth metal), Berkelium (Bk) which is an actinide (meaning it is radioactive) and Vanadium (V) which is found in group 5 of the periodic table (meaning it's a transition metal).
Answer:
Explanation:
Hello there!
In this case, according to the given information of the solubility of copper chloride, as the maximum amount of this salt one can dissolve without having a precipitate, we infer that since just 73 grams are actually dissolved, the following amount will remain solid as a precipitate:
Best regards!
