a compound of hydrogen and carbon, such as any of those which are the chief components of petroleum and natural gas.
"the rain is rich in benzene and hydrocarbons"
Answer: option 1) <span>
2Cr3+(aq)+6Cl-(aq) ------> 2Cr(s)+3Cl2(g)Explanation:1) Write the oxidation half-reaction:
![2Cl^-(aq)---\ \textgreater \ Cl_2(g)+2e^-](https://tex.z-dn.net/?f=2Cl%5E-%28aq%29---%5C%20%5Ctextgreater%20%5C%20Cl_2%28g%29%2B2e%5E-)
2) Write the reduction half-raction:
![Cr^{3+}(aq)+3e^{-}---\ \textgreater \ Cr(s)](https://tex.z-dn.net/?f=Cr%5E%7B3%2B%7D%28aq%29%2B3e%5E%7B-%7D---%5C%20%5Ctextgreater%20%5C%20Cr%28s%29)
3) Multiply each half-reaction by the appropiate coefficient to equal the number of electrons of both half-reactions.
![6Cl^{-}(aq)---\ \textgreater \ 3Cl_2(g)+6e^{-} 2Cr^{3+}(aq)+6e^{-}---\ \textgreater \ 2Cr(s)](https://tex.z-dn.net/?f=6Cl%5E%7B-%7D%28aq%29---%5C%20%5Ctextgreater%20%5C%203Cl_2%28g%29%2B6e%5E%7B-%7D%0A%0A2Cr%5E%7B3%2B%7D%28aq%29%2B6e%5E%7B-%7D---%5C%20%5Ctextgreater%20%5C%202Cr%28s%29)
4) Add both half-reactions
![2Cr^{3+}+6Cl^{-}(aq)---\ \textgreater \ 2Cr(s) +3Cl_2(g)](https://tex.z-dn.net/?f=2Cr%5E%7B3%2B%7D%2B6Cl%5E%7B-%7D%28aq%29---%5C%20%5Ctextgreater%20%5C%202Cr%28s%29%20%2B3Cl_2%28g%29)
And that is the answer. You can count the atoms and charges on every side and check they are equal.
</span>
Answer:
Protons and neutrons are found in the nucleus of the atom.
Explanation:
Protons and neutrons are found in the nucleus of the atom, while electrons are outside of the nucleus.
Answer:
Explanation:
The clue of this question is to find the molar mass of the <em>diprotic acid</em> and compare witht the molars masses of the choices' acid to identify the formula of the diprotic acid.
The procedure is:
- Find the number of moles of the base: LiOH
- Use stoichionetry to infere the number of moles of the acid.
- Use the formula molar mass = mass in grams / number of moles, to find the molar mass of the diprotic acid.
- Compare and conclude.
<u>Solution:</u>
<u>1. Number of moles of the base, LiOH:</u>
- M = n / V in liter ⇒ n = M × V = 0.100 M × 40.0 ml × 1 liter / 1,000 ml = 0.004 mol LiOH.
<u>2. Stoichiometry:</u>
Since this a neutralization reaction of a diprotic acid with a mono hydroxide base (LiOH), the mole ratio at the second equivalence point is: 2 mol of base / 1 mole of acid; because each mole of LiOH releases 1 mol of OH⁻, while each mole of diprotic acid releases 2 mol of H⁺.
Hence, 0.004 mol LiOH × 1 mol acid / 2 mol LiOH = 0.002 mol acid.
<u>3. Molar mass of the acid:</u>
- molar mass = mass in grams / number of moles = 0.300 g / 0.002 mol = 150. g/mol
<u>4. Molar mass of the possible diprotic acids:</u>
a. H₂Se: 2×1.008 g/mol + 78.96 g/mol = 80.976 g/mol
b. H₂Te: 2×1.008 g/mol + 127.6 g/mol = 129.616 g/mol
c. H₂C₂O₄ ≈ 2×1.008 g/mol + 2×12.011 g/mol + 4×15.999 g/mol ≈ 90.034 g/mol
d. H₂C₄H₄O₆ = 6×1.008 g/mol + 4×12.011 g/mol + 6×15.999 g/mol = 150.086 g/mol ≈ 150 g/mol.
<u>Conclusion:</u> since the molar mass of H₂C₄H₄O₆ acid is 150 g/mol, you conclude that is the diprotic acid whose 0.300 g were titrated with 40.0 ml of 0.100 M LiOH solution.