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natta225 [31]
3 years ago
10

____Cl2+____H2=____HCl

Chemistry
1 answer:
Genrish500 [490]3 years ago
6 0

Answer:

<u>   </u>H2 + <u>2</u><u> </u> Cl2 = <u>2</u><u> </u> HCl2

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Kaillie was talking and was distracted as she stirred sugar into her tea. when she finally noticed, there was a few crystals of
AleksandrR [38]

Answer:

Supersaturated

Explanation:

The tea has absorbed and dissolved as much sugar as it could. If there is sugar left at the bottom, it means the solution is supersaturated because it can't absorb any more.

5 0
3 years ago
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Which atomic model states that it is impossible to know the exact location of electrons around the nucleus?
dedylja [7]

Answer:

the Bohr model, an electron's position is known precisely because it orbits the nucleus in a fixed path. In the electron cloud model, the electron's position cannot be known precisely. Only its probable location can be known.

8 0
3 years ago
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Use the molar bond enthalpy data in the table to estimate the value of Δ∘rxn
MakcuM [25]

Answer:

ΔH°rxn = - 433.1 KJ/mol

Explanation:

  • CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)

⇒ ΔH°rxn = 4ΔH°HCl(g) + ΔH°CCl4(g) - 4ΔH°Cl2(g) - ΔH°CH4(g)

∴ ΔH°Cl2(g) = 0 KJ/mol.....pure element in its reference state

∴ ΔH°CCl4(g) = - 138.7 KJ/mol

∴ ΔH°HCl(g) = - 92.3 KJ/mol

∴ ΔH°CH4(g) = - 74.8 KJ/mol

⇒ ΔH°rxn = 4(- 92.3 KJ/mol) + (- 138.7 KJ/mol) - 4(0 KJ/mol) - (- 74.8 KJ/mol)

⇒  ΔH°rxn = - 369.2 KJ/mol - 138.7 KJ/mol - 0 KJ/mol + 74.8 KJ/mol

⇒ ΔH°rxn = - 433.1 KJ/mol

4 0
3 years ago
Read 2 more answers
Why it is advisided to freshly prepare the ferrous sulphate solution for experiment
chubhunter [2.5K]
Here is a site my buddie has to help you. Well co-owner..
https://www.quora.com/Why-is-fresly-prepared-FeSO4-required-for-the-ring-test
5 0
3 years ago
For an alloy that consists of 91.0 g copper, 111 g zinc, and 7.51 g lead, what are the concentrations (a) of Cu (in at%), (b) of
viktelen [127]

Answer:

\% Cu=45.2\%\\\\\% Zn=53.6\%\\\\\% Pb=1.2\%

Explanation:

Hello!

In this case, given the masses of copper, zinc and lead, it is possible to compute the moles via their atomic masses first:

n_{Cu}=\frac{91.0gCu}{63.55g/mol}=1.432mol\\\\ n_{Zn}=\frac{111gZn}{65.41g/mol}=1.697mol\\\\n_{Pb}=\frac{7.51gPb}{207.2g/mol}=0.0362mol\\

Now, we compute the atomic percentages as shown below:

\% Cu=\frac{1.432}{1.432+1.697+0.0362}*100\% =45.2\%\\\\\% Zn=\frac{1.697}{1.432+1.697+0.0362}*100\%=53.6\%\\\\\% Pb=\frac{0.0362}{1.432+1.697+0.0362}*100\%=1.2\%

Best regards!

4 0
3 years ago
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