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3 years ago

9

What is the molarity of ZnCl, that forms when 20.0 g of zinc completely reacts with CuCl2, according to the following reaction?

Assume a final
volume of 285 mL

Zn(s) + CuCl2(aq) → ZnCl2(aq) + Cu(s)

1 answer:

laila [671]3 years ago

8 0

1.61m :) use the photo below

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A standard 10.00 g mass is weighed on an analytical balance 100 times. The average and standard deviation obtained gives 10.12 ±

Mekhanik [1.2K]

Answer:

There was an improvement in accuracy. There was no change in precision.

Explanation:

<em>The average mass after recalibration is closer to the mass of the standard, </em>so the recalibration improved the accuracy<em> </em>(the measurement is closer to an accepted 'true' value).

The standard deviation did not change, so the precision (or how disperse the measurements are) was not affected.

5 0

3 years ago

How does the A Hreaction relate to the A He of molecules involved in a reaction?

igor_vitrenko [27]

Answer:

B. ΔHreaction = ΔH°f reactants- ΔH°f products

Explanation:

<em>Using Hess's law, it is possible to sum ΔH of several related reactions to find ΔH of a particular reaction</em>.

Having in mind Hess's law, ΔH°f is defined as the change in enthalpy during the formation of 1 mole of substance from its constituent elements (That is, pure elements, mono or diatomics, that have a ΔH° = 0).

For example, in ΔH°f of H₂O, the equation is:

H₂(g) + 1/2O₂(g) → H₂O(g)

The constituent elements with ΔH°f = 0 are H₂(g) and O₂(g).

Now, using Hess's law, you can sum the ΔH°f of substance in a reaction as, for example:

NaOH + HCl → H₂O + NaCl. ΔHr

The ΔH°f for each substance in the reaction is:

NaOH: Na + 1/2H₂ + 1/2O₂ → NaOH <em>(1)</em>

HCl: 1/2H₂ + 1/2Cl₂ → HCl <em>(2)</em>

H₂O: H₂ + 1/2O₂ → H₂O <em>(3)</em>

NaCl: Na + 1/2Cl₂ → NaCl <em>(4)</em>

The algebraic sum of (3) + (4) is -(ΔH°f reactants):

H₂ + 1/2O₂ + Na + 1/2Cl₂ → NaCl + H₂O ΔH°f reactants

This reaction - {(1)+(2)} ΔH°f products

NaOH + HCl → H₂O + NaCl.

ΔHr = ΔH°f reactants- ΔH°f products

In the example, we obtain this relationship that can be expanded for all reactions. Thus, right answer is:

<h3>B. ΔHreaction = ΔH°f reactants- ΔH°f products</h3>

8 0

3 years ago

Arrange the following chloroarenes in increasing order of their reactivity in nucleophilic substitution to form their correspond

pantera1 [17]

The question is incomplete. Complete question is attached below
..............................................................................................................................

Correct Answer: Option C i.e. I ~ III < IV < V < II

Reason:
During a nucleophilic subsitution reaction of chloroarenes, Cl- group is replaced by an nucleophile like OH-.

Order of reactivity, during such reactions depends on the electron density on carbon atom that is attached to Cl. Lower the electron density, greater will be the reactivity.

Among the provided chloroarenes, electron density on C atom will be minimum in case of compound II, because of presence of electron withdrawing group (-NO2) at ortho and para position. Due to this, there will be large number of resonating structures. This signifies greater electron de-localization, and hence largest reactivity for nucleophilic substitution reaction.

Followed by this, compound V will show greater reactivity, due to presence of -NO2 group at para and one of the ortho position. Compound IV will have less number of resonating structures as compared to compound II and V, hence it will display poor reactivity towards nucleophilic substitution reaction.

Finally, compound 1 and III will minimum reactivity towards nucleophilic substitution reaction, because -NO2 group present at meta position (compound III) will not participate in resonance.

6 0

3 years ago

Noble gas compounds like KrF, XeCl, and XeBr are used in excimer lasers. Draw an approximate molecular orbital diagram appropria

Aneli [31]

Answer:

Here's what I get.

Explanation:

The MO diagrams of KrBr, XeCl, and XeBr are shown below.

They are similar, except for the numbering of the valence shell orbitals.

Also, I have drawn the s and p orbitals at the same energy levels for both atoms in the compounds. That is obviously not the case.

However, the MO diagrams are approximately correct.

The ground state electron configuration of KrF is

$(1\sigma_{g})^{2}\, (1\sigma_{u}^{*})^{2} \, (2\sigma_{g})^{2} \, (2\sigma_{u}^{*})^{2} \, (3\sigma_{g})^{2} \, (1\pi_{u})^{4} \, (1\pi_{g}^{*})^{4} \, (3\sigma_{g}^{*})^{1}$

KrF⁺ will have one less electron than KrF.

You remove the antibonding electron from the highest energy orbital, so the bond order increases.

The KrF bond will be stronger.

6 0

4 years ago

How many grams of helium must be released to reduce the pressure to 65 atmatm assuming ideal gas behavior? Express the answer in

lyudmila [28]

The question is incomplete, here is the complete question:

How many grams of helium must be released to reduce the pressure to 65 atm assuming ideal gas behavior. Note : 334-mL cylinder for use in chemistry lectures contains 5.209 g of helium at 23°C.

<u>Answer:</u> The mass of helium released is 1.6 grams

<u>Explanation:</u>

We are given:

Mass of helium in the cylinder = 5.209 g

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

Or,

$PV=\frac{w}{M}RT$

where,

P = Pressure of the gas = 65 atm

V = Volume of the gas = 334 mL = 0.334 L (Conversion factor: 1 L = 1000 mL)

w = Weight of the gas = ?

M = Molar mass of helium gas = 4 g/mol

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = Temperature of the gas = $23^oC=[23+273]K=296K$

Putting values in above equation, we get:

$65atm\times 0.334L=\frac{w}{4g/mol}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 296K\\\\w=\frac{65\times 0.334\times 4}{0.0821\times 296}=3.573g$

Mass of helium released = (5.209 - 3.573) g = 1.636 g = 1.6 g

Hence, the mass of helium released is 1.6 grams

5 0

3 years ago