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Arisa [49]
3 years ago
15

30. a. How do the properties of metals differ from those

Chemistry
1 answer:
Ahat [919]3 years ago
3 0

Explanation:

Metals are elements that ionized by loss of electrons.

Ionic and molecular compounds are usually non-metals.

Properties of metals:

  • Metals have free mobile electrons and the metallic bonding ensures that.
  • They are usually electropositive and freely looses their electrons.
  • None of the metal is soluble without a chemical change occurring.
  • They are ductile and malleable.
  • Metals are good conductor or heat and electricity in their free uncombined state.
  • They are lustrous.

B. The specific property of metals accountable for their unusual electrical conductivity is due to the presence of free mobile electrons in their lattices.

learn more:

Metals brainly.com/question/2474874

#learnwithBrainly

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If 20.0 g of NaOH is added to 0.750 L of 1.00 M Cd(NO₃)₂, how many grams of Cd(OH)₂ will be formed in the following precipitatio
bulgar [2K]

Answer:

m_{Cd(OH)_2}=36.6 gCd(OH)_2

Explanation:

Hello.

In this case, for the given chemical reaction, in order to compute the grams of cadmium hydroxide that would be yielded, we must first identify the limiting reactant by computing the yielded moles of that same product, by 20.0 grams of NaOH (molar mass = 40 g/mol) and by 0.750 L of the 1.00-M solution of cadmium nitrate as shown below considering the 1:2:1 mole ratios respectively:

n_{Cd(OH)_2}^{by\ NaOH}=20.0gNaOH*\frac{1molNaOH}{40gNaOH} *\frac{1molCd(OH)_2}{2molNaOH} =0.25molCd(OH)_2\\\\n_{Cd(OH)_2}^{by\ Cd(NO_3)_2}=0.750L*1.00\frac{molCd(NO_3)_2}{L}*\frac{1molCd(OH)_2}{1molCd(NO_3)_2}  =0.75molCd(OH)_2

Thus, since 20.0 grams of NaOH yielded less of moles of cadmium hydroxide, NaOH is the limiting reactant, therefore the mass of cadmium hydroxide (molar mass = 146.4 g/mol) is:

m_{Cd(OH)_2}=0.25molCd(OH)_2*\frac{146.4gCd(OH)_2}{1molCd(OH)_2} \\\\m_{Cd(OH)_2}=36.6 gCd(OH)_2

Best regards.

4 0
3 years ago
Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reaction:P4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°
miskamm [114]

<u>Answer:</u> The \Delta H^o_{rxn} for the reaction is -1835 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

P_4(g)+10Cl_2(g)\rightarrow 4PCl_5(s)      \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) PCl_5(s)\rightarrow PCl_3(g)+Cl_2(g)    \Delta H_1=157kJ   ( × 4)

(2) P_4(g)+6Cl_2(g)\rightarrow 4PCl_3(g)     \Delta H_2=-1207kJ

The expression for enthalpy of the reaction follows:

\Delta H^o_{rxn}=[4\times (-\Delta H_1)]+[1\times \Delta H_2]

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(4\times (-157))+(1\times (-1207))=-1835kJ

Hence, the \Delta H^o_{rxn} for the reaction is -1835 kJ.

6 0
3 years ago
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