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3 years ago

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An aluminum plate is 78 cm wide and 35 cm long. The mass of the plate is 44,226 g. Determine its thickness. [The density of lead

is 2.70g/cm^3]

1 answer:

Artemon [7]3 years ago

6 0

Answer:

i try to solve it for you, and i aplode a similer question for you

Explanation:

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How does the structure of water explain its high boiling point, high heat capacity, and high heat of vaporization?

nikklg [1K]

Water's high heat capacity<span> is a property caused by hydrogen bonding among </span>water<span> molecules. When </span>heat<span> is absorbed, hydrogen bonds are broken and </span>water <span>molecules </span>can<span> move freely. When the temperature of </span>water decreases, the hydrogen bonds are formed and release a considerable amount of energy.
<span>Water's heat of vaporization is around 540 cal/g at </span>100 °C<span>, water's boiling point.

</span>

7 0

3 years ago

When the following aqueous solutions are mixed together, a precipitate forms. Balance the net ionic equation in standard form fo

rewona [7]

<u>Answer:</u>

<u>For (a):</u> The balanced net ionic equation is $2Ag^{+}(aq)+S^{2-}(aq)\rightarrow Ag_2S(s)$ and the sum of coefficients is 4

<u>For (b):</u> The balanced net ionic equation is $Pb^{2+}(aq)+2Cl^{-}(aq)\rightarrow PbCl_2(s)$ and the sum of coefficients is 4

<u>For (c):</u> The balanced net ionic equation is $Ca^{2+}(aq)+CO_3^{2-}(aq)\rightarrow CaCO_3(s)$ and the sum of coefficients is

<u>For (d):</u> The balanced net ionic equation is $Ba^{2+}(aq)+2OH^{-}(aq)\rightarrow Ba(OH)_2(s)$ and the sum of coefficients is 4

<u>For (e):</u> The balanced net ionic equation is $Ag^{+}(aq)+Cl^{-}(aq)\rightarrow AgCl(s)$ and the sum of coefficients is 3

<u>Explanation:</u>

Net ionic equation is defined as the equations in which spectator ions are not included.

Spectator ions are the ones that are present equally on the reactant and product sides. They do not participate in the reaction.

For (a): Sodium sulfide and silver nitrate

The balanced molecular equation is:

$Na_2S(aq)+2AgNO_3(aq)\rightarrow 2NaNO_3(aq)+Ag_2S(s)$

The complete ionic equation follows:

$2Na^{+}(aq)+S^{2-}(aq)+2Ag^+(aq)+2NO_3^{-}(aq)\rightarrow 2Na^+(aq)+2NO_3^-(aq)+Ag_2S(s)$

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$2Ag^{+}(aq)+S^{2-}(aq)\rightarrow Ag_2S(s)$

Sum of the coefficients = [2 + 1 + 1] = 4

For (b): Lead(II) nitrate and sodium chloride

The balanced molecular equation is:

$2NaCl(aq)+Pb(NO_3)_2(aq)\rightarrow 2NaNO_3(aq)+PbCl_2(s)$

The complete ionic equation follows:

$2Na^{+}(aq)+2Cl^{-}(aq)+Pb^{2+}(aq)+2NO_3^{-}(aq)\rightarrow 2Na^+(aq)+2NO_3^-(aq)+PbCl_2(s)$

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$Pb^{2+}(aq)+2Cl^{-}(aq)\rightarrow PbCl_2(s)$

Sum of the coefficients = [2 + 1 + 1] = 4

For (c): Calcium nitrate and potassium carbonate

The balanced molecular equation is:

$K_2CO_3(aq)+Ca(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+CaCO_3(s)$

The complete ionic equation follows:

$2K^{+}(aq)+CO_3^{2-}(aq)+Ca^{2+}(aq)+2NO_3^{-}(aq)\rightarrow 2K^+(aq)+2NO_3^-(aq)+CaCO_3(s)$

As potassium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$Ca^{2+}(aq)+CO_3^{2-}(aq)\rightarrow CaCO_3(s)$

Sum of the coefficients = [1 + 1 + 1] = 3

For (d): Barium nitrate and sodium hydroxide

The balanced molecular equation is:

$2NaOH(aq)+Ba(NO_3)_2(aq)\rightarrow 2NaNO_3(aq)+Ba(OH)_2(s)$

The complete ionic equation follows:

$2Na^{+}(aq)+2OH^{-}(aq)+Ba^{2+}(aq)+2NO_3^{-}(aq)\rightarrow 2Na^+(aq)+2NO_3^-(aq)+Ba(OH)_2(s)$

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions

The net ionic equation follows:

$Ba^{2+}(aq)+2OH^{-}(aq)\rightarrow Ba(OH)_2(s)$

Sum of the coefficients = [2 + 1 + 1] = 4

For (e): Silver nitrate and sodium chloride

The balanced molecular equation is:

$NaCl(aq)+AgNO_3(aq)\rightarrow NaNO_3(aq)+AgCl(s)$

The complete ionic equation follows:

$Na^{+}(aq)+Cl^{-}(aq)+Ag^{+}(aq)+NO_3^{-}(aq)\rightarrow Na^+(aq)+NO_3^-(aq)+AgCl(s)$

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$Ag^{+}(aq)+Cl^{-}(aq)\rightarrow AgCl(s)$

Sum of the coefficients = [1 + 1 + 1] = 3

8 0

3 years ago

If the energy difference between two electronic states is 214.68 kJ / mol , calculate the frequency of light emitted when an ele

atroni [7]

${\qquad\qquad\huge\underline{{\sf Answer}}}$

Here we go ~

Energy difference btween the two electronic states can be expressed as :

${ \qquad \sf \dashrightarrow \: \Delta E = h\nu}$

[ h = planks constant, ${\: \nu }$ = frequency ]

$\qquad \sf \dashrightarrow \:214.68 = 39.79 \times 10 {}^{ - 14} \times \nu$

$\qquad \sf \dashrightarrow \: \nu = \cfrac{214.68}{39.79 \times 10 {}^{ - 4} }$

$\qquad \sf \dashrightarrow \: \nu = \cfrac{214.68}{39.79 } \times 10 {}^{14}$

$\qquad \sf \dashrightarrow \: \nu \approx 5.395 \times10 {}^{14} \:\:hertz$

5 0

1 year ago

Calculate the equilibrium number of vacancies per cubic meter for copper at 1000K. The energy for vacancy formation is 0.9eV/ato

nexus9112 [7]

Answer:

Therefore the equilibrium number of vacancies per unit cubic meter =2.34×10²⁴ vacancies/ mole

Explanation:

The equilibrium number of of vacancies is denoted by $N_v$ .

It is depends on

total no. of atomic number(N)
energy required for vacancy
Boltzmann's constant (k)= 8.62×10⁻⁵ev K⁻¹
temperature (T).

$N_v=Ne^{-\frac{Q_v}{kT} }$

To find equilibrium number of of vacancies we have find N.

$N=\frac{N_A\ \rho}{A_{cu}}$

Here ρ= 8.45 g/cm³ =8.45 ×10⁶m³

$N_A$ = Avogadro Number = 6.023×10²³

$A_{Cu}$ = 63.5 g/mole

$N=\frac{6.023\times 10^{23}\times 8.45\times 10^{6}}{63.5}$

$=8.01\times 10^{28$ g/mole

Here $Q_v$ =0.9 ev/atom , T= 1000k

Therefore the equilibrium number of vacancies per unit cubic meter,

$N_v=( 8.01\times 10^{28}) e^{-(\frac{0.9}{8.62\times10^{-5}\times 1000})$

=2.34×10²⁴ vacancies/ mole

3 0

3 years ago

Object A has a mass of 3,600 kilograms. Object B has a mass of 600 kilograms. The weight of Object B will be ________ times the

Natalija [7]

Hi the answer is 1/6, because 1/6 of 3,600 is 600, so it's 1/6. Have a great day!

7 0

3 years ago