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3 years ago

12

The u.S. Mint produced pure silver quarter commemorative coins in 1992, each coin weighing 5.67 g. How many silver atoms are pre

sent in each coin?

1 answer:

PolarNik [594]3 years ago

4 0

Mass of each silver coin = 5.67 g

Molar mass of silver = 107.87 g/mol

Calculating the moles of silver from given mass and molar mass of silver:

$5.67 g Ag*\frac{1molAg}{107.87g} =0.0526molAg$

Calculating the atoms of silver in 0.05260 mol using the conversion factor, $1mol=6.022*10^{23}atoms$ :

$0.05260 mol*\frac{6.022*10^{23}atoms }{1mol} =3.17*10^{22}atoms$

Therefore, $3.17*10^{22}$ atoms of silver are present in each coin.

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Using the periodic table, choose the more reactive metal.<br> Ag or Te

Bad White [126]

The answer is Ag.................silver

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3 years ago

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Calculate the number of moles of Cl2 produced at equilibrium when 3.98 mol of PCl5 is heated at 283.9 deg celsius in a vessel ha

ddd [48]

Answer:

1.274 moles

Explanation:

The equation for the reaction can be represented as follows:

$PCl_5(g)$ ⇄ $PCl_3(g)$ + $Cl_2(g)$

K = 0.060

K = $\frac{[PCl_3][Cl_2]}{[PCl_5]}$

Concentration of $PCl_5(g)$ = $\frac{numbers of moles}{volume}$

Concentration of $PCl_5(g)$ = $\frac{3.98}{10.0}$

Concentration of $PCl_5(g)$ = 0.398 moles

If we construct an ICE table for the above equation; we have:

$PCl_5(g)$ ⇄ $PCl_3(g)$ + $Cl_2(g)$

Initial 0.398 0 0

Change - x + x + x

Equilibrium (0.398 - x) x x

K = $\frac{[PCl_3][Cl_2]}{[PCl_5]}$

K = $\frac{[x][x]}{[0.398-x]}$

K = $\frac{x^2}{0.398-x}$

0.060 = $\frac{x^2}{0.398-x}$

0.06(0.398-x) = x²

0.02388 - 0.060x = x²

x² + 0.060x - 0.02388 = 0 (quadratic equation)

a = 1; b= 0.06; c= -0.02388

Using quadratic formula;

= $\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

= $\frac{-0.06+/-\sqrt{(0.06)^2-4(1)(-0.02388)} }{(2*1)}$

= $\frac{-0.060+/-\sqrt{0.0036+0.09552} }{2}$

= $\frac{-0.06+/-\sqrt{0.09912} }{2}$

= $\frac{-0.06+/-0.3148}{2}$

= $\frac{-0.060+0.3148}{2}$ or $\frac{-0.060-0.3148}{2}$

= $\frac{0.2548}{2}$ or $\frac{-0.3748}{2}$

= 0.1274 or -0.1874

We go by the positive value which says:

[x] = 0.1274 M

number of moles = 0.1274 × 10.0

= 1.274 moles

∴ the number of moles of Cl₂ produced at equilibrium = 1.274 moles

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3 years ago

The reactants in a chemical equation have 3 carbon atoms, 6 oxygen atoms, and 4 hydrogen atoms. In order to follow the conservat

Anon25 [30]

Answer:

The same amount of atoms (3 carbon atoms, 6 oxygen atoms, and 4 hydrogen atoms).

Explanation:

According to the law of conservation of mass, when a chemical reaction takes place there is no synthesis or degradation of matter, that is, <em>matter can be neither created nor destroyed</em>. It can only be transformed.

Therefore, in a <u>chemical reaction the atoms are conserved</u>. That means that the elements that in definite proportions are forming a compound, will reorganize to form new compounds, and <u>the amount and type of atoms of the products will be the same as the reactants</u>.

In summary, if the reactants in a chemical equation have 3 carbon atoms, 6 oxygen atoms, and 4 hydrogen atoms, the products of the chemical equation will have 3 carbon atoms, 6 oxygen atoms, and 4 hydrogen atoms as well.

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2 years ago

Find the equilibrium value of [CO] if Kc=14.5 : CO (g) + 2H2 (g) ↔ CH3OH (g) Equilibrium concentrations: [H2] = 0.322 M and [CH3

Annette [7]

Answer:

The equilibrium value of [CO] is 1.04 M

Explanation:

Chemical equilibrium is the state to which a spontaneously evolving chemical system, in which a reversible chemical reaction takes place. When this situation is reached, it is observed that the concentrations of substances, both reagents and reaction products, they remain constant over time. That is, the rate of reaction of reagents to products is the same as that of products to reagents.

Reagent concentrations and products in equilibrium are related by the equilibrium constant Kc. Being:

aA + bB ⇔ cC + dD

$Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a} *[B]^{b} }$

Then this constant Kces equals the multiplication of the concentrations of the products raised to their stoichiometric coefficients between the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.

In this case:

$Kc=\frac{[CH_{3}OH ]}{[CO]*[H_{2} ]^{2} }$

You know:

Kc= 14.5
[H₂]= 0.322 M
[CH₃OH] =1.56 M

Replacing:

$14.5=\frac{1.56}{[CO]*0.322^{2} }$

Solving:

$[CO]=\frac{1.56}{14.5*0.322^{2} }$

[CO]= 1.04 M

The equilibrium value of [CO] is 1.04 M

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3 years ago

If you had 6 moles of CaCl2 and 5 moles of Na3PO4, which of these would be the limiting and excess reactant

faltersainse [42]

Answer:

Na3PO4 is excess reactant, CaCl2 is limiting reactant.

Explanation:

3CaCl2 + 2Na3PO4 ---> Ca3(PO4)2 + 6NaCl

from reaction : 3 mol 2 mol

given: 6 mol 5 mol (X)

X = (6*2)/3 = 4 mol Na3PO4

For 6 mol CaCl2 we need 4 mol Na3PO4, but we have 5 mol Na3PO4,

Na3PO4 is excess reactant, so CaCl2 is limiting reactant.

8 0

3 years ago