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Over [174]
3 years ago
9

What mass of znco3 contains 6.11×1022 o atoms? give your answer correctly to three significant digits?

Chemistry
2 answers:
nordsb [41]3 years ago
5 0

Using Avogadros number, we can get that 1 mole of an atom contain 6.022 x 10^23 atoms. Therefore we can use this conversion factor to get the number of moles:

moles ZnCO3 = 6.11 x 10^22 atoms * (1 mole / 6.022 x 10^23 atoms) = 0.10146 moles

 

The molar mass of ZnCO3 is about 125.39 g/mol, therefore the mass is:

mass ZnCO3 = 0.10146 moles * (125.39 g / mol)

<span>mass ZnCO3 = 12.72 g</span>

dexar [7]3 years ago
5 0

Answer is: mass of zinc carbonate is 4.241 grams.

N(O) = 6.11·10²²; number of oxygen atoms.

n(O) = N(O) ÷ Na (Avogadro constant).

n(O) = 6.11·10²² ÷ 6.022·10²³ 1/mol.

n(O) = 0.101 mol; amount of oxygen atoms.

One molecule of ZnCO₃ has three oxygen atoms: n(ZnCO₃) : n(O) = 1 : 3.

n(ZnCO₃) = 0.101 mol ÷ 3.

n(ZnCO₃) = 0.034 mol.

m(ZnCO₃) = n(ZnCO₃) · M(ZnCO₃).

m(ZnCO₃) = 0.034 mol · 125.39 g/mol.

m(ZnCO₃) = 4.241 g.

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Some SO2 and O2 are mixed together in a flask at 1100 K in such a way ,that at the instant of mixing, their partial pressures ar
kakasveta [241]

Answer:

The answer is "\bold{0.525\ \ atm^{-1}}"

Explanation:

Given equation:

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\Delta x = 2-(2+1)\\\\

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\left\begin{array}{cccc}I\ (atm)&1&0.5&0\\C\ (atm)&2x&-x&2x\\E\ (atm)     &1-2x&0.5-x&2x\end{array}\right

calculating the total pressure on equilibrium=  (1-2x)+(0.5-x)+2x \ atm\\\\

                                                                         = 1-2x+0.5-x+2x\\\\= 1+0.5-x\\\\=1.5-x\\

\therefore \\\\\to 1.50-x= 1.35 \\\\\to 1.50-1.35= x\\\\\to 0.15= x\\\\ \to  x= 0.15\\\\

calculating the pressure in  So_2:

= (1-2 \times 0.15)

= 1-0.30 \\\\ =0.70 \ atm

calculating the pressure in  O_2:

= (0.5- 0.15)\\\\= 0.35 \ atm \\

calculating the pressure in  So_3:

= (2 \times 0.15)\\\\= (.30) \ atm \\\\

Calculating the Kp at 1100 K:

= \frac{(Pressure(So_3))^2}{(Pressure(So_2))^2 \times Pressure(O_2)}\\\\= \frac{0.30^2}{0.70^2 \times 0.35}\\\\= \frac{0.30 \times 0.30 }{0.70\times 0.70 \times 0.35}\\\\= \frac{0.09 }{0.49\times 0.35} \\\\= \frac{0.09 }{0.1715} \\\\=  0.5247 \ \  or \ \  0.525 \ \ atm^{-1}  \\\\

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