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3 years ago

Which equation represents the combined gas law?

Chemistry

1 answer:

stealth61 [152]3 years ago

4 0

Answer:

P1V1/T1= P2V2/T2

Explanation:

Combined gas law involves Boyle's law and Charles law altogether with the formula of Boyle's law as P1V1=P2V2

formula for charles law as V1/T1=V2/T2

so when combined form P1V1/T1=P2V2/T2

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Naturally occurring zirconium exists are five isotopes. axe with a mass of 89.905 u(51.45%); Zr with a mass of 90.906 u (11.22%)

Anton [14]

Isotope 1: 89.905 * 51.45 = 4625.61225 / 100 = 46.2561225
Isotope 2: 90.906 * 11.22 = 1019.96532 / 100 = 10.1996532
Isotope 3: 91.905 * 17.15 = 1576.17175 / 100 = 15.7617075
Isotope 4: 93.906 * 17.38 = 1632.08628 / 100 = 16.3208628
Isotope 5: 95.908 * 2.08 = 268.5424 / 100 = 2.685424

46.2561225 + 10.1996532 + 15.7617075 + 16.3208628 + 2.685424 = 91.22377
actual mass Zr = about 91.22

8 0

3 years ago

(will give brainliest) show your work. How many grams of Copper(I) nitrate, CuNO3 are required to produce 88.0 grams of aluminum

ValentinkaMS [17]

Based on the stoichiometry of the reaction, 156.114 g of CuNO3 are required to produce 88.0 grams of aluminum nitrate, Al(NO3)3.

<h3>What is stoichiometry of a reaction?</h3>

The stoichiometry of a reaction is the molar ratio in which reactants combine to form products.

The stoichiometry of the reaction shows that 6 moles of copper (i) nitrate produces 2 moles of aluminium nitrate.

molar mass of Copper(I) nitrate, CuNO3 = 126 g

molar mass of aluminum nitrate, Al(NO3)3 = 213 g

88.0 g of aluminum nitrate, Al(NO3)3 = 88.0/213 moles = 0.413 moles

0.413 moles of Al(NO3)3 will be produced by 0.413 ×6/3 = 1.239 moles of CuNO3

Mass of 1.239 moles of CuNO3 = 1.239 × 126 = 156.114 g of CuNO3

Therefore, 156.114 g of CuNO3 are required to produce 88.0 grams of aluminum nitrate, Al(NO3)3.

Learn more about stoichiometry at: brainly.com/question/16060223

Therefore, 156.114 g of CuNO3

4 0

1 year ago

If 0.251moles of H2O gas are produced, how many liters of oxygen gas was used?

Alborosie

Answer:

o.251 prduces 45.7L of oxogen

Explanation:

hope this helps

4 0

3 years ago

Calculate the electric double layer thickness of a alumina colloid in a dilute (0.1 mol/dm3) CsCI electrolyte solution at 30 °C.

Ad libitum [116K]

Explanation:

The given data is as follows.

Concentration = 0.1 $mol/dm^{3}$

= 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions

= $6.022 \times 10^{25} ions/m^{3}$

T = $30^{o}C$ = (30 + 273) K = 303 K

Formula for electric double layer thickness ( $\lambda_{D}$ ) is as follows.

$\lambda_{D}$ = $\frac{1}{k} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}$

where, $n^{o}$ = concentration = $6.022 \times 10^{25} ions/m^{3}$

Hence, putting the given values into the above equation as follows.

$\lambda_{D}$ = $\sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}$

= $\sqrt \frac{78 \times 8.854 \times 10^{-12} c^{2}/Jm \times 1.38 \times 10^{-23}J/K \times 303 K}{2 \times 6.022 \times 10^{25} ions/m^{3} \times (1)^{2} \times (1.6 \times 10^{-19}C)^{2}}$