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2 years ago

Calculate the molar mass of a gas at 388 torr and 45°c if 206 ng of gas occupies 0.206 ul. (

1 answer:

6 0

To make our solution more systematic, let's convert all units that is consistent with the units of R which is 0.0821 L-atm/mol-K.

For pressure: 760 torr = 1 atm
388 torr * 1 atm/760 torr = 0.5105 atm

For volume, 1 μL = 10⁻⁶ L
0.206 μL * 10⁻⁶ L/1 μL = 2.06×10⁻⁷ L

For temperature,
T = 45 + 273 = 318 K

For mass, 1 ng = 10⁻⁹ g
206 ng * 10⁻⁹ g/1 ng = 2.06×10⁻⁷ g

Assuming ideal gas,

PV=nRT
(0.5105 atm)(2.06×10⁻⁷ L) = n(0.0821 L-atm/mol-K)(318 K)
n = 4×10⁻⁹ mol

Molar mass = Mass/n = 2.06×10⁻⁷ g/4×10⁻⁹ mol
<em>Molar mass = 51.14 g/mol</em>

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consider the reaction between sulfite and a metal anion, X2-, to form the metal, X, and thiosulfate: 2 X2-(aq) + 2 SO32- + 3 H2O

Dafna11 [192]

Answer:

$E_{red}^{0}$ for X is -1.20 V

Explanation:

Oxidation: $2\times$ [ $X^{2-}(aq.)-2e^{-}\rightarrow X(s)$ ]

reduction: $2SO_{3}^{2-}(aq.)+3H_{2}O(l)+4e^{-}\rightarrow S_{2}O_{3}^{2-}(aq.)+6OH^{-}(aq.)$

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overall: $2X^{2-}(aq.)+2SO_{3}^{2-}(aq.)+3H_{2}O(l)\rightarrow 2X(s)+S_{2}O_{3}^{2-}(aq.)+6OH^{-}(aq.)$

So, $E_{cell}^{0}=E_{red}^{0}(SO_{3}^{2-}\mid S_{2}O_{3}^{2-})-E_{red}^{0}(X\mid X^{2-})$

or, $0.63=-0.57-E_{red}^{0}(X\mid X^{2-})$

or, $E_{red}^{0}(X\mid X^{2-})= -1.20$

So, $E_{red}^{0}$ for X is -1.20 V

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2 years ago

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You might be able to roughly estimate mass in a visual manner, but it could be easy to dismiss the fact that a mole or even a gr

qwelly [4]

The balanced equation is

CO(g) + 2H2 (g) -> CH3OH(l)

Which means, every 2 moles of H2 will react with 1 mole of CO.

Find the number of moles, n = m/Mr

CO = (1.5x10^-6)/(12) = 1.25 x 10^-7 mol

H2 = (6.8x10^-6)/(2) = 3.4 x 10^-6 mol

Since we have a mole ratio of 2 to 1, H2 value will be multipled by 2 as 2 moles of H2 react to 1 mole of CO.

3.4 x 10^-6 x 2 = 6.8 x 10^-6

So obviously we can see that H2 is in excess and only some H2 molecules will be left at the end of the reaction.

6.8 x 10^-6 - 1.25 x 10^-7 = 6.675 x 10^-6 Moles of unreacted H2

To find the molecules of this amount of moles, Multiply it by Avogadros number, 6.02 x 10^23

Because for every mole, there are 6.02 x 10^23 molecules

6.575 x 10^-6 x 6.02 x 10^23 = 3.96 x 10^ 18 Molecules of gas

8 0

1 year ago

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Would an alkali metal make a good replacement for tin in a tin can? Explain.

Igoryamba

Answer:

Explanation:

If tin is heated, it can react with alkalis' with the release of hydrogen.

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1 year ago

Up to a point, the elongation of a spring is directly proportional to the force applied to it. Once you extend the spring more t

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2 years ago

A solution is prepared by dissolving 10.0 g of NaBr and 10.0 g of Na2SO4 in water to make a 100.0 mL solution. This solution is

Colt1911 [192]

Answer:

$M_{Na^+}=1.36M$

$M_{Br^-}=1.58M$

Explanation:

Hello,

At first, it turns out convenient to compute the total moles of sodium that will be dissolved into the solution by considering the added amounts of sodium bromide and sodium sulfate:

$n_{Na^+}=n_{Na^+,NaBr}+n_{Na^+,Na_2SO_4}\\n_{Na^+,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molNa^+}{1molNaBr}=0.0971molNa^+\\n_{Na^+,Na_2SO_4}=10.0gNa_2SO_4*\frac{1molNa_2SO_4}{142gNa_2SO_4}*\frac{2molNa^+}{1molNa_2SO_4} =0.141molNa^+\\n_{Na^+}=0.0971molNa^++0.141molNa^+\\n_{Na^+}=0.238molNa^+$

Once we've got the moles we compute the final volume via:

$V=100.0mL+75.0mL=175.0mL*\frac{1L}{1000mL}=0.1750L$

Thus, the molarity of the sodium atoms turn out into:

$M_{Na^+}=\frac{0.238mol}{0.1750L} =1.36M$

Now, we perform the same procedure but now for the bromide ions:

$n_{Br^-}=n_{Br^-,NaBr}+n_{Br^-,AlBr_3}\\n_{Br^-,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molBr^-}{1molNaBr}=0.0971molBr^-\\n_{Br^-,AlBr_3}=0.0750L*0.800\frac{molAlBr_3}{L} *\frac{3molBr^-}{1molAlBr_3}=0.180molBr^- \\n_{Br^-}=0.0971molBr^-+0.180molBr^-\\n_{Br^-}=0.277molBr^-$

Finally, its molarity results:

$M_{Br^-}=\frac{0.277molBr^-}{0.1750L}=1.58M$

Best regards.

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1 year ago