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3 years ago

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Using the ideal gas law, determine the amount of gas, in moles, of a 8.32 L sample of hydrogen gas, under a temperature of 310 K

and a pressure of 1.7 atm.

1 answer:

AleksAgata [21]3 years ago

5 0

Answer:

0.56 moles

Explanation:

Ideal gas equation

PV = nRT

P = 1.7 atm

T = 310K

V = 8.32L

R = 0.08205 L atm / K.mol

PV = nRT

n = PV / RT

n = (1.7 * 8.32) / (0.08205 * 310)

n = 14.144 / 25.4355

n = 0.556

n = 0.56 moles

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The molarity is 6.0 M.

Explanation:

<em>The </em><em>concentration</em><em> of a solution is the amount of solute present in a given amount of solvent, or a given amount of solution.</em>

In this problem, we are asked to convert percent by mass to molarity, which are two different concentration units.

<em>The </em><em>percent by mass </em><em>is the ratio of the mass of a solute to the mass of the solution, multiplied by 100 percent.</em> On the other hand, <em>molarity (M)</em><em> is defined as the number of moles of solute per liter of solution.</em>

If we have 17.15 percent chloride ion, that means that we have 17.15 grams of chloride ion in 100 grams of solution. The density provided is the density of the solution, so we calculate how many mL correspond to 100 grams of solution:

1.23 g ----------- 1 mL

100 g ------------ <u>x= 81.3 mL</u>

Therefore, 17.5 grams of chloride ion are contained in 81.3 mL. Now we need to convert these grams into moles.

The atomic mass of chlorine is 35.5 g/mol:

35.5 g --------------- 1 mol

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This 0.49 moles are contained in 81.13 mL, the definition of molarity says that this moles are contained in a liter (or what is the same, 1000 mL):

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Answer:

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In this case, according to the required half-reaction, we start by setting it up from bismuth (V) oxide ion to bismuth (III) ion:

$BiO_3^-\rightarrow Bi^{3+}$

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Thereafter, we realize three water molecules are needed on the right in order to balance the oxygens and consequently 6 hydrogen atoms on the left to balance hydrogen:

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Now, since the balance is is basic media, we add six molecules of hydroxide ions in order to produce water with the hydrogen ones:

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Then, we accommodate the waters to obtain:

$3H_2O+(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+6OH^-$

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