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erastovalidia [21]
3 years ago
13

How many grams of O2(g) are needed to completely burn 46.0 g C3H8(g)?

Chemistry
1 answer:
Veronika [31]3 years ago
8 0

Answer:

167,3 grams of O2 (g)

Explanation:

According to the following balanced equation:

C3H8 + 5 02 ---> 3 CO2 + 4 H20

We calculate the mass of the moles:

5 mol O2 = 16grams/mol x2x 5= 160 grams/mol

1 mol C3H8= 3 x12 grams/mol + 8 x 1 gram/mol= 44 grams/mol

If 44 grams C3H8 react with--------- 160 grams O2

46 gramsC3H8 react with    --------X = (46x160)/44 =167, 3 grams O2

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Convert 25.4 grams of barium phosphate, Ba3(PO4)2 to formula units.
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To solve this questions you first need to find the number of moles of barium phosphate you have. The molar mass of barium phosphate is 601.93g/mol.
24.4/601.83 = 0.0402 moles barium phosphate
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3 years ago
What is the % dissociation of a solution of acetic acid if at equilibrium the solution has a pH = 4.74 and a pKa = 4.74?
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Answer:

\% diss = 50\%

Explanation:

Hello there!

In this case, when considering weak acids which have an associated percent dissociation, we first need to set up the ionization reaction and the equilibrium expression:

HA\rightleftharpoons H^++A^-\\\\Ka=\frac{[H^+][A^-]}{[HA]}

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Ka=\frac{x^2}{[HA]_0-x} =10^{-4.74}=1.82x10^{-5}

Thus, it is possible to find x given the pH as shown below:

x=10^{-pH}=10^{-4.74}=1.82x10^{-5}M

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\frac{(1.82x10^{-5})^2}{[HA]_0-1.82x10^{-5}} =1.82x10^{-5}\\\\\frac{1.82x10^{-5}}{[HA]_0-1.82x10^{-5}} =1\\\\

[HA]_0=3.64x10^{-5}M

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\% diss=\frac{x}{[HA]_0}*100\% \\\\\% diss=\frac{1.82x10^{-5}M}{3.64x10^{-5}M}*100\% \\\\\% diss = 50\%

Best regards!

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