The density of gold is 19.3 g/cm3. What is the volume of a 575 gram bar of pure gold?

Chemistry

2 answers:

tiny-mole [99]3 years ago

6 0

Answer:

Volume = 29.79 cm³

Explanation:

Gnoma [55]3 years ago

3 0

Answer:

The answer to your question is: Volume = 29.79 cm³

Explanation:

Data

Density of gold = 19.3 g/cm3

Mass = 575 grams

Formula

density = mass / volume

Volume = mass/density

Volume = 575 g / 19.3 g/cm3

Volume = 29.79 cm³

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drek231 [11]

Answer:

Explanation:

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2 years ago

Write a balanced half-reaction for the reduction of bismuth oxide ion to bismuth ion in basic aqueous solution. Be sure to add p

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Answer:

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

Explanation:

Hello there!

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

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

Thus, next realize that the oxidation state of Bi in BiO3^- is 5+ because oxygen is 2- (-2*3+x=-1;x=-1+6;x=+5), so we obtain:

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

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:

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

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

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