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3 years ago

A 5.00 cm^3 sample of gold has a mass of 96.5 g. Calculate the density of gold.

Chemistry

1 answer:

Sholpan [36]3 years ago

8 0

Answer:

$\boxed {\boxed {\sf 19.3 \ g/cm^3}}$

Explanation:

We are asked to calculate the density of gold. The density of a substance is its mass per unit volume. It is calculated using the following formula.

$\rho= \frac{m}{v}$

The mass of the sample is 96.5 grams. The volume of the sample is 5.00 cubic centimeters.

m= 96.5 g
v= 5.00 cm³

Substitute the values into the formula.

$\rho= \frac{ 96.5 \ g}{5.00 \ cm^3}$

Divide.

$\rho= 19.3 \ g/cm^3$

The density of gold is 19.3 grams per cubic centimeter.

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What is the maximum amount of HCl, in grams, that can be produced if 37.5 g of BCl3 and 60.0 g of H2O are reacted according to t

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Ooooh boy alright. So, this may or may not be a limited reactant problem so we need to first find out of it is.

First, how many moles of each substance are there

the molar mass of BCl3 is 117.17 grams so 37.5 g / 117.17 is ~ .32 mol.
The molar mass of H2O is 18.02 so 60 / 18.02 is ~ 3.33 mol.

Now, for every 1 mole of BCl3, there are 3 moles of HCl created. Therefore, BCl3 can create ~ .96 moles.
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But, there is not enough BCl3 to support that 3.33 moles, only enough for .96 moles, therefore BCl3 is the limiting reactant. Now, to answer the question, simply multiply .96 moles by the molar mass of HCl.

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