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Leto [7]
3 years ago
5

A gold nugget has a density of 38.6g/cm3 and a mass of 270.2. what is its volume?​

Chemistry
2 answers:
ArbitrLikvidat [17]3 years ago
8 0

270.2/38.6 = 7

Volume is 7.

aivan3 [116]3 years ago
5 0

Density is given by the equation D=m/V, were D is density, m is mass in grams, and V is volume in cubic centimeters.


In this problem, we have density and we have mass so we can plug into the equation and solve for V.

38.6=270.2/V

<em>*Multiply both sides by V*</em>

38.6V=270.2

<em>*Divide both sides by 38.6*</em>

V=7


The volume of the gold nugget is 7cm3.


Hope this helps!!

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Enter your answer in the provided box. From the data below, calculate the total heat (in J) needed to convert 0.304 mol of gaseo
diamong [38]

Answer:

-35,281.5 J

Explanation:

To convert the gaseous ethanol to liquid ethanol, three steps will occur. First, it will lose heat and the temperature will decrease until its boiling point, so from 300.0°C to 78.5°C. Thus, more heat will be lost, but now, with the temperature constant, so the gas will be converted to liquid. And then, the liquid will lose heat to decrease the temperature from 78.5°C to 25.0°C.

The total heat loss is the sum of the heats of each step. Because the heat is being removed from the system, it's negative. The first and last step occurs with a change in temperature, and so the heat is calculated by:

Q = m*c*ΔT

Where m is the mass, c is the specific heat of the gas (first step) or liquid (last step), and ΔT the temperature variation (final - initial). The mass of ethanol is the molar mass 46.07 g/mol multiplied by the number of moles, so:

m = 46.07 * 0.304 = 14.00 g

The second step occurs without a change in temperature, and the heat is then:

Q = -n*ΔH°vap

Where n is the number of moles, ΔH°vap is the heat of vaporization, and the minus signal indicates that the heat is being lost. Then, the heat of each step is:

Q1 = 14.00*1.43*(78.5 - 300,0) = -4434.43 J

Q2 = -0.304*40.5 = -12.312 kJ = -12312 J

Q3 = 14.00*2.45*(25.0 - 78.5) = -1835.05 J

Q = Q1 + Q2 + Q3

Q = -35,281.5 J

5 0
3 years ago
What is the maximum mass of h2o that can be produced by combining 68.6 g of each reactant?
Sauron [17]
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7 0
3 years ago
Gaseous ICl (0.20 mol) was added to a 2.0 L flask and allowed to decompose at a high temperature:
Ne4ueva [31]

Answer:

The Kc is 1.36 (but this is not an option, may be the options are wrong, or may be I was .. Thanks!)

Explanation:

Let's think all the situation.

               2 ICl(g)   ⇄   I₂(g)    +    Cl₂(g)

Initially      0.20              -               -

Initially I have only 0.20 moles of reactant, and nothing of products. In the reaction, an x amount of compound has reacted.

React          x              x/2               x/2

Because the ratio is 2:1, in the reaction I have the half of moles.

So in equilibrium I will have

           (0.20 - x)          x/2             x/2

Notice that I have the concentration in equilibrium so:

0.20 - x = 0.060

x = 0.14

So in equilibrium I have formed 0.14/2 moles of I₂ and H₂ (0.07 moles)

Finally, we have to make, the expression for Kc and remember that must to be with concentration in M (mol/L).

As we have a volume of 2L, the values must be /2

Kc = ([I₂]/2 . [H₂]/2) / ([ICl]/2)²

Kc = (0.07/2 . 0.07/2) / (0.060/2)²

Kc = 1.225x10⁻³ / 9x10⁻⁴

Kc = 1.36

8 0
3 years ago
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