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olchik [2.2K]
3 years ago
5

If you have a gold brick that is 2cm by 3cm by 4cm and has density of 19.3g/cm3, what is its mass?

Chemistry
1 answer:
natulia [17]3 years ago
4 0
Volume = a x a x a  

V = 2 cm x 3 cm x 4 cm => 24 cm³

Density = 19.3 g/cm³

Mass = ?

Therefore:

m = D x V

m = 19.3 x 24

m = 463.2 g
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Answer:

Adding H₂O(g) to the system.

Explanation:

  • Le Châtelier's principle states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.

<u><em>1) Removing H₂O(g) from the system:</em></u>

  • This will decrease the concentration of the reactants side, so the reaction will be shifted to the left side to suppress the removal of H₂O(g) from the system.

<u><em>2) Adding CH₃CHO(g) to the system :</em></u>

  • This will increase the concentration of the products side, so the reaction will be shifted to the left side to suppress the adding CH₃CHO(g) to the system.

<u><em>3) Removing C₂H₂(g) from the system:</em></u>

  • This will decrease the concentration of the reactants side, so the reaction will be shifted to the left side to suppress the removal of C₂H₂(g) from the system.

<u><em>4) Adding H₂O(g) to the system:</em></u>

  • This will increase the concentration of the reactants side, so the reaction will be shifted to the right side to suppress the addition of H₂O(g) to the system.
  • <u><em>So, it is the right choice.</em></u>
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3 years ago
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What must be changed, temperature or heat energy, during condensation explain in 2-3 sentences
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Answer:

Change occur in heat energy.

Explanation:

Change occur in the heat energy of molecules during condensation process. The molecules of gas releases its heat energy and converted into liquid state. With this heat energy, they are active and escape from each other but when the heat energy is removed from them, the attractive forces between these gas molecules are formed which convert them into liquid state so change in heat energy occur in the condensation process.

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Explanation:

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At sea level, where the pressure was 104 kPa and temperature 21.1 ºC, a certain mass of air occupies 2.0 m3 . To what volume wil
Romashka [77]

Answer:

The volume of air at where the pressure and temperature are  52 kPa, -5.0 ºC is 3.64 m^3.

Explanation:

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 104 kPa

P_2 = final pressure of gas = 52 kPa

V_1 = initial volume of gas = 2.0m^3

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 21.1^oC=273+21.1=294.1K

T_2 = final temperature of gas = -5.0^oC=273+(-5.0)=268 K

Now put all the given values in the above equation, we get:

\frac{104 kPa\times 2.0m^3}{294.1 K}=\frac{52 kPa\times V_2}{268 K}

V_2=3.64 m^3

The volume of air at where the pressure and temperature are  52 kPa, -5.0 ºC is 3.64 m^3.

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How did mendeleev orginally organize the element in his periodic table
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