The density of mercury, the only liquid metal at room temperature, is 13.6 g/ml.

aluminum has a density or 2.7g/cm3. copper has a density of 8.96/cm3. which metal would you choose to build a model airplane?

erik [133]

Answer: Aluminium

Explanation: Aluminium metal has a lower density than copper. So, for the same volume of metal used to build a model airplane, the aluminium plane would be very lightweight while that of copper would be heavy. The lightweight airplane will fly easily.

7 0

3 years ago

Based on the wintertime La Niña weather map, what do you think Florida's temperature and precipitation would be like during a wi

nika2105 [10]

Answer:

Temperature decreases and precipitation increases.

Explanation:

Temperature of Florida decreases and precipitation increases during a winter El Niño event. Due to El Niño event, precipitation occurs above average in Florida during Fall-Winter-Spring season. Storminess also increases during this El Niño event which increases the threat of severe weather in Florida during El Niño winters.

7 0

3 years ago

What is the volume of 18.9 g of a liquid that has a density of 0.956 g/ml

Alex_Xolod [135]

The volume is 19.76987448 by taking the known variables mass=18.9g and density=0.956g/ml
To get volume you divide the mass by the density which gives you about
19.77 ml in volume

6 0

3 years ago

At 298 K, what is the Gibbs free energy change (ΔG) for the following reaction?

9966 [12]

Answer:

(a). The Gibbs free energy change is 2.895 kJ and its positive.

(b). The Gibbs free energy change is 34.59 J/mole

(c). The pressure is 14924 atm.

(d). The Gibbs free energy of diamond relative to graphite is 4912 J.

Explanation:

Given that,

Temperature = 298 K

Suppose, density of graphite is 2.25 g/cm³ and density of diamond is 3.51 g/cm³.

$\Delta H\ for\ diamond = 1.897 kJ/mol$

$\Delta H\ for\ graphite = 0 kJ/mol$

$\Delta S\ for\ diamond = 2.38 J/(K mol)$

$\Delta S\ for\ graphite = 5.73 J/(K mol)$

(a) We need to calculate the value of $\Delta G$ for diamond

Using formula of Gibbs free energy change

$\Delta G=\Delta H-T\Delta S$

Put the value into the formula

$\Delta G= (1897-0)-298\times(2.38-5.73)$

$\Delta G=2895.3$

$\Delta G=2.895\ kJ$

The Gibbs free energy change is positive.

(b). When it is compressed isothermally from 1 atm to 1000 atm

We need to calculate the change of Gibbs free energy of diamond

Using formula of gibbs free energy

$\Delta S=V\times\Delta P$

$\Delta S=\dfrac{m}{\rho}\times\Delta P$

Put the value into the formula

$\Delta S=\dfrac{12\times10^{-6}}{3.51}\times999\times10130$

$\Delta S=34.59\ J/mole$

(c). Assuming that graphite and diamond are incompressible

We need to calculate the pressure

Using formula of Gibbs free energy

$\beta= \Delta G_{g}+\Delta G+\Delta G_{d}$

$\beta=V(-\Delta P_{g})+\Delta G+V\Delta P_{d}$

$\beta=\Delta P(V_{d}-V_{g})+\Delta G$

Put the value into the formula

$0=\Delta P(\dfrac{12\times10^{-6}}{3.51}-\dfrac{12\times10^{-6}}{2.25})\times10130+2895.3$

$0=-0.0194\Delta P+2895.3$

$\Delta P=\dfrac{2895.3}{0.0194}$

$\Delta P=14924\ atm$

(d). Here, $C_{p}=0$

So, The value of $\Delta H$ and $\Delta S$ at 900 k will be equal at 298 K

We need to calculate the Gibbs free energy of diamond relative to graphite

Using formula of Gibbs free energy

$\Delta G=\Delta H-T\Delta S$

Put the value into the formula

$\Delta G=(1897-0)-900\times(2.38-5.73)$

$\Delta G=4912\ J$

Hence, (a). The Gibbs free energy change is 2.895 kJ and its positive.

(b). The Gibbs free energy change is 34.59 J/mole

(c). The pressure is 14924 atm.

(d). The Gibbs free energy of diamond relative to graphite is 4912 J.

7 0

3 years ago

How many atoms of argon occupy 30.0L

Airida [17]

5.18mL i hope this helps i hope this does to!

6 0

3 years ago