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2 years ago

How does altitude affect the freezing, melting, and boiling points of water?

2 answers:

8 0

Increasing the temperature above the boiling point (212 degrees Fahrenheit) causes water to change from liquid to gas. the melting/ freezing and boiling points changed with pressure.

suter [353]2 years ago

6 0

Answer:

At higher altitudes, air pressure is lower. When atmospheric pressure is lower, such as at a higher altitude, it takes less energy to bring water to the boiling point. Less energy means less heat, which means water will boil at a lower temperature at a higher altitude.

In an ordinary container (exposed to atmospheric pressure). The freezing point will increase with altitude (and corresponding decrease in atmospheric pressure), but the difference is very slight.

Explanation:

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How is data not actually obtained from the experiment represented in a line graph? with a double line with only dots with a colo

Fittoniya [83]

The way how <span>data is not actually obtained from the experiment represented in a line graph is defnitely that </span><span>a colored line with a broken line. It is a well known fact that to obtain the actual data from the experiment you there should be plotted points on the line. Hope it will help you! Regards.</span>

7 0

2 years ago

Read 2 more answers

A student checks the air in her bicycle tires early in the morning when it is cool outside. If she measures it again later in th

Dmitry_Shevchenko [17]

Answer:

She will observe that the pressure on the tire is higher.

Explanation:

By the ideal gas law, the pressure and the temperature are directly proportional, so, if the temperature increases the pressure increases too:

PV = nRT (P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature).

The temperature is a measure of the average kinetic energy of the gas molecules, so when the temperature increases, the energy also increases, and the gas molecules will move more quickly, so they will collide more often between themselves and in the wall. Those collisions will be with more force because the velocity is higher.

So, the pressure will be higher, because it is the result of collisions of the gas molecules with the walls of the tire.

6 0

2 years ago

To find the Ce4+ content in a solid sample, 4.3718 g of the solid sample were dissolved and treated with excess iodate to precip

gulaghasi [49]

Answer:

3.43 %

Explanation:

We need to calculate first the number of moles of CeO2 produced in the combustion. Given its formula we know how many moles of Ce atom are present. From there calculate the mass this number of moles this represent and then one can calculate the percentage.

0.1848 g CeO2 x 1 mol CeO2/172.114g = 0.00107 mol CeO2

0.00107 mol CeO2 x 1 mol Ce/ 1 mol CeO2 = 0.00107 mol Ce

.00107 mol Ce x 140.116 g Ce/ mol = 0.150 g Ce

0.150 g Ce/ 4.3718 g sample x 100 = 3.43 %

5 0

2 years ago

Read 2 more answers

For the reaction PCl5(g) <--> PCl3(g) Cl2(g) at equilibrium, which statement correctly describes the effects of increasing

xenn [34]

The given question is incomplete. The complete question is :

For the reaction $PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)$ at equilibrium, which statement correctly describes the effects of increasing pressure and adding $PCl_5$ , respectively

a) Increasing pressure causes shift to reactants, adding $PCl_5$ causes shift to products.

b) Increasing pressure causes shift to products ,adding $PCl_5$ causes shift to reactants.

c) Increasing pressure causes shift to products, adding $PCl_5$ causes shift to products.

d) Increasing pressure causes shift to reactants,adding $PCl_5$ causes shift to reactants

Answer: Increasing pressure causes shift to reactants, adding $PCl_5$ causes shift to products.

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

For the given equation:

$PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)$

a) If the pressure is increased, the volume will decrease according to Boyle's Law. Now, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where decrease in pressure is taking place. As the number of moles of gas molecules is lesser at the reactant side. So, the equilibrium will shift in the left direction. i.e. towards reactants.

b) If $PCl_5$ is added, the equilibrium will shift in the direction where $PCl_5$ is decreasing. So, the equilibrium will shift in the right direction. i.e. towards products.

8 0

2 years ago

Using your data above, draw conclusions about the d-splitting for each ligand (H2O, en, phen). Order the complexes from least to

Delvig [45]

Answer:

H2O<en<phen

Explanation:

The degree of d- splitting is observed from the intensity of colour. The order of d splitting from least to greatest is H2O<en<phen. Phen shows the greatest d-splitting. The degree of splitting of d- orbitals by ligands depends on their relative positions in the spectrochemical series. The spectrochemical series is an experimentally determined series. The series separates the ligands into strong field and weak field ligands. Strong field ligands are found towards the end of the series. Strong field ligands such as en and phen can participate in metal to ligand or ligand to metal pi-bonding. Hence they cause more d-splitting. Ethylendiamine and phenanthroline occur towards the end of the spectrochemical series hence the higher order of d-splitting.

7 0

2 years ago