What does the latent heat of vaporization measure?

A 47.0 mL aliquot of a 0.400 M stock solution must be diluted to 0.100 M. Assuming the volumes are additive, how much water shou

mel-nik [20]

Answer:

The answer is 0.188L

Explanation:

I did the problem on paper and put the answer on the pictures. I'm sorry if I didn't explain it well but I hope I helped.

5 0

3 years ago

Mount Everest rises to a height of 8,850 m above sea level. At a base camp on the mountain the atmospheric pressure is measured

yan [13]

Answer:

Water will boil at $76^{0}\textrm{C}$ .

Explanation:

According to clausius-clapeyron equation for liquid-vapor equilibrium:

$ln(\frac{P_{2}}{P_{1}})=\frac{-\Delta H_{vap}^{0}}{R}[\frac{1}{T_{2}}-\frac{1}{T_{1}}]$

where, $P_{1}$ and $P_{2}$ are vapor pressures of liquid at $T_{1}$ (in kelvin) and $T_{2}$ (in kelvin) temperatures respectively.

Here, $P_{1}$ = 760.0 mm Hg, $T_{1}$ = 373 K, $P_{2}$ = 314.0 mm Hg

Plug-in all the given values in the above equation:

$ln(\frac{314.0}{760.0})=\frac{-40.7\times 10^{3}\frac{J}{mol}}{8.314\frac{J}{mol.K}}\times [\frac{1}{T_{2}}-\frac{1}{373K}]$

or, $T_{2}=349 K$

So, $T_{2}=349K=(349-273)^{0}\textrm{C}=76^{0}\textrm{C}$

Hence, at base camp, water will boil at $76^{0}\textrm{C}$

6 0

3 years ago

9. How many grams of potassium sulfate are needed to make 250 mL of a 0.150 M

antiseptic1488 [7]

Answer:

6.53g of K₂SO₄

Explanation:

Formula of the compound is K₂SO₄

Given parameters:

Volume of K₂SO₄ = 250mL = 250 x 10⁻³L

= 0.25L

Concentration of K₂SO₄ = 0.15M or 0. 15mol/L

Unknown:

Mass of K₂SO₄ =?

Methods:

We use the mole concept to solve this kind of problem.

>>First, we find the number of moles using the expression below:

Number of moles= concentration x volume

Solving for number of moles:

Number of moles = 0.25 x 01.5

= 0.0375mole

>>Secondly, we use the number of moles to find the mass of K₂SO₄ needed. This can be obtained using the expression below:

Mass(g) = number of moles x molar mass

Solving:

To find the molar mass of K₂SO₄, we must know the atomic mass of each element in the compound. This can be obtained using the periodic table.

For:

K = 39g

S = 32g

O = 16g

Molar mass of K₂SO₄ = (39x2) + 32 + (16x4)

= 78 +32 + 64

= 174g/mol

Using the expression:

Mass(g) = number of moles x molar mass

Mass of K₂SO₄ = 0.0375 x 174 = 6.53g

5 0

3 years ago

Zinc blende, or sphalerite, can be described as having a face-centered cubic unit cell with sulfur atoms at each lattice point a

g100num [7]

Answer:

ZnS

Explanation:

1. Number of Zn atoms

4 internal atoms = 4 Zn atoms

2. Number of S atoms

8 corners × ⅛ S atom/corner + 6 faces × ½ S atom/face = 1 S atom + 3 S atoms = 4 S atoms

3. Empirical formula

The atomic ratio is

4Zn:4S = 1Zn:1S

The empirical formula is ZnS.

7 0

3 years ago

1.42 g H2 is allowed to react with 10.4 g N2 , producing 2.14 g NH3 . Part A What is the theoretical yield in grams for this rea

Bad White [126]

Taking into account the reaction stoichiometry, the theorical yield for the reaction is 8.0467 grams of NH₃.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

3 H₂ + N₂ → 2 NH₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

H₂: 3 moles
N₂: 1 mole
NH₃: 2 moles

The molar mass of the compounds is:

H₂: 2 g/mole
N₂: 28 g/mole
NH₃: 17 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

H₂: 3 moles ×2 g/mole= 6 grams
N₂: 1 mole ×28 g/mole= 28 grams
NH₃: 2 moles ×17 g/mole= 34 grams

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 28 grams of N₂ reacts with 6 grams of H₂, 10.4 grams of N₂ reacts with how much mass of H₂?

$mass of H_{2} =\frac{10.4 grams of N_{2}x 6 grams of H_{2} }{28 grams of N_{2}}$

<u><em>mass of H₂= 2.2286 grams</em></u>

But 2.2286 grams of H₂ are not available, 1.42 grams are available. Since you have less mass than you need to react with 10.4 grams of N₂, H₂ will be the limiting reagent.

<h3>Definition of theorical yield</h3>

The theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.

<h3>Theoretical yield in this case</h3>

Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 6 grams of H₂ form 34 grams of NH₃, 1.42 grams of H₂ form how much mass of NH₃?

$mass of NH_{3} =\frac{1.42 grams of H_{2} x 34 grams of NH_{3}}{6grams of H_{2} }$

<u><em>mass of NH₃= 8.0467 grams</em></u>

Then, the theorical yield for the reaction is 8.0467 grams of NH₃.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

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7 0

1 year ago