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Tamiku [17]
3 years ago
9

How would you convert 500cc of 2M H2so4 in g/l and normality?​

Chemistry
1 answer:
Ann [662]3 years ago
6 0

Concentration :

196 g/L and 4 N

<h3> </h3><h3>Further explanation</h3>

The concentration of a substance can be expressed in several quantities such as moles, percent (%) weight / volume,), molarity, molality, parts per million (ppm) or mole fraction. The concentration shows the amount of solute in a unit of the amount of solvent.

500 cc of 2M H₂SO₄

V = 500 cc = 0.5 L

mol H₂SO₄

\tt mol=M\times V=2\times 0.5=1

mass H₂SO₄ (MW = 98 g/mol)

\tt mass=mol\times MW=1\times 98=98~g

concentration in g/L :

\tt \dfrac{98}{0.5}=196~g/L

concentration in normality

Relationship between normality and molarity

N = M x n (n=valence , amount of H⁺ or OH⁻)

so :

\tt N=2\times 2(n=2~for~H_2SO_4\rightarrow 2H^+)\\\\N=4

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Define an independent variable in an experiment. Explain the term with respect to the solubility of salt at different temperatur
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Independent variable would be salt since you can't change it in this experiment.

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3 years ago
Molten gallium reacts with arsenic to form the semiconductor, gallium arsenide, GaAs, used in light-emitting diodes and solar ce
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Answer:

a) 1.2g of arsenic b) 0.64g of arsenic c) 3.481g of gallium d) 3.806g of gallium e) 2.61g arsenic

Explanation:

The balanced equation is:

Ga + As = GaAs, 1:1 mole ratio

a) mass (gallium)/ molar mass of Ga = 4/ 69.723 = 0.0574mol

Mass (arsenic)/ molar mass of As = 5.5/74.9216 = 0.0734, subtracting the moles from each other (knowing already that the ratio is 1:1), arsenic is in excess by 1.2g

b) repeating the procedure (changing the values)

It will be 0.0574 to 0.06593

Arsenic is in excess by 0.00854

0.00854* mass of arsenic (such must be done for the first remaining mole) = 0.64g

c) the mole ratio is 0.0574: 0.00747

Gallium is in excess by 0.05

Mass of excess gallium = 0.05* 69.723 = 3.481g

d) using the mass given, the new ratio is

0.128: 0.0734

Gallium is in excess by 0.054mol

Mass of excess gallium = 0.054*69.723 = 3.806g

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7 0
3 years ago
What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.3×1015Hz?
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Answer:

We know that

ħf = ф + Ekmax  

where  

ħ = planks constant = 6.626x10^-34 J s  

f = frequency of incident light = 1.3x10^15 /s (1 Hz =

1/s)

ф = work function of the cesium = 2.14 eV  

Ekmax = max  kinetic energy of the emmitted electron.  

We distinguish that:

1 eV = 1.602x10^-19 J

So:

2.14 eV x (1.602x10^-19 J / 1 eV) = 3.428x10^-19 J

So,

Ekmax = (6.626x10^-34 J s) x (1.3x10^15 / s) - 3.428x10^-19 J

= 8.6138x10^-19 J - 3.428x10^-19 J = 5.1858x10^-19 J  

Answer:

5.19x10^-19 J  

Kinetic energy:

In physics, the kinetic energy of an object is the energy that it owns due to its motion. It is defined as the work required accelerating a body of a given mass from rest to its specified velocity. Having expanded this energy during its acceleration, the body upholds this kinetic energy lest its speed changes.

Answer details:

Subject: Chemistry

Level: College

Keywords:

• Energy  

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Answer:

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Explanation:

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<em>∵ The freezing point if water is 0.0 °C and it is depressed by - 4.39 °C.</em>

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