If a student weighs out 0.744 g Fe ( NO 3 ) 3 ⋅ 9 H 2 O , what is the final concentration of the ∼0.2 M Fe ( NO 3 ) 3 solution that the student makes?

Answers

Answer 1
Answer:

Answer:

Molar concentration of Fe(NO3)3 . 9H2O = 0.12M

Explanation:

Fe(NO3).9H2O --> Fe(NO3)3 + 9H2O

By stoichiometry,

1 mole of Fe(NO3)3 will be absorb water to form 1 mole of Fe(NO3)3 . 9H2O

Therefore, calculating the mass concentration of Fe(NO3)3;

Molar mass of Fe(NO3)3 = 56 + 3*(14 + (16*3))

= 242 g/mol

Mass concentration of Fe(NO3)3 = molar mass * molar concentration

= 242 * 0.2

= 48.4 g/L

Molar mass of Fe(NO3)3 . 9H2O = 56 + 3*(14 + (16*3)) + 9* ((1*2) + 16)

= 242 + 162 g/mol

= 404g/mol

Concentration of Fe(NO3)3 . 9H2O = mass concentration/molar mass

= 48.4 /404

= 0.12 mol/l

Molar concentration of Fe(NO3)3 . 9H2O = 0.12M


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The correct electron configuration for magnesium is: 1s 22s 22p 63s 3 True False

Answers

Answer:

False

Explanation:

Magnesium is the element of second group and third period. The electronic configuration of magnesium is - 2, 8, 2 or 1s^22s^22p^63s^2

There are 2 valence electrons of magnesium.

Only the valence electrons are shown by dots in the Lewis structure.  

As, stated above, there are only two valence electrons of magnesium, so in the Lewis structure, two dots are made around the magnesium symbol.

Given that the electronic configuration is:- 1s^22s^22p^63s^3.

Orbital s cannot accommodate 3 electrons and also in magnesium it has 3s^2. Hence, the statement is false.

Guys i need a long inforation about....
"Formation of colloids"
Please and thank you ​

Answers

Answer:

Colloids

There are two basic methods of forming a colloid: reduction of larger particles to colloidal size, and condensation of smaller particles (e.g., molecules) into colloidal particles. Some substances (e.g., gelatin or glue) are easily dispersed (in the proper solvent) to form a colloid; this spontaneous dispersion is called peptization. A metal can be dispersed by evaporating it in an electric arc; if the electrodes are immersed in water, colloidal particles of the metal form as the metal vapor cools. A solid (e.g., paint pigment) can be reduced to colloidal particles in a colloid mill, a mechanical device that uses a shearing force to break apart the larger particles. An emulsion is often prepared by homogenization, usually with the addition of an emulsifying agent. The above methods involve breaking down a larger substance into colloidal particles. Condensation of smaller particles to form a colloid usually involves chemical reactions—typically displacement, hydrolysis, or oxidation and reduction.

When you convert feet to inches, how do you decide which portion of the conversion factor should be in the numerator and which in the denominator?

Answers

Answer : The conversion used is, 1\text{ feet}=12\text{ inches}

Explanation :

The conversion used :

1\text{ feet}=12\text{ inches}

For example : To convert 10 feet into inches.

1\text{ feet}=12\text{ inches}

10\text{ feet}=10\text{ feet}* \frac{12\text{ inches}}{1\text{ feet}}

                    =120\text{ inches}

From this we conclude that, the conversion factor used in the numerator and denominator for 1 feet should be, \frac{12\text{ inches}}{1\text{ feet}}

Final answer:

When converting from feet to inches, use the conversion factor 12 inches/1 foot. This way, the unit 'foot' cancels out and leaves 'inches', performing the conversion. An example is converting 5 feet to inches is 5 times 12, which equals to 60 inches.

Explanation:

When you're converting units, such as feet to inches, the main guideline is to set up your conversion factor in a way that cancels out the unit you want to convert from and leaves you with the unit you want to convert to. In this case, since you're converting from feet to inches, you'll be using the fact that there are 12 inches in 1 foot as your conversion factor. Therefore, when converting, the conversion factor should be set up as 12 inches/1 foot. This in essence means you're multiplying by the number 1, which doesn't change the value, just the units.

For instance, if you have 5 feet and you want to convert this to inches, you'll set your conversion factor up as 12 inches/1 foot, with inches in the numerator and feet in the denominator, to cancel out feet. Multiply 5 feet by the conversion factor (12 inches/1 foot), the result would be 60 inches. Here, the 'feet' units cancel leaving the answer in inches, completing the conversion.

Learn more about Unit Conversion here:

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How is Hess's law used to measure enthalpy of a desired reaction?A. The enthalpy is obtained from the enthalpy of an intermediate
step.
B. The enthalpy is determined from the enthalpy of similar reactions.
C. The enthalpy from the final equation in a series of reactions is
used
D. Intermediate equations with known enthalpies are added together.

Answers

Hess's law is used to measure the enthalpy of a desired chemical reaction because: D. Intermediate equations with known enthalpies are added together.

What is Hess's Law?

Hess's Law is also known as Hess's law of constant heat summation (enthalpy) and it was named after a Swiss-born Russian chemist called Germain Hess.

Hess's Law states that the energy change (enthalpy) experienced in a desired chemical reaction is equal to the sum of the energy changes (enthalpies) in each chemical reactions that it is made up of or contains.

Read more on Hess's Law here: brainly.com/question/9328637

A rock contains 0.37 mg of Pb-206 and 0.95 mg of U-238. Approximately how many U-238 atoms were in the rock when it was formed billions of years ago? (The half-life for 238U  206Pb is 4.5  109 yr.)

Answers

Half-life of a radioactive element

The Half-Life of a radioactive element osbtime taken for half the nucleus of the atom of the element to decay

Calculating the original amount of U-238

  • The number of moles present in each element is first determined:
  • Number of moles = mass/molar mass

For Pb-206:

mass = 0.37 mg = 0.37 * 10⁻³ g

molar mass = 206 g/mol

number of moles = 0.37 * 10⁻³ g/206 g/mol

number of moles of Pb-206 = 1.79 * 10⁻⁶ moles

For U-238:

mass = 0.95 mg = 0.95 * 10⁻³ g

molar mass = 238 g/mol

number of moles = 0.95 * 10⁻³ g/238 g/mol

number of moles = 3.99 * 10⁻⁶ moles

Assuming that all the Pb-206 were formed from U-238

  • Initial moles of U-238 = Moles of present U-238 + molesw of present Pb-208

Initial moles of U-238 = 3.99 * 10⁻⁶ moles + 1.79 * 10⁻⁶ moles

Initial moles of U-238 = 5.78 * 10⁻⁶ moles

One mole of U-238 contains = 6.02 * 10²³ atoms

5.78 * 10⁻⁶ moles of U-238 will contain 6.02 * 10²³ * 5.78 * 10⁻⁶ atoms

Number of atoms of U-238 initially present = 3.48 * 10¹⁸ atoms

Therefore, the number of atoms of U-238 initially present is 3.48 * 10¹⁸ atoms

Learn more about Half-life at: brainly.com/question/4702752

Why does ammonium nitrate (NH4NO3) dissolve readily in water even though the dissolution process is endothermic by 26.4 kJ/mol? Why does ammonium nitrate (NH4NO3) dissolve readily in water even though the dissolution process is endothermic by 26.4 kJ/mol? The vapor pressure of the water decreases upon addition of the solute. The osmotic properties of the system lead to this behavior. The overall enthalpy of the system decreases upon addition of the solute. The overall entropy of the system increases upon dissolution of this strong electrolyte. The overall enthalpy of the system increases upon dissolution of this strong electrolyte.

Answers

Answer: Option (c) is the correct answer.

Explanation:

Entropy is defined as the degree of randomness that is present within the particles of a substance.

As NH_(4)NO_(3) is ionic in nature. Hence, when it is added to water then it will readily dissociate into ammonium ions (NH^}{+}_(4)) and nitrate ions (NO^(-)_(3)).

Therefore, it means that ions of ammonium nitrate will be free to move from one place to another. Hence, there will occur an increase in entropy.

Thus, we can conclude that ammonium nitrate (NH_(4)NO_(3)) dissolve readily in water even though the dissolution process is endothermic by 26.4 kJ/mol because the overall entropy of the system increases upon dissolution of this strong electrolyte.