How many valence electrons are available for bonding in bromine (Br)?

To measure the amount of chlorine in a well-boring fluid, an analytical chemist adds silver nitrate solution to a sample of the

anyanavicka [17]

Answer:

c. precipitation

Explanation:

Let's consider the following balanced net ionic equation.

Cl⁻(aq) + AgNO₃(ag) ⇒ AgCl(s) + NO₃⁻ (ag)

What kind of reaction is this?

a. acid-base. No, because the reactants are not acids and bases.

b. redox. No, because all the species have the same oxidation numbers in the reactants and the products

c. precipitation. Yes, because an insoluble compound (AgCl) is produced, which then precipitates.

3 0

3 years ago

Making methanol the element hydrogen is not abundant in nature, but it is a useful reagent in, for example, the potential synthe

stealth61 [152]

ΔG⁰ = ΔH⁰ - TΔS
ΔH⁰ = Hf,(CH₃OH) - Hf,(CO) = -238.7 + 110.5 = -128.2 kJ/mol
ΔS = S(CH₃OH) - S(CO) - 2S(H₂) = 126.8 - 197.7 - 2 x 130.6 = -332.1 J/mol.K
So
ΔG⁰ = - 128200 + 332.1 T
For the reaction to be spontaneous:
ΔG⁰ < 0
So: -128200 + 332.1 T < 0
332.1 T < 128200
T < 386.028 K

5 0

3 years ago

If you live in the southern hemisphere, what season would you be experiencing in position C this diagram?

lisabon 2012 [21]

The answer is spring it would be fall in the north and spring in the south

7 0

3 years ago

Read 2 more answers

Calculate the freezing point of a solution containing 5. 0 grams of kcl and 550. 0 grams of water. the molal-freezing-point-depr

lutik1710 [3]

The freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is - 0.45°C

Using the equation,

Δ $T_{f}$ = i $K_{f}$ m

where:

Δ $T_{f}$ = change in freezing point (unknown)

i = Van't Hoff factor

$K_{f}$ = freezing point depression constant

m = molal concentration of the solution

Molality is expressed as the number of moles of the solute per kilogram of the solvent.

Molal concentration is as follows;

MM KCl = 74.55 g/mol

molal concentration = $\frac{5.0g*\frac{1mol}{74.55g} }{550.0g*\frac{1kg}{1000g} }$

molal concentration = 0.1219m

Now, putting in the values to the equtaion Δ $T_{f}$ = i $K_{f}$ m we get,

Δ $T_{f}$ = 2 × 1.86 × 0.1219

Δ $T_{f}$ = 0.4536°C

So, Δ $T_{f}$ of solution is,

Δ $T_{f_{solution} }$ = 0.00°C - 0.45°C

Δ $T_{f_{solution} }$ = - 0.45°C

Therefore,freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is - 0.45°C

Learn more about freezing point here;

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7 0

2 years ago

Write the isotopic symbol for the following (show your work) a) An isotope of iodine whose atoms have 78 neutrons b) An isotope

morpeh [17]

Answer:

For a: The isotopic representation of iodine is $_{53}^{131}\textrm{I}$

For b: The isotopic representation of cesium is $_{55}^{137}\textrm{Cs}$

For c: The isotopic representation of strontium is $_{38}^{52}\textrm{Sr}$

Explanation:

The isotopic representation of an atom is: $_Z^A\textrm{X}$

where,

Z = Atomic number of the atom

A = Mass number of the atom

X = Symbol of the atom

For a:

We are given:

Number of neutrons = 78

Atomic number of iodine = 53 = Number of protons

Mass number = 53 + 78 = 131

Thus, the isotopic representation of iodine is $_{53}^{131}\textrm{I}$

For b:

We are given:

Number of neutrons = 82

Atomic number of cesium = 55 = Number of protons

Mass number = 55 + 82 = 137

Thus, the isotopic representation of cesium is $_{55}^{137}\textrm{Cs}$

For c:

We are given:

Number of neutrons = 52

Atomic number of strontium = 38 = Number of protons

Mass number = 38 + 52 = 90

Thus, the isotopic representation of strontium is $_{38}^{52}\textrm{Sr}$

3 0

3 years ago