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3 years ago

Which anion will form a precipitate with Ca2+

2 answers:

ioda3 years ago

6 0

Answer : Precipitation reactions occur when particular species of cations and anions in an aqueous solution combines to form an insoluble ionic solid which can be called as precipitate.

Here, the anions that will mostly form precipitate with $Ca^{+2}$ are OH anions, Cl ions, Br ions, $SO_{4}$ ions, $C_{2} H_{3} O{2}$ ions.

Calcium has valency of 2 electrons in its valence shell. So, any anion that can satisfy the valency of calcium can form a compound with calcium.

Anika [276]3 years ago

4 0

One example of an anion that will form a precipitate with Ca 2+ is CO3 2-
The substance formed would be CaCO3 of calcium carbonate.

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Romashka-Z-Leto [24]

Answer:

Explanation:

3 0

3 years ago

What’s the distance from the nucleus of an atom to its theoretical outer edge of electron orbits

ollegr [7]

Answer:

The range of atoms = (30-300 pm) depending upon the element

Explanation:

The Atomic radii of the atom is the distance from the center of the circle to the outermost orbital.

The center of the circle is the nucleus and the radii is the outermost boundary.

The actual size of the atom is decided on the basis of the Zeff . Also known as effective nuclear charge.

Zeff: It is the net positive charge felt by the outermost electron by the nucleus.

The value of Zeff depends upon the shielding constant. More the shielding less will be the Zeff . Hence the size of the atom increases.

Due to shielding the outermost electrons feel less pull of nucleus.

The greater the Zeff , the smaller the radius of the atom.

The formula used to calculate the atomic mass is :

$r_{n}=\frac{52.9n^{2}}{Z}$ pm

Here "pm"= picometers

$1 pm = 10^{-12}m$

The size of the smallest atom H-atom = 120 pm

The range of atoms = (30-300 pm)

4 0

3 years ago

Hg-178 undergoes 2 alpha decay and positron emission. Determine the new isotope.

Illusion [34]

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4 0

3 years ago

A 1.00 g sample of octane (C8H18) is burned in a bomb calorimeter with a heat capacity of 837J∘C that holds 1200. g of water at

lubasha [3.4K]

Answer:

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

Explanation:

Step 1: Data given

Mass of octane = 1.00 grams

Heat capacity of calorimeter = 837 J/°C

Mass of water = 1200 grams

Temperature of water = 25.0°C

Final temperature : 33.2 °C

Step 2: Calculate heat absorbed by the calorimeter

q = c*ΔT

⇒ with c = the heat capacity of the calorimeter = 837 J/°C

⇒ with ΔT = The change of temperature = T2 - T1 = 33.2 - 25.0 : 8.2 °C

q = 837 * 8.2 = 6863.4 J

Step 3: Calculate heat absorbed by the water

q = m*c*ΔT

⇒ m = the mass of the water = 1200 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = The change in temperature = T2 - T1 = 33.2 - 25 = 8.2 °C

q = 1200 * 4.184 * 8.2 = 41170.56 J

Step 4: Calculate the total heat

qcalorimeter + qwater = 6863.4 + 41170. 56 = 48033.96 J = 48 kJ

Since this is an exothermic reaction, there is heat released. q is positive but ΔH is negative.

Step 5: Calculate moles of octane

Moles octane = 1.00 gram / 114.23 g/mol

Moles octane = 0.00875 moles

Step 6: Calculate heat combustion for 1.00 mol of octane

ΔH = -48 kJ / 0.00875 moles

ΔH = -5485.7 kJ/mol

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

8 0

3 years ago

A nucleus emits an alpha particle. What happens to the mass

babunello [35]

Answer:

the mass number decreases by 4 and the atomic number decreases by 2

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3 years ago