All categories

1 year ago

identify the precipitate(s) of the reaction that occurs when a silver nitrate solution is mixed with a sodium chloride solution.

1 answer:

6 0

The precipitate of reaction that occurs when silver nitrate solution is mixed with sodium chloride solution is of silver chloride (AgCI).

The chemical equation for this precipitation reaction is as following:

AgNO3(aq) + NaCl(aq) --> AgCl(s) [precipitate]+NaNO3(aq)

Silver Chloride is a chemical compound which is poorly soluble in water because the chemical bond between Ag+ and Cl is very strong as the two ions are nearly the same size and have similar electronic densities.
If the chemical bond is strong, it is difficult for water molecules to break this bond to surround the individual ions, thus solubility will decrease.

learn more about precipitate at

Brainly.com brainly.com/question/13877944

#SPJ4

You might be interested in

How many kilometers are in a 7.21. mm

Readme [11.4K]

Answer:

0.00000721 Kilometer

6 0

3 years ago

How can i find nuclear charge and overall charge while only knowing the atomic mumber, number of protons and electrons?

Mazyrski [523]

Explanation:

An atom is made up of:

Protons which are the positively charged particles
Electrons are negatively charged
Neutrons carry no charges.

Atomic number is the number of protons in an atom.

To find the nuclear charge:

We use the effective nuclear charge formula:

Effective nuclear charge = Atomic number - number of shielding electrons(non-valence electrons)

The shielding electrons are the non-valence electrons in the inner orbitals.

For example:

Ne:

atomic number = 10

Effective nuclear charge = 10 - (2) = 8+

Number of non-valence electrons is 2

The effective nuclear charge is 8+

Overall charge:

Charge = number of protons - number of electrons

For neutral atoms, the overall charge is zero as the number of protons and electrons are the same.

Learn more:

Effective nuclear charge brainly.com/question/5441986

#learnwithBrainly

7 0

3 years ago

If 420 joules of heat energy is added to 25 grams of water at 25 degrees celcius, what will be the final temperature of the wate

natka813 [3]

Answer:

T2 = 29°C

Explanation:

Given data:

Heat added = 420 j

Mass of water = 25 g

Initial temperature = 25°C

Final temperature = ?

Solution;

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Specific heat capacity of water = 4.18 j/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Now we will put the values.

420 j = 25 g ×4.18 j/g.°C × (Final temperature - initial temperature)

420 j = 25 g ×4.18 j/g.°C × (T2 - 25°C)

420 j = 104.5 j/°C × (T2 - 25°C)

420 j /104.5 j/°C = T2 - 25°C

4°C + 25°C = T2

T2 = 29°C

4 0

4 years ago

If sea water has a density of 1.025 kg/L and 3.5% of the mass in sea water is salt, determine the mass of salt dissolved in 500

Alja [10]

68 kg. There are 58 kg salt in 500 gal seawater.

Step 1. Convert gallons to litres

1 US gal = 3.79 L (1 Imp gal = 4.55 L)

Step 2. Find the volume of the seawater

Volume = 500 gal × (3.79 L/1 gal) = 1895 L

Step 3. Find the mass of the seawater

Mass = 1895 L × (1.025 kg/1 L) = 1942 kg

Step 4. Find the mass of the salt

Mass of salt = 1942 kg seawater × (3.5 kg salt/100 kg seawater) = 68 kg salt

6 0

4 years ago

The questions are attached

Sladkaya [172]

Answer:

Explanation:

a) Balanced chemical equation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

b) Mass of oxygen gas needed = ?

Mass of propane = 25 g

Solution:

Number of moles of propane = mass/molar mass

Number of moles = 25 g/ 44.1 g/mol

Number of moles = 0.57 mol

now we will compare the moles of oxygen and propane:

C₃H₈ : O₂

1 : 5

0.57 : 5/1×0.57 = 2.85 mol

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 2.85 mol × 32 g/mol

Mass = 91.2 g

c) water vapors produced = ?

Mass of propane = 25 g

Solution:

Number of moles of propane = mass/molar mass

Number of moles = 25 g/ 44.1 g/mol

Number of moles = 0.57 mol

now we will compare the moles of water and propane:

C₃H₈ : H₂O

1 : 4

0.57 : 4/1×0.57 = 2.28 mol

Mass of water vapors:

Mass = moles × molar mass

Mass = 2.28 mol × 18 g/mol

Mass = 41.0 g

mass of water vapors produced = 1.5 kg (1500 g)

Mass of propane reacted = ?

Solution:

Number of moles of water vapors:

Number of moles = mass/molar mass

Number of moles = 1500 g/ 18 g/mol

Number of moles = 83.33 mol

now we will compare the moles of water and propane.

H₂O : C₃H₈

4 : 1

83.3 : 1/4×83.3 = 20.8 mol

Mass of propane:

Mass = number of moles ×molar mass

Mass = 20.8 mol × 44.1 g/mol

Mass = 917 g

3 0

3 years ago