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finlep [7]
2 years ago
8

One of the products when aqueous Na, CO, reacts with aqueous

Chemistry
1 answer:
mixas84 [53]2 years ago
6 0

Answer:

NaNO3

balanced equation - Na2CO3 + Sn(NO3)2 → 2NaNO3 + Sn(CO3)

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Students in a lab used a hot plate to boil water. Which of the following illustrations best represents the change in the movemen
Digiron [165]

The second illustration is the best representation of the change in the movement of particles as the temperature of the water changes.

<u>Explanation:</u>

The second option perfectly represents the boiling of water. As when the temperature is increased, the water molecules gain energy to move faster, thus their kinetic energy of the atoms will be more. This will lead to more freely movement of all the atoms of the water.

And as boiling leads to transformation from liquid state to gaseous state, so the increase in the distance between atoms and molecules occurs in the gaseous state. Thus, the second illustration is best suitable for representing the boiling of water.

As on increasing temperature of the water, the distance between the molecules is increasing in the second illustration while the other illustration shows the decrease in the distance between the molecules. So, the second illustration is the best representation of the change in the movement of particles as the temperature of the water changes.

6 0
3 years ago
Calculate the molar mass for oxygen gas
Sladkaya [172]

For example, the atomic mass of an oxygen atom is 16.00 amu; that means the molar mass of an oxygen atom is 16.00 g/mol. Further, if you have 16.00 grams of oxygen atoms, you know from the definition of a mole that your sample contains 6.022 x 10^23 oxygen atoms.

8 0
3 years ago
Balance each of these equations.
mafiozo [28]
The answer I would choose is the third one
4 0
2 years ago
How many particles are in 23 g of H 2 O?
Sedaia [141]
1 mole of any substance contains 6.022 × 1023 particles.

⚛ 6.022 × 1023 is known as the Avogadro Number or Avogadro Constant and is given the symbol NA

N = n × NA

· N = number of particles in the substance

· n = amount of substance in moles (mol)

· NA = Avogardro Number = 6.022 × 10^23 particles mol-1


For H2O we have:

2 H at 1.0 each = 2.0 amu
1 O at 16.0 each = 16.0 amu
Total for H2O = 18.0 amu, or grams/mole

It takes 18 grams of H2O to obtain 1 mole, or 6.02 x 1023 molecules of water. Think about that before we answer the question. We have 25.0 grams of water, so we have more than one mole of water molecules. To find the exact number, divide the available mass (25.0g) by the molar mass (18.0g/mole). Watch how the units work out. The grams cancel and moles moves to the top, leaving moles of water. [g/(g/mole) = moles].

Here we have 25.0 g/(18.0g/mole) = 1.39 moles water (3 sig figs).

Multiply 1.39 moles times the definition of a mole to arrive at the actual number of water molecules:

1.39 (moles water) * 6.02 x 1023 molecules water/(mole water) = 8.36 x 1023 molecules water.

That's slightly above Avogadro's number, which is what we expected. Keeping the units in the calculations is annoying, I know, but it helps guide the operations and if you wind up with the unit desired, there is a good chance you've done the problem correctly.

N = n × (6.022 × 10^23)


1 grams H2O is equal to 0.055508435061792 mol.

Then 23 g of H2O is 1.2767 mol


To calculate the number of particles, N, in a substance:

N = n × NA

N = 1.2767 × (6.022 × 10^23)

N= 176.26

N=
3 0
2 years ago
How will adding NaCl affect the freezing point of a solution?
lord [1]

Answer is: adding NaCl will lower the freezing point of a solution.

A solution (in this example solution of sodium chloride) freezes at a lower temperature than does the pure solvent (deionized water).

The higher the solute concentration (sodium chloride), freezing point depression of the solution will be greater.

Equation describing the change in freezing point:  

ΔT = Kf · b · i.

ΔT - temperature change from pure solvent to solution.

Kf - the molal freezing point depression constant.

b -  molality (moles of solute per kilogram of solvent).

i - Van’t Hoff Factor.

Dissociation of sodium chloride in water: NaCl(aq) →  Na⁺(aq) + Cl⁻(aq).

3 0
2 years ago
Read 2 more answers
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