The molar mass of Na₂SO₄ -
2 x Na - 2 x23 = 46
1 x S - 1 x 32 = 32
4 x O - 4 x 16 = 64
total = 46 + 32 + 64 = 142 g/mol
the molarity of solution - 2.0 M
in 1 L of solution , 2.0 moles
Therefore in 2.5 L - 2 mol/L x 2.5 L = 5 mol
then the mass of Na₂SO₄ required = 142 g/mol x 5 mol = 710 g
Answer:
C). The Bohr-Rutherford model
Explanation:
The 'Bohr-Rutherford model' of the atom failed to elaborate on the attraction between some substances. It essentially targeted hydrogen atoms and failed to explain its stability across multi-electrons. The nature and processes of the chemical reactions remained unillustrated and thus, this is the key drawback of this model. Thus, <u>option C</u> is the correct answer.
<span>Find
the speed of the car in which it travels at 150 km in 7200 seconds.
This is an easy question, you just need to follow the given formula as always.
Since the time and the distance is already given, we are now looking for the speed
=> time = 7200 seconds
=> distance = 150km</span><span>
d = speed x time
S = distance / time
s = 150 km / 7200 seconds
s = 0.021 km / seconds.
Thus, the car travels for about 0.021 km per seconds.
</span>
Answer:
1.263 moles of HF
Explanation:
The balance chemical equation for given single replacement reaction is;
Sn + 2 HF → SnF₂ + H₂
Step 1: <u>Calculate Moles of Tin as;</u>
As we know,
Moles = Mass / A.Mass ----- (1)
Where;
Mass of Tin = 75.0 g
A.Mass of Tin = 118.71 g/mol
Putting values in eq. 1;
Moles = 75.0 g / 118.71 g/mol
Moles = 0.6318 moles of Sn
Step 2: <u>Find out moles of Hydrogen Fluoride as;</u>
According to balance chemical equation,
1 mole of Sn reacted with = 2 moles of HF
So,
0.6318 moles of Sn will react with = X moles of HF
Solving for X,
X = 0.6318 mol × 2 mol / 1 mol
X = 1.263 moles of HF
Answer:
2.32 m
Explanation:
So, according to definition of mole fraction:
Mole fraction = 0.176
Applying values as:
So,
Also, Molar mass of toluene = 92.14 g/mol
Thus,
The formula for the calculation of moles is shown below:
Also, 1 g = 0.001 kg
So,
Molality is defined as the moles of the solute present in 1 kg of the solvent.
It is represented by 'm'.
Thus,
<u>Molality of benzene = 2.32 m</u>
