Answer:
viscosity
Explanation:
one one left is mercury and the other one is honey right?
Answer:
Explanation:
Initial burette reading = 1.81 mL
final burette reading = 39.7 mL
volume of NaOH used = 39.7 - 1.81 = 37.89 mL .
37.89 mL of .1029 M NaOH is used to neutralise triprotic acid
No of moles contained by 37.89 mL of .1029 M NaOH
= .03789 x .1029 moles
= 3.89 x 10⁻³ moles
Since acid is triprotic , its equivalent weight = molecular weight / 3
No of moles of triprotic acid = 3.89 x 10⁻³ / 3
= 1.30 x 10⁻³ moles .
Answer:
310.53 g of Cu.
Explanation:
The balanced equation for the reaction is given below:
CuSO₄ + Zn —> ZnSO₄ + Cu
Next, we shall determine the mass of CuSO₄ that reacted and the mass Cu produced from the balanced equation. This can be obtained as follow:
Molar mass of CuSO₄ = 63.5 + 32 + (16×4)
= 63.5 + 32 + 64
= 159.5 g/mol
Mass of CuSO₄ from the balanced equation = 1 × 159.5 = 159.5 g
Molar mass of Cu = 63.5 g/mol
Mass of Cu from the balanced equation = 1 × 63.5 = 63.5 g
Summary:
From the balanced equation above,
159.5 g of CuSO₄ reacted to produce 63.5 g of Cu.
Finally, we shall determine the mass of Cu produced by the reaction of 780 g of CuSO₄. This can be obtained as follow:
From the balanced equation above,
159.5 g of CuSO₄ reacted to produce 63.5 g of Cu.
Therefore, 780 g of CuSO₄ will react to produce = (780 × 63.5)/159.5 = 310.53 g of Cu.
Thus, 310.53 g of Cu were obtained from the reaction.
Explanation:
Charles' law gives the relationship between the volume and the temperature of the gas. Mathematically,
Volume ∝ Temperature
i.e. 
We have, V₁ = 1.6 L, T₁ = 278 K, T₂ = 253, V₂=?
So, the new volume is 1.45 L.
Answer: 
Explanation:
Firstly, we have to find the Molecular mass of potassium oxide (
):
atomic mass: 39 u
atomic mass: 16 u
molecular mass: 
This means that in 1 mole of there are 
and we need to find how many moles there are in 
:
1 mole of ----- of
----- of
This is the quantity of moles in 73.9 g of potassium oxide
Now we can calculate the number of atoms in 73.9 g of potassium oxide by the following relation:
Where:
is the number of atoms in 73.9g of potassium oxide
is the Avogadro's number, which is determined by the number of particles (or atoms) in a mole.
Then:
This is the quantity of atoms in 73.9g of potassium oxide
