Answer:
Products
Explanation:
During a chemical reaction, there are the reactants (left side), and the products (right side).
To find moles in this sample, you would divide grams by molar mass of ethyl alcohol
(18.0g)/(46.07g/mol) = 0.391mol C2H6O
The mass of hydrogen atoms that is measured at 54 u given the relationship is 89.64×10¯²⁴ g
<h3>Conversion scale </h3>
1 u = 1.66×10¯²⁴ g
<h3>How to determine the mass of hydrogen atoms </h3>
- Mass of Hydrogen (u) = 54 u
- Mass of Hydrogen (g) =?
1 u = 1.66×10¯²⁴ g
Therefore
54 u = 54 × 1.66×10¯²⁴ g
54 u = 89.64×10¯²⁴ g
Thus, the mass of the hydrogen atoms measured at 54 u is 89.64×10¯²⁴ g
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Answer:
Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water <em><u>to the same extent</u></em> by adding 1 mol of
to 1 kg of water.
Explanation:
1) Moles of NaCl ,
Mass of water = m= 1 kg = 1000 g
Moles of water = 
Vapor pressure of the solution = 
Vapor pressure of the pure solvent that is water = 
Mole fraction of solute(NaCl)= 



The vapor pressure for the NaCl solution at 17.19 Torr.
2) Moles of sucrose ,
Mass of water = m = 1 kg = 1000 g
Moles of water = 
Vapor pressure of the solution = 
Vapor pressure of the pure solvent that is water = 
Mole fraction of solute ( glucose)= 



The vapor pressure for the glucose solution at 17.19 Torr.
p = p' = 17.19 Torr
Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent by adding 1 mol of
to 1 kg of water.
Answer:
V₁ = 10 mL
Explanation:
Given data:
Initial volume of HCl = ?
Initial molarity = 3.0 M
Final molarity = 0.10 M
Final volume = 300.0 mL
Solution:
Formula:
M₁V₁ = M₂V₂
M₁ = Initial molarity
V₁ = Initial volume of HCl
M₂ =Final molarity
V₂ = Final volume
Now we will put the values.
3.0 M ×V₁ = 0.10 M×300.0 mL
3.0 M ×V₁ = 30 M.mL
V₁ = 30 M.mL /3.0 M
V₁ = 10 mL