It wasn’t as powerful because your sense of smell enhances your sense of taste
Answer:
[HClO₄] = 11.7M
Explanation:
First of all we need to know, that a weight percent represents, the mass of solute in 100 g of solution.
Let's convert the mass to moles → 70.5 g . 1mol/100.45 g = 0.702 moles
Now we can apply the density to calculate the volume.
Density always refers to solution → Solution density = Solution mass / Solution volume
1.67 g/mL = 100 g / Solution volume
Solution volume = 100 g / 1.67 g/mL → 59.8 mL
To determine molarity (mol/L) we must convert the mL to L
59.8 mL . 1L/1000mL = 0.0598 L
Molarity → Moles of solute in 1L of solution → 0.702 mol / 0.0598 L = 11.7M
Answer:
V₂ =31.8 mL
Explanation:
Given data:
Initial volume of gas = 45 mL
Initial temperature = 135°C (135+273 =408 K)
Final temperature = 15°C (15+273 =288 K)
Final volume of gas = ?
Solution:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 45 mL × 288 K / 408 k
V₂ = 12960 mL.K / 408 K
V₂ =31.8 mL
The balanced chemical reaction is written as:
4Al + 3O2 = 2Al2O3
To determine the mass of oxygen gas that would react with the given amount of aluminum metal, we use the initial amount and relate this amount to the ratio of the substances from the chemical reaction. We do as follows:
moles Al = 16.4 g ( 1 mol / 26.98 g ) = 0.61 mol Al
moles O2 = 0.61 mol Al ( 3 mol O2 / 4 mol Al ) = 0.46 mol O2
mass O2 = 0.46 mol O2 ( 32.0 g / mol ) = 14.59 g O2
Therefore, to completely react 16.4 grams of aluminum metal we need a minimum of 14.59 grams of oxygen gas.
