The answer is particle accelerators.
N(CH₃OH)=3,62·10²⁴/6·10²³ 1/mol = 6,033 mol
m(CH₃OH) = 6,033 mol · 32 g/mol (molar mass) = 193,06 g.
I think is False! Because it is not example of a disaccharide it is not saccharin-aspartame molecule.
Answer:
0.1410 M
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
HCl + NaOH —> NaCl + H₂O
From the balanced equation above,
The mole ratio of the acid, HCl (nA) = 1
The mole ratio of the base, NaOH (nB) = 1
Next, the data obtained from the question. This include:
Volume of acid, HCl (Va) = 25 mL
Volume of base, NaOH (Vb) = 34.55 mL
Concentration of base, NaOH (Cb) = 0.1020 M
Concentration of acid, HCl (Ca) =?
CaVa / CbVb = nA/nB
Ca × 25 / 0.1020 × 34.55 = 1/1
Ca × 25 / 3.5241 = 1/1
Cross multiply
Ca × 25 = 3.5241 × 1
Ca × 25 = 3.5241
Divide both side by 25
Ca = 3.5241 / 25
Ca = 0.1410 M
Therefore, the concentration of the acid, HCl is 0.1410 M
The answer for the following problem is mentioned below.
- <u><em>Therefore the final volume of the gas is 52.7 ml.</em></u>
Explanation:
Given:
Initial pressure (
) = 290 kPa
Final pressure (
) = 104 kPa
Initial volume (
) = 18.9 ml
To find:
Final volume (
)
We know;
From the ideal gas equation;
P × V = n × R × T
where;
P represents the pressure of the gas
V represents the volume of gas
n represents the no of the moles
R represents the universal gas constant
T represents the temperature of the gas
So;
P × V = constant
P ∝ 
From the above equation;
represents the initial pressure of the gas
represents the final pressure of the gas
represents the initial volume of the gas
represents the final volume of the gas
Substituting the values of the above equation;
= 
= 52.7 ml
<u><em>Therefore the final volume of the gas is 52.7 ml.</em></u>
