Answer:
16% v/v es la nueva concentración de alcohol en la solución
Explanation:
El porcentaje volumen/volumen (% v/v) es definido como 100 veces la relación entre el volumen de soluto (Alcohol en este caso) y el volumen total de la solución (Agua + Alcohol). Para resolver esta pregunta necesitamos hallar el volumen de alcohol y el de agua:
<em>Volumen alcohol:</em>
320cc * (20cc etanol / 100cc) = 64cc etanol
<em>Volumen agua:</em>
80cc + (320cc-64cc) = 336cc agua
<em>% v/v:</em>
64cc / (336cc + 64cc) * 100
= 16% v/v es la nueva concentración de alcohol en la solución
<h3>
Answer:</h3>
70.906 g
<h3>
Explanation:</h3>
We are given;
- Atoms of Chlorine = 1.2 × 10^24 atoms
We are required to calculate the mass of Chlorine
- We know that 1 mole of an element contains atoms equivalent to the Avogadro's number, 6.022 × 10^23.
- That is , 1 mole of an element = 6.022 × 10^23 atoms
- Therefore; 1 mole of Chlorine = 6.022 × 10^23 atoms
But since Chlorine gas is a molecule;
- 1 mole of Chlorine gas = 2 × 6.022 × 10^23 atoms
But, molar mass of Chlorine gas = 70.906 g/mol
Then;
70.906 g Of chlorine gas = 2 × 6.022 × 10^23 atoms
= 1.20 × 10^24 atoms
Thus;
For 1.2 × 10^24 atoms ;
= ( 70.906 g/mol × 1.2 × 10^24 atoms ) ÷ (1.20 × 10^24 atoms)
<h3>= 70.906 g </h3>
Therefore, 1.20 × 10^24 atoms of chlorine contains a mass of 70.906 g
=
Answer:
see explanation for answer
Explanation:
salivary gland: 1
stomach: 2
small intestine: 6
liver: 4
gallbladder: 5
large intestine: 3
The answers correspond with the numbers on the text boxes, so you would drag number 1 to the salivary gland and so on.
P₄O₁₀ + 6H₂O → 4H₃PO₄
The equation shows us that the molar ratio of
P₄O₁₀ : 6H₂O = 1:6
We also know that one mole of a substance contains 6.02 x 10²³ particles. We can use this to calculate the moles of water.
moles(H₂O) = (5.51 x 10²³) / (6.02 x 10²³)
= 0.92 mole
That means moles of P₄O₁₀ = 0.92 / 6
= 0.15
Each mole of P₄O₁₀ contains 4 moles of P.
moles(P) = 4 x 0.15 = 0.6 mol
Mr of P = 207 grams per mol
Mass of P = 207 x 0.6
= 124.2 grams
Answer:
The molality of isoborneol in camphor is 0.53 mol/kg.
Explanation:
Melting point of pure camphor= T =179°C
Melting point of sample = 
= 165°C
Depression in freezing point = 
Depression in freezing point is also given by formula:
= The freezing point depression constant
m = molality of the sample
i = van't Hoff factor
We have: = 40°C kg/mol
i = 1 ( organic compounds)
The molality of isoborneol in camphor is 0.53 mol/kg.
