Answer:
1.58gcm⁻³
Explanation:
Given parameters:
Volume of sample = 1.00L
Mass of sample = 1.58kg
unknown:
Density of the sample
Solution:
The density of a substance is expressed as its mass per unit volume.
The units of the parameters are given for volume as liters and for mass as kg.
We need to convert to cm³ for volume and into g for the mass:
1L = 1000cm³:
Therefore, the volume of the sample is 1000cm³
For the mass:
1kg = 1000g
1.58kg gives 1.58 x 1000g = 1580g
The density of the sample will be:
Density =
Density = = 1.58gcm⁻³
Answer:
As there are two objects applying forces to each other they must be equal and opposite. Thus the force on your foot will be precisely 26.3N in a direction 180∘ to the direction of travel of the ball.
Explanation:
The answer to this question would be: true
Adiabatic is a process of thermodynamic where the system temperature is increased without taking it from the surrounding. The temperature is increased from the increased pressure by compressing or expanding the air.
Two atoms far apart being brought closer to each other, their potential energy decreases until the nuclei repel each other (the repulsion). The nucleus of one atom is attracted to the electrons of the other atom. As the distance gets less, the potential energy gets less.
Answer:
Explanation:
The clue of this question is to find the molar mass of the <em>diprotic acid</em> and compare witht the molars masses of the choices' acid to identify the formula of the diprotic acid.
The procedure is:
- Find the number of moles of the base: LiOH
- Use stoichionetry to infere the number of moles of the acid.
- Use the formula molar mass = mass in grams / number of moles, to find the molar mass of the diprotic acid.
- Compare and conclude.
<u>Solution:</u>
<u>1. Number of moles of the base, LiOH:</u>
- M = n / V in liter ⇒ n = M × V = 0.100 M × 40.0 ml × 1 liter / 1,000 ml = 0.004 mol LiOH.
<u>2. Stoichiometry:</u>
Since this a neutralization reaction of a diprotic acid with a mono hydroxide base (LiOH), the mole ratio at the second equivalence point is: 2 mol of base / 1 mole of acid; because each mole of LiOH releases 1 mol of OH⁻, while each mole of diprotic acid releases 2 mol of H⁺.
Hence, 0.004 mol LiOH × 1 mol acid / 2 mol LiOH = 0.002 mol acid.
<u>3. Molar mass of the acid:</u>
- molar mass = mass in grams / number of moles = 0.300 g / 0.002 mol = 150. g/mol
<u>4. Molar mass of the possible diprotic acids:</u>
a. H₂Se: 2×1.008 g/mol + 78.96 g/mol = 80.976 g/mol
b. H₂Te: 2×1.008 g/mol + 127.6 g/mol = 129.616 g/mol
c. H₂C₂O₄ ≈ 2×1.008 g/mol + 2×12.011 g/mol + 4×15.999 g/mol ≈ 90.034 g/mol
d. H₂C₄H₄O₆ = 6×1.008 g/mol + 4×12.011 g/mol + 6×15.999 g/mol = 150.086 g/mol ≈ 150 g/mol.
<u>Conclusion:</u> since the molar mass of H₂C₄H₄O₆ acid is 150 g/mol, you conclude that is the diprotic acid whose 0.300 g were titrated with 40.0 ml of 0.100 M LiOH solution.