Answer: Thus the mass, in grams, of 27 moles of smooth muscle myosin is 12960000 grams
Explanation:
According to avogadro's law, 1 mole of every substance weighs equal to the molecular mass, occupies 22.4 L at STP contains avogadro's number
of particles.
Molecular mass of protein = 480,000 g/mol
Thus 1 mole of protein weighs = 480,000 g
So 27 moles of protein weighs = 
Thus the mass, in grams, of 27 moles of smooth muscle myosin is 12960000 grams
16 g/ml
<h3>Further explanation</h3>
Given
mass = 32.40 g
volume = 2.0mL
Required
Density
Solution
Density is a quantity derived from the mass and volume
Density is the ratio of mass per unit volume
Density formula:

ρ = density
m = mass
v = volume
Input the value :
ρ = 32.40 g / 2.0 ml
ρ = 16.2 g/ml
Rules for division: The least number of significant figures in the problem determines the number of significant figures in the answer.
32.40 = 4 sig.fig
2.0 = 2 sig fig
The answer must be 2 sig fig
So the answer = 16 g/ml
Is this an actual question?
Answer:
Explanation:
<u>1. Number of moles of gasoline</u>
a) Convert 60.0 liters to grams
- mass = 0.77kg/liter × 60.0 liter = 46.2 kg
- 46.2kg × 1,000g/kg = 46,200g
b) Convert 46,200 grams to moles
- molar mass of C₈H₁₈ = 114.2 g/mol
- number of moles = mass in grams / molar mass
- number of moles = 46,200g / (114.2 gmol) = 404.55 mol
<u>2. Number of moles of carbon dioxide, CO₂ produced</u>
a) Balanced chemical equation (given):
- C₈H₁₈ (l) + ²⁵/₂ O₂ (g) → 8 CO₂ (g) + 9 H₂O (g)
b) mole ratio:
- 1 mol C₈H₁₈ / 8 mol CO₂ = 404.55 mol C₈H₁₈ / x
Solve for x:
- x = 404.55mol C₈H₁₈ × 8 mol CO₂ / 1mol C₈H₁₈ = 3,236.4 mol CO₂
<u> 3. Convert the number of moles of carbon dioxide to volume</u>
Use the ideal gas equation:
- R = 0.08206 (mol . liter)/ (K . mol)
Substitute and compute:
- V =3,236.4 mol × 0.08206 (mol . liter) / (K . mol) 298.15K / 1 atm
Round to two significant figures (because the density has two significant figures): 79,000 liters ← answer
Answer:
D
Explanation:
because bio means life and biology would be the study of life.