Answer:
0.055g/mL
Explanation:
Data obtained from the question include:
Molar Mass of the gass sample = 71g/mol
Volume of the gas sample = 1300 mL
Density =?
The density of a substance is simply mass per unit volume. It is represented mathematically as:
Density = Mass /volume.
With the above equation, we can easily obtain the density of sample of gas as illustrated below:
Density = 71g / 1300 mL
Density = 0.055g/mL
Therefore, the density of the gas sample is 0.055g/mL
Answer:
Oxidative phosphorylation proceeds with the formation of energy laden molecules i.e; carbondioxide and water.
Therefore, Total CO₂ production is directly related to VCO₂ = R x VO₂
where, R is the respiratory quotient varing among 0.7 to 1.0 according to the energy intake (ATP) ie 0.25 of the total diet consumed .
VO₂ is, as mentioned above arterial venous oxygen difference = 6.2ml/dl
therefore, VCO₂ = 0.25 x 6.2
= 1.55 ml/dl
ie; VO₂ : VCO₂ = 6.2 : 1.55.
Explanation:
2. B. Weathering breaks rocks into minerals, and plants die and decay.
3. D. sand
4. B. silt
5. B. sand
Hope this helps!
Answer:
The atoms of different elements have different numbers of protons and electrons.
Answer : The mass of sodium bromide added should be, 18.3 grams.
Explanation :
Molality : It is defined as the number of moles of solute present in kilograms of solvent.
Formula used :
Solute is, NaBr and solvent is, water.
Given:
Molality of NaBr = 0.565 mol/kg
Molar mass of NaBr = 103 g/mole
Mass of water = 315 g
Now put all the given values in the above formula, we get:
Thus, the mass of sodium bromide added should be, 18.3 grams.
