Answer:
0.1 g/dl
Explanation:
The standard curve is a graph that relates the absorbance at 400 nm with the concentration of haemoglobin in mg/dl. To obtain the concentration from the absorbance value, we enter in the x-axis (absorbance at 400 nm) with the value 0.40 (the line between 0.2 and 0.6), we extrapolate the line to the curve and read the correspondent value on y-axis (concentration in mg/dl): 100 mg/dl.
So, we convert the concentration from mg/dl to g/dl by dividing into 1000:
100 mg/dl x 1 g/1000 mg = 0.1 g/dl
Therefore, the concentration of haemoglobin of the patient is 0.1 g/dl.
Na2C2O4(aq) + CaCl2(aq) -----> 2NaCl(aq) + CaC2O4(s)
Here, CaC2O4(s) is a precipitate in the reaction as a result of precipitation reaction or double displacement reaction.
As we know that double displacement reaction two metal ions displaces each other from their salt solutions.
As we know that precipitation reaction is a reaction in which precipitate is formed.
Your answer would be a change in odor! Hope this helps! ;D
Answer:
A. fluorine, 1.79 moles
Explanation:
Given parameters:
Mass of carbon = 87.7g
Mass of fluorine gas = 136g
Unknown:
The limiting reactant and the maximum amount of moles of carbon tetrafluoride that can be produced = ?
Solution:
Equation of the reaction:
C + 2F₂ → CF₄
let us find the number of the moles the given species;
Number of moles = 
C; molar mass = 12;
Number of moles = 
= 7.31moles
F; molar mass = 2(19) = 38g/mol
Number of moles = 
= 3.58moles
So;
From the give reaction:
1 mole of C requires 2 moles of F₂
7.31 moles of C will then require 2 x 7.31 moles of F₂ = 14.62moles
But we have 3.58 moles of the F₂;
Therefore, the reactant in short supply is F₂ and it is the limiting reactant;
So;
2 moles of F₂ will produce mole of CF₄
3.58 moles of F₂ will then produce 
= 1.79moles of CF₄
