It would most likely be fossil fuels.
Answer: 64.1 ml of concentrated is taken and (250-64.4)= 185.6 ml f water is added to make 250.00 mL of 3.00 M HCl .
Explanation:
Given : 36 g of is dissolved in 100 g of solution.
Density of solution = 1.18 g/ml
Thus volume of solution =
where,
n= moles of solute
= volume of solution
According to the dilution law,
where,
= molarity of stock solution = 11.7M
= volume of stock solution = V ml
= molarity of diluted solution = 3.00 M
= volume of diluted solution = 250.0 ml
Thus 64.1 ml of stock is taken and (250-64.4)= 185.6 ml of water is added to make 250.00 mL of 3.00 M HCl .
Answer:
8.934 g
Step-by-step explanation:
We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 192.12 44.01
H₃C₆H₅O₇ + 3NaHCO₃ ⟶ Na₃C₆H₅O₇ + 3H₂O + 3CO₂
m/g: 13.00
For ease of writing, let's write H₃C₆H₅O₇ as H₃Cit.
(a) Calculate the <em>moles of H₃Cit
</em>
n = 13.00 g × (1 mol H₃Cit /192.12 g H₃Cit)
n = 0.067 67 mol H₃Cit
(b) Calculate the <em>moles of CO₂
</em>
The molar ratio is (3 mol CO₂/1 mol H₃Cit)
n = 0.067 67 mol H₃Cit × (3 mol CO₂/1 mol H₃Cit)
n = 0.2030 mol CO₂
(c) Calculate the <em>mass of CO₂
</em>
m = 0.2030 mol CO₂ × (44.01 g CO₂/1 mol CO₂)
m = 8.934 g CO₂
I can help but more Context is needed. Like exactly what your looking at for the Melting point, boiling point and the appearance.