Answer:
What mass (g) of barium iodide is contained in 188 mL of a barium iodide solution that has an iodide ion concentration of 0.532 M?
A) 19.6
B) 39.1
C) 19,600
D) 39,100
E) 276
The correct answer to the question is
B) 39.1 grams
Explanation:
To solve the question
The molarity ratio is given by
188 ml of 0.532 M solution of iodide.
Therefore we have number of moles = 0.188 × 0.532 M = 0.100016 Moles
To find the mass, we note that the Number of moles = 
from which we have
Mass = Number of moles × molar mass
Where the molar mass of Barium Iodide = 391.136 g/mol
= 0.100016 moles ×391.136 g/mol = 39.12 g
We are given
n = 87.3 moles He
no = 52.0 moles
Po = 9.63 atm
Using the relationship between Pressure and number of moles
Po/no = P/n
Substituting
9.63 atm / 52.0 mol = P / 87.3 mol
Solve for P<span />
First one is False. The second is true.
To solve for the enthalpy of reaction, we apply the Hess's Law.
ΔHrxn = ∑(ν×Hf of products) - ∑(ν×Hf of reactants)
where
v is the stoichiometric coefficient determined from the balanced reaction
Hf is the standard heat of formation; these are empirical values:
*For CH₄: Hf = <span>−74.87 kJ/mol
*For O</span>₂: Hf = 0
*For CO₂: <span>-393.5 kJ/mol
*For H</span>₂O: <span>-241.82 kJ/mol
</span>ΔHrxn = [(2*-241.82 kJ/mol)+(1*-393.5 kJ/mol)] - [(1*−74.87 kJ/mol)+(2*0 kJ/mol)] =<em> -802.27 kJ/mol</em>
Answer:
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Explanation:
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