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
Answer:
Abiotic is non living thing, while biotic is a living thing
hope this info helps
Answer:
q = 38,5 kJ
Explanation:
In its melting point, at 0°C, water is liquid. The boiling point of water is 100°C. It is possible to estimate the heat you required to raise the temperature of water from 0°C to 100°C using:
q = C×m×ΔT
Where C is specific heat of water (4,184J/g°C), m is mass of water (92,0g) and ΔT is change in temperature (100°C-0°C = 100°C)
Replacing:
q = 4,184J/g°C×92,0g×100°C
q = 38493 J, in kilojoules:
<em>q = 38,5 kJ</em>
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I hope it helps!
Answer:
4.25% is the final concentration of phosphoric acid.
Explanation:
Initial concentration of phosphoric acid = 
Initial volume of phosphoric acid = 
Final concentration of phosphoric acid = 
Final volume of phosphoric acid = 
( 1L = 1000 mL)
4.25% is the final concentration of phosphoric acid.
Answer:
12.62 L
Explanation:
First, we have to calculate the moles corresponding to 18.0 g of oxygen gas (MW 32.0).
18.0 g × (1 mol/32.0 g) = 0.563 mol
Then, we can find the volume occupied by 0.563 moles of oxygen at STP (273,15 K, 1.00 atm) using the ideal gas law.
P × V = n × R × T
V = n × R × T / P
V = 0.563 mol × 0.0821 atm.L/mol.K × 273.15 K / 1.00 atm
V = 12.62 L
