It looks like we are solving for a pressure. All that is required is some algebraic manipulation to find our pressure in mmHg.
Given:
(5.0 m³)(7.5 mmHg) = (P)(4.0m³)
Multiply:
37.5 = 4.0P
Divide:
9.375 = P
P = 9.4 mmHg (remember sig figs)
<h3>
Answer:</h3>
9.4 mmHg
<span>the balanced chemical equation for the reaction is as follows;
C</span>₃H₈ + 5O₂ ---> 3CO₂ + 4H₂<span>O
stoichiometry of </span> C₃H₈ to O₂ is 1:5
number of moles of C₃H₈ reacted - 0.025 g / 44.1 g/mol = 0.000567 mol according to molar ratio of 1:5
number of O₂ moles required are 5 times the amount of C₃H₈ moles reacted therefore number of O₂ moles required - 0.000567 x 5 = 0.00284 mol .
mass of O₂ required - 0.00284 mol x 32.00 g/mol = 0.091 mol .
answer is 0.091 mol
Answer: I THINK C
Explanation: It has a 25% chance of being right so I'd wait if I were you just sayin
Answer:
The boiling point of a 8.5 m solution of Mg3(PO4)2 in water is<u> 394.91 K.</u>
Explanation:
The formula for molal boiling Point elevation is :
= elevation in boiling Point
= Boiling point constant( ebullioscopic constant)
m = molality of the solution
<em>i =</em> Van't Hoff Factor
Van't Hoff Factor = It takes into accounts,The abnormal values of Temperature change due to association and dissociation .
In solution Mg3(PO4)2 dissociates as follow :
Total ions after dissociation in solution :
= 3 ions of Mg + 2 ions of phosphate
Total ions = 5
<em>i =</em> Van't Hoff Factor = 5
m = 8.5 m
= 0.512 °C/m
Insert the values and calculate temperature change:
Boiling point of pure water = 100°C = 273.15 +100 = 373.15 K
= 373.15 K[/tex]
21.76 = T - 373.15
T = 373.15 + 21.76
T =394.91 K
Answer:
Here are a few more examples:
Smoke and fog (Smog)
Dirt and water (Mud)
Sand, water and gravel (Cement)
Water and salt (Sea water)
Potassium nitrate, sulfur, and carbon (Gunpowder)
Oxygen and water (Sea foam)
Petroleum, hydrocarbons, and fuel additives (Gasoline)
Heterogeneous mixtures possess different properties and compositions in various parts i.e. the properties are not uniform throughout the mixture.
Examples of Heterogeneous mixtures – air, oil, and water, etc.
Examples of Homogeneous mixtures – alloys, salt, and water, alcohol in water, etc.
Explanation:
