Q unknown = -Q H2O
Q H2O= (100g x 4.18 J/g C x (25C - 20C)
Q H2O = 2090J
-2090J = (50g x c x (25C-90C)
-2090J = -3250c
.64 = c
I am a high school student and have done this problem a few months ago for my test, hope this helps
Always remember that a compound can be separated into simpler substances by chemical methods/reactions. While elements cannot be broken down into simpler substances by chemical reactions. You can do a flame test and spectrum analysis to determine whether a solid material is an element or a compound. Check the boiling and/or melting point, color or density. Also check the solid material’s reaction with oxygen, hydrogen, calcium, or various acids. Examine and study its physical chemistry. The element(s) that may be present may be identified by checking the absorption edges from an x-ray spectrum.
Answer:
The Key difference between average vs weighted average is that simple average is nothing but simply adding up all the observation values and dividing the same by the total number of observations to calculate the average whereas weighted average is an average where each observation value will have a frequency assigned.
Explanation:
Answer:
y1 = 0.3162
y2 = 0.6838
Explanation:
ok let us begin,
first we would be defining the parameters;
at 25°C;
1-propanol P1° = 20.90 Torr
2-propanol P2° = 45.2 Torr
From Raoults law:
P(1-propanol) = P⁰ × X(1-propanol)
P(1-propanol) = 20.9 torr × 0.45 = 9.405
P(1-propanol) = 9.405 torr
Also P(2-propanol) = P⁰ × X(2-propanol)
P(2-propanol) = 45.2 torr × 0.45
P(2-propanol) = 20.34 torr
but the total pressure = sum of individual pressures
total pressure = 9.405 + 20.34
total pressure = 29.745 torr
given that y1 and y2 represent the mole fraction of each in the vapor phase
y1 = P1 / total pressure
y1 = 9.405/29.745
y1 = 0.3162
Since y1 + y2 = 1
y2 = 1 - y1
∴ y2 = 1 - 0.3162
y2 = 0.6838
cheers, i hope this helps.
Answer:
1.36 × 10³ mL of water.
Explanation:
We can utilize the dilution equation. Recall that:
Where <em>M</em> represents molarity and <em>V</em> represents volume.
Let the initial concentration and unknown volume be <em>M</em>₁ and <em>V</em>₁, respectively. Let the final concentration and required volume be <em>M</em>₂ and <em>V</em>₂, respectively. Solve for <em>V</em>₁:
Therefore, we can begin with 0.640 L of the 2.50 M solution and add enough distilled water to dilute the solution to 2.00 L. The required amount of water is thus:
Convert this value to mL:
Therefore, about 1.36 × 10³ mL of water need to be added to the 2.50 M solution.
