Answer:
158 L.
Explanation:
What is given?
Pressure (P) = 1 atm.
Temperature (T) = 112 °C + 273 = 385 K.
Mass of methane CH4 (g) = 80.0 g.
Molar mass of methane CH4 = 16 g/mol.
R constant = 0.0821 L*atm/mol*K.
What do we need? Volume (V).
Step-by-step solution:
To solve this problem, we have to use ideal gas law: the ideal gas law is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas. The formula is:
Where P is pressure, V is volume, n is the number of moles, R is the constant and T is temperature.
So, let's find the number of moles that are in 80.0 g of methane using its molar mass. This conversion is:
So, in this case, n=5.
Now, let's solve for 'V' and replace the given values in the ideal gas law equation:
The volume would be 158 L.
Answer:
The final temperature is 348.024°C.
Explanation:
Given data:
Specific heat of copper = 0.385 j/g.°C
Energy absorbed = 7.67 Kj (7.67×1000 = 7670 j)
Mass of copper = 62.0 g
Initial temperature T1 = 26.7°C
Final temperature T2 = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = T2 - T1
Q = m.c. ΔT
7670 J = 62.0 g × 0.385 j/g °C ×( T2- 26.7 °C
)
7670 J = 23.87 j.°C ×( T2- 26.7 °C
)
7670 J / 23.87 j/°C = T2- 26.7 °C
T2- 26.7 °C = 321.324°C
T2 = 321.324°C + 26.7 °C
T2 = 348.024°C
The final temperature is 348.024°C.
A simple chemical reaction with a single substrate shows a linear relationship between the rate of formation of product and the concentration of substrate, as shown below: ... The relationship between rate of reaction and concentration of substrate depends on the affinity of the enzyme for its substrate.
Answer:
a) Unsaturated
b) Supersaturated
c) Unsaturated
Explanation:
A saturated solution contains the <u>maximum amount of a solute that will dissolve in a given solvent at a specific temperature</u>.
An unsaturated solution contains <u>less solute than it has the capacity to dissolve. </u>
A supersaturated solution, <u>contains more solute than is present in a saturated solution</u>. Supersaturated solutions are not very stable. In time, some of the solute will come out of a supersaturated solution as crystals.
According to these definitions and considering that the solubility of KCl in 100 mL of H₂O at <u>20 °C is 34 g</u>, and at <u>50 °C is 43 g</u> we can label the solutions:
a) 30 g in 100 mL of H₂O at 20 °C ⇒ unsaturated
b) 65 g in 100 mL of H₂O at 50 °C ⇒ supersaturated
c) 42 g in 100 mL of H₂O at 50 °C and slowly cooling to 20 °C to give a clear solution <u>with no precipitate</u> ⇒ unsaturated (if it were saturated it would have had precipitate)
C, because a homogeneous mixture is evenly mixed.
All those others would be heterogeneous (not evenly mixed).
