Answer:
81°C.
Explanation:
To solve this problem, we can use the relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat released from water (Q = - 1200 J).
m is the mass of the water (m = 20.0 g).
c is the specific heat capacity of water (c of water = 4.186 J/g.°C).
ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = final T - 95.0°C).
∵ Q = m.c.ΔT
∴ (- 1200 J) = (20.0 g)(4.186 J/g.°C)(final T - 95.0°C ).
(- 1200 J) = 83.72 final T - 7953.
∴ final T = (- 1200 J + 7953)/83.72 = 80.67°C ≅ 81.0°C.
<em>So, the right choice is: 81°C.</em>
Answer:
Three of the five oxides are expected to form acidic solutions in water
Explanation:
We have different types of oxides : Acidic oxides, Basic oxides, Amphoteric oxides, Peroxides and Higher oxides.
Only acidic oxides will dissolve in water to give an acidic solution.
Considering the given oxides carefully,
- SO2 will dissolve in water to produce H2SO3 which is acidic.
- Y2O3 will dissolve in water to produce Yttrium(III) hydroxide which is basic.
- MgO will dissolve in water only to produce Mg(OH)2 which is also basic.
- Cl2O dichlorine mono oxide will dissolve in water to produce HClO which is acidic.
- N2O5 will dissolve in water to produce HNO3 which is also acidic.
Answer:
- Molar mass = 608.36 g/mol
Explanation:
It seems the question is incomplete. However a web search us shows this data:
" Reserpine is a natural product isolated from the roots of the shrub Rauwolfia serpentina. It was first synthesized in 1956 by Nobel Prize winner R. B. Woodward. It is used as a tranquilizer and sedative. When 1.00 g reserpine is dissolved in 25.0 g camphor, the freezing-point depression is 2.63 °C (Kf for camphor is 40 °C·kg/mol). Calculate the molality of the solution and the molar mass of reserpine. "
The <em>freezing-point depression</em> is expressed by:
We put the data given by the problem and <u>solve for m</u>:
- 2.63 °C = 40°C·kg/mol * m
For the calculation of the molar mass:<em> Molality</em> is defined as moles of solute per kilogram of solvent:
- 0.06575 m = Moles reserpine / kg camphor
- 25.0 g camphor ⇒ 25.0/1000 = 0.025 kg camphor
We<u> calculate moles of reserpine:</u>
- 0.06575 m = Moles reserpine / 0.025 kg camphor
- Moles reserpine = 1.64x10⁻³ mol
Finally we use the mass of reserpine and the moles to calculate <u>the molar mass</u>:
- 1.00 g reserpine / 1.64x10⁻³ mol = 608.36 g/mol
<em>Keep in mind that if the data in your problem is different, the results will be different. But the solving method remains the same.</em>
Answer:
Final pressure = 2.3225 atm
Amontons’s law states that
At constant volume and number of molecules, the pressure of a given mass of gas is directly proportional to its temperature
Explanation:
Temperature causes increased excitement of gas molecules increasing the number of collisions with the walls of the container which is sensed as increase in pressure
Amontons’s law: P/T = Constant at constant V and n
That is P1/T1 = P2/T2
Where temperature is given in Kelvin
Hence T1 of 10°C = 273.15 + 10 = 283.15K
Also temperature T2 of 40°C = 313.15 K
Hence
P2 = (P1/T1)×T2 = (2.1/283.15)×313.15 = 2.3225 atm
