Answer:
537.68 torr.
Explanation:
- We can use the general law of ideal gas:<em> PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n and V are constant, and have different values of P and T:
<em>(P₁T₂) = (P₂T₁).</em>
P₁ = 485 torr, T₁ = 40°C + 273 = 313 K,
P₂ = ??? torr, T₂ = 74°C + 273 = 347 K.
∴ P₂ = (P₁T₂)/(P₁) = (485 torr)(347 K)/(313 K) = 537.68 torr.
You have to find the stoichiometric ratio between AlCl₃ and BaCl₂. The common element between them is Cl. So, the ratio of Cl in BaCl₂ to AlCl₃ is 2/3. The molar mass of AlCl₃ is 133.34 g/mol. The solution is as follows:
Mass of AlCl₃ = (6 mol BaCl₂)(2 mol Cl/1 mol BaCl₂)(1 mol AlCl₃/3 mol Cl)(133.34 g/mol) = 533.36 g AlCl₃
Hydrocarbons are carbon and hydrogen. Methane is CH4, and propane is C3H8. Methene is CH3, and propene is C3H6. Carbohydrates are hydrates of carbon. They <span>have the general formula (CH2O)x. thats how it differs
</span>
<h3>
Answer:</h3>
2.47 × 10^24 molecules
<h3>
Explanation:</h3>
One mole of a compound contains molecules equivalent to the Avogadro's number, 6.022 × 10^23.
That is, 1 mole of a compound = 6.022 × 10^23 molecules
Therefore,
1 mole of Na₂CO₃ = 6.022 × 10^23 molecules
Thus, we can calculate the number of molecules in 4.1 moles of Na₂CO₃
we get,
= 4.1 moles × 6.022 × 10^23 molecules
= 2.47 × 10^24 molecules
Hence, 4.1 moles of Na₂CO₃ contains 2.47 × 10^24 molecules
Answer:
The amount of heat that is released is -925.2 cal
Explanation:
Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
Sensible heat is the amount of heat that a body can receive or release without affecting its molecular structure, that is, it does not change the state (solid, liquid, gaseous). In other words, sensible heat is the amount of heat that a body absorbs or releases without any changes in its physical state.
The equation that allows to calculate heat exchanges is:
Q = c * m * ΔT
Where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature.
In this case:
- c= 1

- m= 25.7 g
- ΔT= Tfinal - Tinitial= 49 °C - 85 °C= -36 °C
Replacing:
Q= 1
*25.7 g* (-36 C)
Solving:
Q= -925.2 cal
<u><em>The amount of heat that is released is -925.2 cal</em></u>