Answer:
when gas condenses to liquid the quantity of energy converts.
Explanation:
two phase changes where the heat in energy is released
during storms with high wind and heavy
rains. Summarize evidence to support the
claim and explain your reasoning.
Answer:
the claim is that when people avoid storms they hide in any secret place in there house
Explanation:
when people do that they don't even have to worry about a single thing that will happen to them
B:Some offspring are more likely to survive a disease.
C:Less energy is required to reproduce.
D:The population can increase from only one parent.
E:The population can increase quickly.
Answer:
sorry if I get this wrong I think it is C
Explanation:
O 3.04 x 1028
O 5.76 x 1029
O 6.02 x 1023
O 3.13
There are 5.78 × 10^20 sulfide ions in sodium sulfide
The formula of the compound is Na2S. The molar mass of the compound is calculated as follows; 2(23) +32 = 46 + 32 = 78 g/mol
Number of moles of Na2S = 0.15 g/ 78 g/mol = 0.0019 moles
Since there is 1 mole of Na^+ and 2 moles of S^2- in Na2S, the number of S^2- ions in 0.19 moles of Na2S is 0.00096 moles of S^2-.
If 1 mole of S^2- contains 6.02 × 10^23
0.095 moles of S^2- contains 0.00096 moles × 6.02 × 10^23/ 1 mole
= 5.78 × 10^20 sulfide ions
Learn more: brainly.com/question/1309057
Answer:
-68.4 kJ
Explanation:
The standard enthalpy of vaporization = 23.3 kJ/mol
which means the energy required to vaporize 1 mole of ammonia at its boiling point (-33 °C).
To calculate heat released when 50.0 g of ammonia is condensed at -33 °C.
This is the opposite of enthalpy of vaporization which means that same magnitude of heat is released.
Thus, Q = -23.3 kJ/mol
Where negative sign signifies release of heat
Given: mass of 50.0 g
Molar mass of ammonia = 17.034 g/mol
Moles of ammonia = 50.0 /17.034 moles = 2.9353 moles
Also,
1 mole of ammonia when condenses at -33 °C releases 23.3 kJ
2.9412 moles of ammonia when condenses at -33 °C releases 23.3×2.9353 kJ
Thus, amount of heat released when 50 g of ammonia condensed at -33 °C= -68.4 kJ, where negative sign signifies release of heat.
The heat released when 50.0 g of ammonia condenses at its boiling point is -68.4 kJ. This is calculated by multiplying the moles of ammonia by the enthalpy of vaporization and recognizing that heat is released in condensation.
To solve this problem, we need to understand the concept of enthalpy of vaporization, which is the heat needed to convert 1 mole of a substance from a liquid to a gas at constant pressure and temperature. For ammonia (NH3), which boils at -33 °C, the enthalpy of vaporization is 23.3 kJ/mol. However, we want the heat released when 50.0 g (around 2.94 moles) of ammonia condenses, which is the reverse process of vaporization. Thus, the energy would be released rather than absorbed.
Now, let's calculate this value. We multiply the number of moles of ammonia by the enthalpy of vaporization:
2.94 moles x 23.3 kJ/mol = 68.4 kJ
Since this is the reverse of the process of vaporization, heat is released, so the enthalpy change is negative (-68.4 kJ). Therefore, the correct answer is -68.4 kJ.
#SPJ11
Answer:
10.8 mm
Explanation:
Step 1: Calculate the volume (V) of the copper cylinder
The cylinder has a mass of 94.6754 g and a density of 8.96 g/cm³. The volume of the cylinder is:
Step 2: Calculate the length (h) of the copper cylinder
The diameter (d) of the copper cylinder is 3.52 cm. We can calculate the length using the following expression.
Step 3: Convert the length to millimeters
We will use the relationship 1 cm = 10 mm.
(ii) 2-bromo-2-methylpentane.
Answer:
The mechanisms for the elimination reactions between NaOH in ethanol and the halogenoalkanes are demonstrated in the figure attached.
Explanation:
(i) 1-bromobutane will suffer elimination to for an alkene. The mechanism will be E2, which means that the attack and the elimination will occur simultaneously. This is the preferred mechanism because the bromine is in a primary carbon.
(ii) 2-bromo-2-methylpentane will suffer elimination to for an alkene. The mechanism will be E1, which means that the attack and the elimination will occur in two different steps. The bromine will be eliminated in the first step with the formation of a carbocation and in a second step the double bond will be formed after the anionic attack. This is the preferred mechanism because the bromine is in a terciary carbon which is able to stabilize the carbocation formed.