Answer:
In order to initiate most fission reactions, an atom is bombarded by a neutron to produce an unstable isotope, which undergoes fission. When neutrons are released during the fission process, they can initiate a chain reaction of continuous fission which sustains itself.
If you do not wait until the crucible is at room temperature, its density will vary constantly until it reaches this temperature, since density is a property that varies with temperature. For the above reason, weighing a hot crucible will not allow the measurement of a constant weight on the balance, since <u>the weight of the object will be constantly changing, which will not allow a constant reading.</u>
Answer:
Ag⁺(aq) + I⁻(aq) → AgI(s)
Explanation:
Net ionic equation is a way to write a chemical equation in which you are listing only the species that are participating in the reaction.
In the reaction:
AgNO₃(aq) + NaI(aq) → AgI(s) + NaNO₃(aq).
The ionic equation is:
Ag⁺(aq) + NO₃⁻(aq) + Na⁺(aq) + I⁻(aq) → AgI(s) + Na⁺(aq) + NO₃⁻(aq).
Now, listing only the species that are participating in the reaction:
<h3>Ag⁺(aq) + I⁻(aq) → AgI(s)</h3>
This is equivalent to having a standard enthalpy change of reaction equal to 10.611 kJ
<u>Explanation</u>:
The standard enthalpy change of reaction, Δ
H
∘
, is given to you in kilojoules per mole, which means that it corresponds to the formation of one mole of carbon dioxide.
C
(s] + O
2(g]
→
CO
2(g]
Remember, a negative enthalpy change of reaction tells you that heat is being given off, i.e. the reaction is exothermic.
First to convert grams of carbon into moles,
use carbon's molar mass(12.011 g).
Moles of C = mass in gram / molar mass
= 0.327 g / 12.011 g
Moles of C = 0.027 moles
Now, in order to determine how much heat is released by burning of 0.027 moles of carbon to form carbon-dioxide.
= 0.027 moles C 
393 kJ
Heat released = 10.611 kJ.
So, when 0.027 moles of carbon react with enough oxygen gas, the reaction will give off 10.611 kJ of heat.
This is equivalent to having a standard enthalpy change of reaction equal to 10.611 kJ
The masses can be found by substractions:
- Mass of CaSO₄.H2O (hydrate):
16.05 g - 13.56 g = 2.49 g
15.07 g - 13.56 g = 1.51 g
- The mass of water is equal to the difference between the mass of the hydrate and the mass of the anhydrate:
2.49 g - 1.51 g = 0.98 g
- The percent of water is found by the formula:
massWater ÷ massHydrate * 100%
0.98 g ÷ 2.49 g * 100% = 39.36%
- The mole of water is calculated using water's molecular weight (18g/mol):
0.98 g ÷ 18 g/mol = 0.054 mol water
- A similar procedure is made for the mole of salt (CaSO₄ = 136.14 g/mol)
1.51 g ÷ 136.14 g/mol = 0.011 mol CaSO₄
- The ratio of mole of water to mole of anhydrate is:
0.054 mol water / 0.011 mol CaSO₄ = 0.49
In other words the molecular formula for the hydrate salt is CaSO₄·0.5H₂O
