Answer:
The formula for lithium acetate is CH3COOLi
Explanation:
The formula for lithium acetate is obtained by replacing the hydrogen atom bonding to the oxygen atom in acetic acid with Li as shown below:
CH3COOH + LiOH —> CH3COOLi + H2O
Answer:
A
Explanation:
Element A has 3 electrons in outermost shell so its valency is 3. It will loose 3 electrons to attain stability, as loss of 3 electrons is easier than gain of 5 electrons.
Valency of B is −2 as it will gain 2 electrons to attain stability and combine with other atom.
Valency of A⟶+3
Valency of B⟶−2
(Refer to Image)
Cross multiply valency of A and B
∴A2B3 compound will be formed.
Answer:
The amount of heat required to raise the temperature of a 32g sample of water from 8°C to 22°C is 1,874.432 J
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 absorbs or releases without any changes in its physical state (phase change).
Between heat and temperature there is a direct proportional relationship. The constant of proportionality depends on the substance that constitutes the body and its mass, and is the product of the specific heat and the mass of the body. So, 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= 4.184
- m= 32 g
- ΔT= Tfinal - Tinitial= 22°C - 8°C= 14°C
Replacing:
Q= 32 g* 4.184 *14 °C
Solving:
Q= 1,874.432 J
<u><em>The amount of heat required to raise the temperature of a 32g sample of water from 8°C to 22°C is 1,874.432 J</em></u>
Answer: 22.4L
Gas at STP means gas at standard temperature and pressure of one mole of an ideal gas
Ideal gas Law PV=nRT
Standard pressure is 1atm or 22.4L
Standard temperature is 273.15K
Any questions feel free to ask. Thanks
Answer:
The ΔH°rxn for this reaction is -905.9 kJ
Explanation:
<u>Step 1:</u> Balanced reaction:
4NH3(g) + 5O2(g) --> 4NO(g) + 6H2O(g)
<u />
<u>Step 2</u>: Given data
ΔH°f [NH3(g)] = –45.9 kJ/mol
ΔH°f [NO(g)] = 90.3 kJ/mol
ΔH°f [H2O(g)] = –241.8 kJ/mol
ΔH°f of O2 is defined to be zero
<u>Step 3</u>: Calculate ΔH°rxn
ΔH°rxn = ΔH°products - ΔH°reactants
ΔH°rxn = ((4*90.3) + 6*-(241.8)) - 4*(-45.9) = -905.9 kJ
The ΔH°rxn for this reaction is -905.9 kJ
