Alkanes can be prepared from carboxylic acid via the removal of carbon dioxide. This process is known as decarboxylation. It produces alkane with a carbon atom lesser than that present in the carboxylic acid.
Answer:
The answer to your question is 2.21 %
Explanation:
Data
mass of Na₂SO₄ = 10.9 g
mass of H₂O = 482 g
Formula
Percent by mass = 
x 100
Substitution
x 100
Simplification and result
x 100
Percent by mass = 0.022 x 100
Percent by mass = 2.21
Answer:
ΔS = -0.1076 kJ /kg*K
Explanation:
Step 1: Data given
Initial state = 0.8 m³/kg and 25 °C = 298.15 K
Final state = 0. 3³/kg and 287 °C = 560.15 K
Cv = 0.686 kJ/kg*K
Step 2: Calculate the average temperature
The average temperature = (25°C + 287 °C)/2 =156 °C ( = 429 K)
Step 3: Calculate the ΔS
ΔS =(Cv, average) * ln(T2/T1) + R*ln(V2/V1)
ΔS = 0.686 * ln(560.15/298.15) + 0.2598*ln( 0.1/0.8)
ΔS = -0.1076 kJ /kg*K
A. 6 NaOH + 2(NH4)3 PO4 -----> 2Na PO4 + 6H2O + 6NH3
b. C2 H6 O + 3O2 ----> 2CO2 + 3H2O
In 3.8 moles of li, there are about <span>0.144071459 grams. I assume that you are talking about lithium.</span>
