1) We need to find mass of the solution
m=D*V
D= <span>1.725 g/mL
V= 250 mL
m=</span>1.725 g/mL*250 mL= 431.25 g
2) 75% = 0.75
0.75*431.25 ≈ 323 g of NH4NO3
Answer:
Explanation:
Whenever a question asks you, "What is the concentration after a given time?" or something like that, you must use the appropriate integrated rate law expression.
The reaction is 2nd order, because the units of k are L·mol⁻¹s⁻¹.
The integrated rate law for a second-order reaction is
Data:
k = 2.4 × 10⁻²¹ L·mol⁻¹s⁻¹
[A]₀ = 0.0100 mol·L⁻¹
[A] = 0.009 00 mol·L⁻¹
Calculation
:
Answer:
61.6 g of CO₂ were produced
Explanation:
Let's think the combustion
C + O₂ → CO₂
First of all we determine the moles:
16.8 g . 1mol / 12 g = 1.4 moles
As the question states the oxygen is remained unreacted, the oxygen is the reactant in excess, so the limiting is the C.
1 mol of C produces 1 mol of CO₂, so 1.4 moles in the reactant will produce the same amount products.
We convert the moles to mass.
1.4 mol . 44 g / 1mol = 61.6 g
A neutral atom in its ground state contains 28 electrons. this element is considered a <span>transition </span>element, and has 8 electrons in orbitals with l = 2.
Electron configuration: ₂₈X 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸ 4s².
l is azimuthal quantum number, l = 2 describes d orbital. There is eight electrons in 3d orbital.
Answer:
<h2>Acetylene to ethene</h2>
First, acetylene is converted into ethene ( CH2 = CH2 ) using hydrogen gas and Lindlar catalyst. Ethene is an an alkene compound.
<h2>Lindlar catalyst</h2>
Alkynes are converted to alkenes by in the presence of hydrogen gas and lindlar catalyst.
lindlar reagent contains CaCO3, Pd and various forms of sulfur and lead
<h2>Ethene to ethanol</h2>
Then ethanol is prepared from the reaction of ethene with dilute sulfuric acid.