Answer:
7650
Explanation:
formula- multiply the volume value by 1e+6
Answer:
Cell culture is one of the most important techniques in cellular and molecular biology since it provides a platform to investigate the biology, biochemistry, physiology (e.g., aging) and metabolism of wild-type cells and diseased cells.
Molecular equation
Hg₂(NO₃)₂ (aq) + KI(aq) ⇒Hg₂I₂(s) + 2KNO₃(aq)
Total Ionic equation
Hg²⁺(aq) + 2NO³⁻(aq) + 2K⁺aq) ⇒Hg₂I₂(s) + 2K⁺(aq) + NO³⁻ (aq)
Net Ionic equation
Hg²⁺(aq) + 2I⁻(aq) ⇒ Hg₂I₂(s)
<h3>What is the molecular equation?</h3>
Sometimes, a balanced equation is all that is used to refer to a chemical equation. Any ionic substances or acids are represented using their chemical formulas as neutral compounds in a molecular equation. Each substance's state is described in parenthesis after the formula. A complete ionic equation also contains the spectator ions, whereas a net ionic equation just displays the chemical species that are involved in a reaction.
The steps listed below can be used to determine the net ionic equation for a specific reaction:
Include the states of each chemical in the balanced molecular equation for the reaction.
To know more about the molecular equation, visit:
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Answer:
A) 8.00 mol NH₃
B) 137 g NH₃
C) 2.30 g H₂
D) 1.53 x 10²⁰ molecules NH₃
Explanation:
Let us consider the balanced equation:
N₂(g) + 3 H₂(g) ⇄ 2 NH₃(g)
Part A
3 moles of H₂ form 2 moles of NH₃. So, for 12.0 moles of H₂:
Part B:
1 mole of N₂ forms 2 moles of NH₃. And each mole of NH₃ has a mass of 17.0 g (molar mass). So, for 4.04 moles of N₂:
Part C:
According to the <em>balanced equation</em> 6.00 g of H₂ form 34.0 g of NH₃. So, for 13.02g of NH₃:
Part D:
6.00 g of H₂ form 2 moles of NH₃. An each mole of NH₃ has 6.02 x 10²³ molecules of NH₃ (Avogadro number). So, for 7.62×10⁻⁴ g of H₂:
Here we apply the Clausius-Clapeyron equation:
ln(P₁/P₂) = ΔH/R x (1/T₂ - 1/T₁)
The normal vapor pressure is 4.24 kPa (P₁)
The boiling point at this pressure is 293 K (P₂)
The heat of vaporization is 39.9 kJ/mol (ΔH)
We need to find the vapor pressure (P₂) at the given temperature 355.3 K (T₂)
ln(4.24/P₂) = 39.9/0.008314 x (1/355.3 - 1/293)
P₂ = 101.2 kPa