Answer:
113 g NaCl
Explanation:
The Ideal Gas Law equation is:
PV = nRT
In this equation,
> P = pressure (atm)
> V = volume (L)
> n = number of moles
> R = 8.314 (constant)
> T = temperature (K)
The given values all have to due with the conditions fo F₂. You have been given values for all of the variables but moles F₂. Therefore, to find moles F₂, plug each of the values into the Ideal Gas Law equation and simplify.
(1.50 atm)(15.0 L) = n(8.314)(280. K)
2250 = n(2327.92)
0.967 moles F₂ = n
Using the Ideal Gas Law, we determined that the moles of F₂ is 0.967 moles. Now, to find the mass of NaCl that can react with F₂, you need to (1) convert moles F₂ to moles NaCl (via the mole-to-mole ratio using the reaction coefficients) and then (2) convert moles NaCl to grams NaCl (via molar mass from periodic table). It is important to arrange the ratios/conversions in a way that allows for the cancellation of units (the desired unit should be in the numerator).
1 F₂ + 2 NaCl ---> Cl₂ + 2NaF
Molar Mass (NaCl): 22.99 g/mol + 35.45 g/mol
Molar Mass (NaCl): 58.44 g/mol
0.967 moles F₂ 2 moles NaCl 58.44 g
---------------------- x ----------------------- x ----------------------- = 113 g NaCl
1 mole F₂ 1 mole NaCl
Vanillin is the common name for 4-hydroxy-3-methoxy-benzaldehyde.
See attached figure for the structure.
Vanillin have 3 functional groups:
1) aldehyde group: R-HC=O, in which the carbon is double bonded to oxygen
2) phenolic hydroxide group: R-OH, were the hydroxyl group is bounded to a carbon from the benzene ring
3) ether group: R-O-R, were hydrogen is bounded through sigma bonds to carbons
Now for the hybridization we have:
The carbon atoms involved in the benzene ring and the red carbon atom (from the aldehyde group) have a <u>sp²</u> hybridization because they are involved in double bonds.
The carbon atom from the methoxy group (R-O-CH₃) and the blue oxygen's have a <u>sp³</u> hybridization because they are involved only in single bonds.
Answer: m = 24.31 g/mol · 1.13 mol
Explanation: 2 mol HCl use 1 mol Mg.
Magnesium is used 0.5 · 2.26 mol = 1.13 mol
M(Mg) = 24.31 g/ mol
Answer:
<em>Alkali metals are among the most reactive metals. This is due in <u>part to their larger atomic radii and low ionization energies.</u> They tend to donate their electrons in reactions and have an oxidation state of +1. ... All these characteristics can be attributed to these elements' large atomic radii and weak metallic bonding.</em>
Explanation:
<em>I </em><em>hope</em><em> it</em><em> will</em><em> help</em><em> you</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>
<em>#</em><em>C</em><em>A</em><em>R</em><em>R</em><em>Y</em><em>O</em><em>N</em><em>L</em><em>E</em><em>R</em><em>A</em><em>N</em><em>I</em><em>N</em><em>G</em>
Answer:
2.067 L ≅ 2.07 L.
Explanation:
- The balanced equation for the mentioned reaction is:
<em>CS₂(g) + 3O₂(g) → CO₂(g) + 2SO₂(g),</em>
It is clear that 1.0 mole of CS₂ react with 3.0 mole of O₂ to produce 1.0 mole of CO₂ and 2.0 moles of SO₂.
- At STP, 3.6 L of H₂ reacts with (?? L) of oxygen gas:
It is known that at STP: every 1.0 mol of any gas occupies 22.4 L.
<u><em>using cross multiplication:</em></u>
1.0 mol of O₂ represents → 22.4 L.
??? mol of O₂ represents → 3.1 L.
∴ 3.1 L of O₂ represents = (1.0 mol)(3.1 L)/(22.4 L) = 0.1384 mol.
- To find the no. of moles of SO₂ produced from 3.1 liters (0.1384 mol) of hydrogen:
<u><em>Using cross multiplication:</em></u>
3.0 mol of O₂ produce → 2.0 mol of SO₂, from stichiometry.
0.1384 mol of O₂ produce → ??? mol of SO₂.
∴ The no. of moles of SO₂ = (2.0 mol)(0.1384 mol)/(3.0 mol) = 0.09227 mol.
- Again, using cross multiplication:
1.0 mol of SO₂ represents → 22.4 L, at STP.
0.09227 mol of SO₂ represents → ??? L.
∴ The no. of liters of SO₂ will be produced = (0.09227 mol)(22.4 L)/(1.0 mol) = 2.067 L ≅ 2.07 L.
