Answer:
Explanation:
Mitochondria are a part of eukaryotic cells. The main job of mitochondria is to perform cellular respiration. This means it takes in nutrients from the cell, breaks it down, and turns it into energy. This energy is then in turn used by the cell to carry out various functions.
Answer: 8.28g Na
Explanation: use ideal gas law
PV= nRT
Solve for moles of Cl2
n= PV/ RT
Substitute:
= 1 atm x 4.0 L / 0.08205 L.atm/ mol. K x 273 K
= 0.18 moles Cl2
Do stoichiometry to solve for m of Na
2 Na + Cl2 => 2 NaCl2
=0.18 moles Cl2 x 2 mol Na/ 1 mol Cl2 x 23g Na / 1 mol Na
= 8.28 g Na.
Answer:
Down below
Explanation:
The following uses nickel(II) chloride
2AgNO3(aq) + NiCl2(aq) ==> Ni(NO3)2(aq) + 2AgCl(s) Molecular
Carbons starting from the left end:
- sp²
- sp²
- sp²
- sp
- sp
Refer to the sketch attached.
<h3>Explanation</h3>
The hybridization of a carbon atom depends on the number of electron domains that it has.
Each chemical bond counts as one single electron domain. This is the case for all chemical bonds: single, double, or triple. Each lone pair also counts as one electron domain. However, lone pairs are seldom seen on carbon atoms.
Each carbon atom has four valence electrons. It can form up to four chemical bonds. As a result, a carbon atom can have up to four electron domains. It has a minimum of two electron domains, with either two double bonds or one single bond and one triple bond.
- A carbon atom with four electron domains is sp³ hybridized;
- A carbon atom with three electron domains is sp² hybridized;
- A carbon atom with two electron domains is sp hybridized.
Starting from the left end (H₂C=CH-) of the molecule:
- The first carbon has three electron domains: two C-H single bonds and one C=C double bond; It is sp² hybridized.
- The second carbon has three electron domains: one C-H single bond, one C-C single bond, and one C=C double bond; it is sp² hybridized.
- The third carbon has three electron domains: two C-C single bonds and one C=O double bond; it is sp² hybridized.
- The fourth carbon has two electron domains: one C-C single bond and one C≡C triple bond; it is sp hybridized.
- The fifth carbon has two electron domains: one C-H single bond and one C≡C triple bond; it is sp hybridized.
Answer:
ΔH = -20kJ
Explanation:
The enthalpy of formation of a compound is defined as the change of enthalpy during the formation of 1 mole of the substance from its constituent elements. For H₂S(g) the reaction that describes this process is:
H₂(g) + S(g) → H₂S(g)
Using Hess's law, it is possible to sum the enthalpies of several reactions to obtain the change in enthalpy of a particular reaction thus:
<em>(1) </em>H₂S(g) + ³/₂O₂(g) → SO₂(g) + H₂O(g) ΔH = -519 kJ
<em>(2) </em>H₂(g) + ¹/₂O₂(g) → H₂O(g) ΔH = -242 kJ
<em>(3) </em>S(g) + O₂(g) → SO₂(g) ΔH = -297 kJ
The sum of -(1) + (2) + (3) gives:
<em>-(1) </em>SO₂(g) + H₂O(g) → H₂S(g) + ³/₂O₂(g) ΔH = +519 kJ
<em>(2) </em>H₂(g) + ¹/₂O₂(g) → H₂O(g) ΔH = -242 kJ
<em>(3) </em>S(g) + O₂(g) → SO₂(g) ΔH = -297 kJ
<em>-(1) + (2) + (3): </em><em>H₂(g) + S(g) → H₂S(g) </em>
<em>ΔH =</em> +519kJ - 242kJ - 297kJ = <em>-20 kJ</em>
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I hope it helps!