Answer:
0.7 %
Explanation:
In a star like the Sun, or of smaller mass, the most frequent chain of nuclear reactions constitutes the cycle called p-p (proton-proton). This cycle begins with the fusion of two hydrogen nuclei (two protons) into one of deuterium, with the emission of a positron (the positron is the antiparticle of the electron, it is a positive electron) and a neutrino.
In this reaction 0.7% of the mass of the protons of the initial state is converted into energy. This number, 0.7%, is huge compared to that of a chemical reaction (a combustion, for example), which is typically of the order of 0.000000001%. Deuterium can in turn be fused with another hydrogen, this with helium.
From an energy point of view, what happens as a result of all these reactions is that four hydrogen nuclei fuse together, producing one of helium, two positrons, 2 neutrinos and 24.68 MeV of kinetic energy.
Answer:
Dissociation Equations Worksheet
Write balanced chemical equations to represent the slight dissociation or the
complete dissociation for 1 mole of the following compounds. In the case of slight
dissociation use a double arrow and for complete dissociation use a single arrow.
Include phase notation in the equations.
1) silver chloride
2) sodium acetate
3) ammonium sulfate
4) calcium carbonate
5) potassium carbonate
6) sodium hydroxide
7) silver chlorate
8) iron(II) sulfate
9) lead(II) phosphate
10) lead(II) chromate
11) iron(III) chloride
12) calcium nitrate
13) iron(III) oxide
14) copper(II) sulfate
15) mercury(II) sulfide
16) zinc chloride
17) lead(II) acetate
18) aluminum phosphate
Solutions
1) AgCl(s) ↔ Ag+
(aq) + Cl-
(aq)
2) NaC2H3O2(s) Æ Na+
(aq) + C2H3O2
-
(aq)
3) (NH4)2SO4(s) Æ 2NH4
+
(aq) + SO4
2-(aq)
4) CaCO3(s) ↔ Ca2+(aq) + CO3
2-(aq)
5) K2CO3(s) Æ 2K+
(aq) + CO3
2-(aq)
6) NaOH(s) Æ Na+
(aq) + OH-
(aq)
7) AgClO3(s) Æ Ag+
(aq) + ClO3
-
(aq)
8) FeSO4(s) Æ Fe2+(aq) + SO4
2-(aq)
9) Pb3(PO4)2(s) ↔ 3Pb2+(aq) + 2PO4
3-(aq)
10) PbCrO4(s) ↔ Pb2+(aq) + CrO4
2-(aq)
11) FeCl3(s) Æ Fe3+(aq) + 3Cl-
(aq)
12) Ca(NO3)2(s) Æ Ca2+(aq) + 2NO3
-
(aq)
13) Fe2O3(s) ↔ 2Fe3+(aq) + 3O2-(aq)
14) CuSO4(s) Æ Cu2+(aq) + SO4
2-(aq)
15) HgS(s) ↔ Hg2+(aq) + S2-(aq)
16) ZnCl2(s) Æ Zn2+(aq) + 2Cl-
(aq)
17) Pb(C2H3O2)2(s) Æ Pb2+(aq) + C2H3O2
-
(aq)
18) AlPO4(s) ↔ Al3+(aq) + PO4
3-(aq)
Explanation:
Answer:
Explanation:
1. Gather the information
We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.
Mᵣ: 16.04 32.00 44.01
CH₄ + 2O₂ ⟶ CO₂ + 2H₂O
m/g: 60
1. Moles of CO₂
2. Mass of CH₄
(a) Moles of CH₄
(b) Mass of CH₄
3. Mass of O₂ required
(a) Moles of O₂
(b) Mass of O₂
Hello. You have not presented the image that presents the two elements mentioned in the question. This makes it impossible for your question to be answered. However, I will try to help you as best I can.
To say whether the compounds obey the law of multiple proportions, you should observe whether one of the elements, formed from the compounds, maintains a fixed mass, while the other element presents the mass in a varied amount of small, whole numbers, spread across the formed compounds .
This is because the law of multiple proportions states that an element must have a fixed mass when reacting with another element to create compounds. This reaction will allow this element to keep the mass fixed, while the other element will generate different compounds, where each one presents a part of the mass of the forming element, in small and whole numbers.