If the sun was not there the earth would travel in a straight line
Answer:-
Alpha decay
Explanation:-
Uranium 238 has atomic number 92 and mass number 238.
Thorium 234 has atomic number 90 and mass number 234.
So, the change in atomic number as uranium 238 disintegrates into thorium234 = 92 – 90 = 2
So, the change in mass number as uranium 238 disintegrates into thorium234= 238 – 234 = 4
We know that when an alpha particle is emitted, the mass number decreases by 4 and the atomic number decreases by 2.
So when an atom of uranium 238 undergoes radioactive decay to form an atom of thorium-234, alpha decay has occurred.
Answer:
Maintenance of homeostasis usually involves negative feedback loops. These loops act to oppose the stimulus, or cue, that triggers them.
Sulfuric acid is prepared industrially by means of the response of water with sulfur trioxide which in turn is made by way of a chemical combination of sulfur dioxide and oxygen both by way of the touch system or the chamber system.
H2SO4 (l) H2O (g) + SO3 (g).
The reaction is highly exothermic as an enormous amount of heat is liberated.
The usual approach is to dilute the sulfur trioxide in sulphuric acid. This produces oleum. SO3 (g) + H2SO4 → H2S2O7 (1) Oleum can be in addition diluted in water to acquire concentrated sulphuric acid.
An acid catalyst is added to protonate the carbonyl carbon. How does this catalyze the response, robust acid catalysts catalyze the hydrolysis and transesterification of esters which enables the mechanism with a view to boom the electrophilicity of the carbonyl carbon to assist protonate the carbonyl oxygen.
Learn more about sulfuric acid here:-brainly.com/question/11857977
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Answer:
81.97 g of NaAl(OH)₄
Solution:
The reaction for the preparation of Sodium Aluminate from Aluminium Metal and NaOH is as follow,
2 NaOH + 2 Al + 6 H₂O → 2 NaAl(OH)₄ + 3 H₂
According to this equation,
2 Moles of NaOH produces = 163.94 g (2 mole) of NaAl(OH)₄
So,
1 Mole of NaOH will produce = X g of NaAl(OH)₄
Solving for X,
X = (1 mol × 163.94 g) ÷ 2 mol
X = 81.97 g of NaAl(OH)₄
