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Nastasia [14]
3 years ago
11

Uranium-236 is _____. created by fusion reactions an unstable isotope of uranium created from four hydrogen atoms used in the H-

bomb
Chemistry
2 answers:
nata0808 [166]3 years ago
8 0
<span>Uranium-236 is intermediate nuclei. created by fusion reactions an unstable isotope of uranium created from four hydrogen atoms used in the H-bomb.

Following is the reaction involved in above process:

</span>^{235}U+ ^{1}n →  ^{236}U→ ^{144}Ba + ^{89}Kr + 3 ^{1}n<span> + 177 MeV
</span>
Here, 
^{235}U = Fission material,
^{1}n = projectile,
^{236}U = intermediate nuclei,
^{144}Ba and ^{89}Kr = Fission product



Shkiper50 [21]3 years ago
8 0
The asnwer in the blank would be intermediate nuclei. Uranium-236 is intermediate nuclei created by fusion reactions an unstable isotope of uranium created from four hydrogen atoms used in the H-bomb. This is use in radioactivity and  generates the heat in nuclear power reactors, and produces the fissile material for nuclear weapons.
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2 years ago
How many moles of carbon in 6.64 moles of CCl2 F
Bad White [126]

Answer: 6.64 moles of carbon.

Explanation:

Given data:

Number  of moles of C = ?

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Solution:

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Read more on Brainly.com - brainly.com/question/15602143#readmore

3 0
3 years ago
What is the pH of a 3.40 mM of acetic acid, 500 mL in which 50 mL of 1.00 M NaOAc has been added? Ka HOAc is 1.8 x 10-5? What pe
Deffense [45]

Answer:

a) pH = 4.213

b) % dis = 2 %

Explanation:

Ch3COONa → CH3COO- + Na+

CH3COOH ↔ CH3COO- + H3O+

∴ Ka = 1.8 E-5 = ([ CH3COO- ] * [ H3O+ ]) / [ CH3COOH ]

mass balance:

⇒ <em>C</em> CH3COOH + <em>C</em> CH3COONa = [ CH3COOH ] + [ CH3COO- ]

<em>∴ C </em>CH3COOH = 3.40 mM = 3.4 mmol/mL * ( mol/1000mmol)*(1000mL/L)

∴ <em>C</em> CH3COONa = 1.00 M = 1.00 mol/L = 1.00 mmol/mL

⇒ [ CH3COOH ] = 4.4 - [ CH3COO- ]

charge balance:

⇒ [ H3O+ ] + [ Na+ ] = [ CH3COO- ] + [ OH- ]....is negligible [ OH-], comes from water

⇒ [ CH3COO- ] = [ H3O+ ] + 1.00

⇒ Ka = (( [ H3O+ ] + 1 )* [ H3O+ ]) / ( 3.4 - [ H3O+])) = 1.8 E-5

⇒ [ H3O+ ]² + [ H3O+ ] = 6.12 E-5 - 1.8 E-5 [ H3O+ ]

⇒ [ H3O+ ]² + [ H3O+ ] - 6.12 E-5 = 0

⇒  [ H3O+ ] = 6.12 E-5 M

⇒ pH = - Log [ H3O+ ] = 4.213

b) (% dis)* mol acid = <em>C</em> CH3COOH = 3.4

∴ mol CH3COOH = 500*3.4 = 1700 mmol = 1.7 mol

⇒ % dis = 3.4 / 1.7 = 2 %

7 0
3 years ago
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