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slava [35]
3 years ago
12

This nuclear reaction requires a high temperature environment.

Chemistry
1 answer:
dusya [7]3 years ago
3 0
B) nuclear fission, mark me brainliest if that helps
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(pls help)
lara [203]

Answer:

1

Explanation:

you needa do the last question cause i am not learning this but 1 is right

7 0
3 years ago
Read 2 more answers
What property of lasers causes coherent light to be emitted?
Andrew [12]
The question is a.different intensities

4 0
3 years ago
Strontium metal reacts with aluminum chlorate to produce strontium
Fudgin [204]

Answer: single replacement reaction, 3Sr+2Al(ClO_3)_3\rightarrow 3Sr(ClO_3)_2+2Al

Explanation:

A single replacement reaction is one in which a more reactive element displaces a less reactive element from its salt solution.

A general single displacement reaction can be represented as :

XY+Z\rightarrow XZ+Y

As strontium metal is added to aluminum chlorate , strontium being more reactive than aluminium, displaces aluminium atom its salt solution and lead to formation of strontium  chlorate and aluminum metal.

3Sr+2Al(ClO_3)_3\rightarrow 3Sr(ClO_3)_2+2Al

3 0
3 years ago
How many grams of KO2 are needed to form 8.0 g of O2?
dem82 [27]
Using the atomic weights, K = 39, O=16 so KO2= 39+32=71. So K=39/71= 0.549 of KO2 and O2=32=0.45KO2 and O2=8g and 0.45KO2=8g so KO2=8/0.45=17.77g. So to check, K=0.549 x 17.77g=9.76 and so K+O2= 9.76+8g=17.76.
4 0
4 years ago
Read 2 more answers
When N,N-Dimethylaniline is treated with bromine, ortho and para products are observed. However, when N,N-Dimethylaniline is tre
lilavasa [31]

Answer:

See explanation below

Explanation:

To get a better understanding watch the picture attached.

In the case of the reaction with Bromine, the -N(CH₃)₂ is a strong ring activator, therefore, it promotes a electrophilic aromatic sustitution, so, in the mechanism of reaction, the lone pair of the Nitrogen, will move to the ring by resonance and activate the ortho and para positions. That's why the bromine wil go to the ortho and para positions, mostly the para position, because the -N(CH₃)₂ cause a steric hindrance in the ortho position.

In the case of the reaction with HNO₃/H₂SO₄, the acid transform the -N(CH₃)₂ in a protonated form, the anilinium ion, which is a deactivating of the ring, and also a strong electron withdrawing, so, the electrophile will go to the meta position instead.

Hope this helps.

6 0
3 years ago
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