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natali 33 [55]
3 years ago
13

when a substance undergoes combustion and carbon completely it produces carbon monoxide and water true or false ​

Chemistry
1 answer:
eimsori [14]3 years ago
6 0

False

complete combustion produces carbon dioxide + water

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When a nucleus undergoes nuclear decay by gamma rays the atomic number of the element?
kifflom [539]
I believe that when a nucleus undergoes a nuclear decay by gamma rays the atomic number of element remains the same. Atomic number is the number of protons of a given atom of an element. Gamma decay unlike alpha and beta decay does not have an effect on the mass number and atomic number of an atom.
8 0
3 years ago
How are oxidation-reduction reactions related to how you use energy?​
Alexxandr [17]

Answer:

In oxidation reduction reactions, one species gets reduced by taking on electron(s) and another species gets oxidized by losing electrons. The movement of electrons can be used to do work. ... The electron flow can be run through a wire and these electrons can be used to do work (like run a battery). Hope this helps.

3 0
2 years ago
Calculate the ph of 0,24 m of kch3coo.? ​
Galina-37 [17]

Answer:

Correct option is A)

[H

+

]=

KaC

=

1.8×10

−6

=1.34×10

−3

pH=−log[H

+

]

=2.88

Explanation:

here is your answer if you like my answer please follow

6 0
3 years ago
Which statement best describes the nuclear model of the atom. A) negative charges dispersed in a positively charged cloud B) pos
Svetradugi [14.3K]

Answer:

D.) small, dense negative nucleus surrounded by a diffuse positively charged cloud

Explanation:

I am not 100% sure but the reason I think this is the answer is because Protons and Neutrons are in the nucleus. Protons being positively charged and neutrons being negatively charged. Again, I do not know for sure, but I think D is the answer.

4 0
3 years ago
The retina of a human eye can detect light when radiant energy incident on it is at least 4.00 × 10−17 J. For light of 535−nm wa
VLD [36.1K]

Answer:

1080 photons

Explanation:

ΔE = hc/λ => Energy per photon

h = 6.63 x 10⁻³⁴ j·s

c = 3 x 10⁸ m/s

λ= 535 nm = 5.35 x 10⁻⁷ m

ΔE/photon = (6.63 x 10⁻³⁴ j·s)(c = 3 x 10⁸ m/s)/(5.35 x 10⁻⁷ m) = 3.71 x 10⁻¹⁹ j/photon

For #photons in 4.00 x 10⁻⁷ j =  (4.00 x 10⁻⁷ j) / (3.71 x 10⁻¹⁹ j/photon) = 1076 photons ≅ 1080 photons (3 sig. figs.)

7 0
3 years ago
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