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dem82 [27]
3 years ago
5

Please help!

Chemistry
1 answer:
Dahasolnce [82]3 years ago
6 0

Answer:

reproduction

Explanation:

reproduction, process by which organisms replicate themselves

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Aluminum has a density of 2.70 g/mL. A sample of aluminum weighs 67.5 g. If this sample is put into a graduated cylinder that ha
romanna [79]

Answer:

60 ml

Explanation:

Now, we know that density = mass/volume

Mass of aluminium = 67.5 g

volume of aluminium =?

density of aluminium = 2.70 g/mL

volume of aluminium = mass/density = 67.5/2.70 = 25 ml

Since volume of water in the cylinder = 35 ml

The water rises to ; 35 ml + 25 ml = 60 ml

4 0
3 years ago
Electromagnetic energy that travels in transverse waves which produces visible light is
Lubov Fominskaja [6]
Nuclear is the answer
8 0
3 years ago
Define symbole and valeny.​
Alexxx [7]

Answer:

Chemical symbol

This is used to identify an element without having to write the full name and as such is a designation for the element. It is used in chemical reactions and is usually either one or two letters. Some symbols are, Iron - Fe, Hydrogen - H and Oxygen - O.

Valency

This refers to the capacity of an element to combine with another to form a compound. It is measured by the number of electrons that an atom of an element gains, looses or shares when a chemical reaction involving it takes place.

3 0
3 years ago
Fe2O3+2Al=Al2O3+2Fe
True [87]
We determine the limiting reactant by using the moles present in the equation and the actual moles.
According to equation, ratio of Fe₂O₃ : Al = 1 : 2
Actual moles of Fe₂O₃ = 187.3 / (56 x 2 + 16 x 3)
= 1.17
Actual moles of Al = 94.51 / 27
= 3.5
Fe₂O₃ is limiting. Fe₂O₃ required:
(moles Al)/2 = 3.5/2 = 1.75
Moles to be added = 1.75 - 1.17
= 0.58
Mass to be added = moles x Mr
= 0.58 x (56 x 2 + 16 x 3)
= 92.8 grams
6 0
3 years ago
Read 2 more answers
A sample of a chromium-containing alloy weighing 3.450 g was dissolved in acid, and all the chromium in the sample was oxidized
Zina [86]

Answer:

H_2O + 2CrO_4^2- + 3SO_3^2- -> 3SO_3^2- + 2CrO_2^- + 2OH^-

Explanation:

Reduction half reaction

2H_2O + CrO_4^2- + 3e -> CrO_2^- + 4OH^-

Oxidation half reaction

2OH^- + SO_3^2- -> SO_4^2- + H_2O + 2e

Balanced overall equation

H_2O + 2CrO_4^2- + 3SO_3^2- -> 3SO_3^2- + 2CrO_2^- + 2OH^-

 

5 0
3 years ago
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