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Kaylis [27]
3 years ago
10

The most dangerous form of skin cancer which is often caused by over exposure to ultraviolet radiation is what?

Chemistry
1 answer:
katrin [286]3 years ago
8 0
UV exposure that leads to sunburn is proven in developing melanoma, the most dangerous of the three most common types of skin cancer.
You might be interested in
If aluminum has a density of 2.7 g/cm3, what is the volume, in cubic
Pachacha [2.7K]

Answer:

20cm3

Explanation:

density = mass/volume, so volume = mass/density

volume = (54g)/(2.7g/cm3)= 20cm3

6 0
3 years ago
What is a characteristic of a lewis base?
statuscvo [17]
A Lewis base is characterized as any species that can donate a lone pair of electrons the Lewis acid (which has a vacant orbital for the lone pair to bond with)

8 0
3 years ago
How many grams of NaOH is needed to neutralize 90 mL of 1.5 N HCl?
AnnZ [28]

mol = conc × v

= 1.5 × 0.09

= 0.135 moles of HCl

HCl + NaOH > NaCl + H2O

1 mole HCl = 1 mole NaOH

0.135 mol HCl = x

x = 0.135 mol NaOH

mass = mol × molar mass

= 0.135 × 40

= 5.4 g

NaOH = 23 + 16 + 1 = 40 g/mol

I'm not a 100% sure if it's correct

6 0
3 years ago
What type of consumer are bears?
sashaice [31]
Bears are omnivores (and scavengers), meaning they eat both plants and animals.
it can be both a primary consumer (if it eats plants) and secondary & tertiary consumers (if it eats a plant-eating animal). so it depends on what it eats.
5 0
3 years ago
if 3.0 grams of aluminum and 6.0 grams of bromine react to form AlBr3, how many grams of product would theoretically be produced
Gelneren [198K]
1) Chemical reaction: 2Al + 3Br₂ → 2AlBr₃.
m(Al) = 3,0 g.
m(Br₂) = 6,0 g.
n(Al) = m(Al) ÷ M(Al).
n(Al) = 3,0 g ÷ 27 g/mol.
n(Al) = 0,11 mol.
n(Br₂) = n(Br₂) ÷ m(Br₂).
n(Br₂) = 6 g ÷ 160 g/mol.
n(Br₂) = 0,0375 mol; limiting reagens.
n(Br₂) : n(AlBr₃) = 3 : 2.
n(AlBr₃) = 0,025 mol.
m(AlBr₃) = 0,025 mol · 266,7 g/mol.
m(AlBr₃) = 6,67 g.

2) m(Br₂) - all bromine reacts, so mass of bromine after reaction is zero grams (m(Br₂) = 0 g).
n(Al) = 0,11 mol - 0,025 mol = 0,085 mol.
m(Al) = 0,085 mol · 27 g/mol.
m(Al) = 2,295 g.
m(AlBr₃) = 6,67 g · 0,72 (yield of reaction).
m(AlBr₃) = 4,8 g.
n - amount of substance.
M - molar mass.

4 0
3 years ago
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