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Lilit [14]
3 years ago
6

What element would be found in period 5 that has similar chemical properties to Al?

Chemistry
1 answer:
Iteru [2.4K]3 years ago
6 0

Answer:

Indium

Explanation:

That is Indium (In) which is in same group as Al

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an atmosphere is considered hazardous if it contains a hazardous gas in excess of 10 percent of the hazardous material's:
svlad2 [7]

Lower flammable limit means the lowest concentration of a material that will propagate a flame.

What is hazardous atmosphere?

It is an atmosphere that may expose employees to risk of death, incapacitation, impairment of ability to self-rescue, injury, or acute illness from one or more of following causes

  • Flammable gas, vapor, or mist in excess of 10 percent of lower flammable limit (LFL)
  • Airborne combustible dust at concentration that meets or exceeds its LFL

What is lower flammable limit?

  • It means the lowest concentration of a material that will propagate a flame.
  • The LFL is usually expressed as percent by volume of material in air (or other oxidant)
  • Atmospheres with concentration of flammable vapors at or above 10 percent of lower explosive limit (LEL) are considered hazardous when located in confined spaces.
  • However, atmospheres with flammable vapors below 10 percent of LEL are not necessarily safe. Such atmospheres are too lean to burn

Learn more about lower flammable limit at brainly.com/question/2456135

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6 0
1 year ago
State two essential conditions for rusting of iron. List two methods of protecting iron from rusting?​
gizmo_the_mogwai [7]
Conditions for rusting
• moisture
•oxygen

Prevention
•Painting
• Galvanisation
• Making Alloys
• greasing
Hope this helps you ;)(:

8 0
3 years ago
During lab, a student used a Mohr pipet to add the following solutions into a 25 mL volumetric flask. They calculated the final
kompoz [17]

Answer:

(FeSCN⁺²) = 0.11 mM

Explanation:

Fe ( NO3)3 (aq) [0.200M] + KSCN (aq) [ 0.002M] ⇒ FeSCN+2

M (Fe(NO₃)₃  = 0.200 M

V (Fe(NO₃)₃ =  10.63 mL

n (Fe(NO₃)₃ = 0.200*10.63 = 2.126 mmol

M (KSCN) =  0.00200 M

V (KSCN) = 1.42 mL

n (KSCN) =  0.00200 * 1.42 = 0.00284 mmol

Total volume = V (Fe(NO₃)₃  + V (KSCN)

                       = 10.63 + 1.42

                       = 12.05 mL

Limiting reactant = KSCN

So,

FeSCN⁺² = 0.00284 mmol

M (FeSCN⁺²) = 0.00284/12.05

                     = 0.000236 M

Excess reactant = (Fe(NO₃)₃

n(Fe(NO₃)₃ =  2.126 mmol -  0.00284 mmol

                  =2.123 mmol

For standard 2:

n (FeSCN⁺²) = 0.000236 * 4.63

                    =0.00109

V(standard 2) = 4.63 + 5.17

                       = 9.8 mL

M (FeSCN⁺²)  = 0.00109/9.8

                      = 0.000111 M = 0.11 mM

Therefore, (FeSCN⁺²) = 0.11 mM

7 0
3 years ago
6. Would you describe each of these temperatures as warm, hot, or cold?
mash [69]

Answer:

b: Hot

a: Cold

c: Cold

d: Warm

e: Warm

f: Cold

g: Hot

Explanation:

:)

8 0
3 years ago
What is the charge on every atom? why is this the charge
inn [45]
Protons, neutron, and elecrons

8 0
3 years ago
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