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Nataly [62]
3 years ago
7

List any two uses of Brass , Bronze , Sulphur , Iodine

Chemistry
1 answer:
NikAS [45]3 years ago
5 0

BRASS :It is easy to form into various shapes, a good conductor of heat, and generally resistant to corrosion from salt water. 1 pipes and 2 tubes, 3 screws, 4 cartridge casings for firearms.

BRONZE :for bearings because of its friction properties, and as 1 musical instruments ,2 and medals

Sulphur :1 making car batteries, 2 fertilizer

IODINE :1 Iodine regulates skin moisture levels and aids in the healing of cuts and scars through cellular regeneration. 2 Iodine also regulates the hormones responsible for acne breakouts.3 Treating thyroid cancer.

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It might be to long but please help. I did search but couldn't find answer
Inessa05 [86]
The events that occur is based on the type of hormone being released. Like puberty. It's an event coming from different types of hormones being sent to a tissue
7 0
3 years ago
In an oxidation-reduction reaction, 0.0450 mol of aqueous FeSO4 (source of Fe2+) reacts completely with 120.0 mL of an acidified
igor_vitrenko [27]

Answer:

0.075 M

Explanation:

5Fe²⁺(aq) + MnO₄⁻(aq) + 8H⁺(aq) → 5Fe³⁺(aq) + Mn²⁺(aq) + 4H₂O(ℓ)

Using the moles of Fe²⁺ that reacted, we can <em>calculate the reacting moles of MnO₄⁻</em>:

0.0450 mol Fe⁺² * \frac{1molMnO_{4}^{-}}{5molFe^{+2}} = 0.0090 mol MnO₄⁻

Now we divide the moles of MnO₄⁻ by the volume in order to <u>calculate the molarity of the solution</u>, keeping in mind that 120.0 mL = 0.120 L.

0.0090 mol MnO₄⁻ / 0.120 L = 0.075 M

7 0
4 years ago
Read 2 more answers
PLEASE HELP ASAP!!! I REALLY NEED HELP ON THIS TEST!!!
Oksana_A [137]

Answer:

1. Hyphothesis 2.Theory 3. law

Explanation:

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5 0
3 years ago
Chemical compounds that contain the –NH2 functional groups are _____.
Illusion [34]
The answer is amides 
6 0
3 years ago
Read 2 more answers
Consider the following reaction:
noname [10]

Answer:

26.6

Explanation:

Step 1: Calculate the molar concentrations

We will use the following expression.

M = mass solute / molar mass solute × liters of solution

[CO]i = 26.6 g / (28.01 g/mol) × 5.15 L = 0.184 M

[H₂]i = 2.36 g / (2.02 g/mol) × 5.15 L = 0.227 M

[CH₃OH]e = 8.63 g / (32.04 g/mol) × 5.15 L = 0.0523 M

Step 2: Make an ICE chart

        CO(g) + 2 H₂(g) ⇄ CH₃OH(g)

I        0.184      0.227           0

C         -x           -2x             +x

E     0.184-x   0.227-2x        x

Since [CH₃OH]e = x, x = 0.0523

Step 3: Calculate all the concentrations at equilibrium

[CO]e = 0.184-x = 0.132 M

[H₂]e = 0.227-2x = 0.122 M

[CH₃OH]e = 0.0523 M

Step 4: Calculate the equilibrium constant (Kc)

Kc = [CH₃OH] / [CO] [H₂]²

Kc = 0.0523 / 0.132 × 0.122² = 26.6

8 0
3 years ago
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