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Inessa05 [86]
3 years ago
6

What determines the property of hardness in a mineral?

Chemistry
1 answer:
Rama09 [41]3 years ago
5 0

Answer:

It's determined by the ability of one mineral to scratch another mineral.

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Yes it was appropriate according to this article”Vinegar and lemon juice are great options if you're looking for a good home remedy for bee stings and wasp stings too. Both contain a type of acid that will help to neutralise the sting and provide a soothing sensation.”
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3 years ago
HELP PLZ!!!!!
Molodets [167]

The answer would be c as the cart is not in motion therefor ruling out kinetic and it is completely at rest making all of it energy potential

6 0
3 years ago
How do balanced chemical equations show the conservation of mass
antiseptic1488 [7]

Answer: Matter cannot be created or destroyed

Explanation: Balanced equations are set equations we cannot change one element or compound in the equation without changing the entire equation. So balanced equation show the conservation of mass because while other substances may be formed from the synthesis or decomposition of compounds new elements are never introduced and are not created out of thin air    :)

4 0
3 years ago
Read 2 more answers
Mercury(II) oxide (HgO) decomposes to form mercury (Hg) and oxygen (O2). The balanced chemical equation is shown below.
IceJOKER [234]
2HgO=2Hg + O2
433,18 g. = 32 g
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4 0
3 years ago
The combustion reaction described in part (b) occurred in a closed room containing 5.56 10g of air
ziro4ka [17]

Answer:

Explanation:

Combustion reaction is given below,

C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)

Provided that such a combustion has a normal enthalpy,

ΔH°rxn = -1270 kJ/mol

That would be 1 mol reacting to release of ethanol,

⇒ -1270 kJ of heat

Now,

0.383 Ethanol mol responds to release or unlock,

(c) Determine the final temperature of the air in the room after the combustion.

Given that :

specific heat c = 1.005 J/(g. °C)

m = 5.56 ×10⁴ g

Using the relation:

q = mcΔT

- 486.34 =  5.56 ×10⁴  × 1.005 × ΔT

ΔT= (486.34 × 1000 )/5.56×10⁴  × 1.005

ΔT= 836.88 °C

ΔT= T₂ - T₁

T₂ =  ΔT +  T₁

T₂ = 836.88 °C + 21.7°C

T₂ = 858.58 °C

Therefore, the final temperature of the air in the room after combustion is 858.58 °C

4 0
3 years ago
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