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Luba_88 [7]
3 years ago
14

Rust results from iron’s reaction to oxygen. An iron nail gains mass when it rusts. How does this reaction support the law of co

nservation of mass? "The mass of the rusted nail equals the mass of iron and the oxygen from the air it reacted with to form the rust." "The mass of the rusted nail increases because iron attracts more protons from the air." "The increased mass of the rusted nail is an exception to the law of conservation of mass since the rusted nail’s mass increases." "The increased mass of the rusted nail results from the rearrangement of protons and neutrons within oxygen, according to the law of conservation of mass."
Chemistry
1 answer:
Olegator [25]3 years ago
5 0

Answer:

Option (A). The mass of the rusted nail equals the mass of iron and the oxygen from the air it reacted with to form the rust.

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B) increases, expand, rises

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7 0
3 years ago
If you have 0.50 mol of ca, how many atoms are present?
Evgen [1.6K]
Answer is: there are 3.011·10²³ atoms of calcium.

n(Ca) = 0.50 mol; amount of substance(calcium).
Na = 6.022·10²³ 1/mol;  Avogadro's constant or number.
N(Ca) = n(Ca) · Na.
N(Ca) = 0.50 mol · 6.022·10²³ 1/mol.
N(Ca) = 3.011·10²³; number of calcium atoms.
The mole is an SI unit which measures the number of particles in substance. One mole is equal to <span><span>6.022</span></span>·<span><span><span>10</span></span></span>²³<span> atoms.</span>
6 0
3 years ago
30.0 ml of an hf solution were titrated with 22.15 ml of a 0.122 m koh solution to reach the equivalence point. what is the mola
a_sh-v [17]
Answer is: molarity of hydrofluoric solution is 0.09 M.

Chemical reaction: HF(aq) + KOH(aq) → KF(aq) + H₂O(l).
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V(KOH) = 22.15 mL:
c(HF) = ?.
From chemical reaction: n(HF) : n(KOH) = 1 : 1.
n(HF) = n(KOH).
c(HF) · V(HF) = c(KOH) · V(KOH).
c(HF) = c(KOH) · V(KOH) ÷ V(HF).
c(HF) = 0.122 M · 22.15 mL ÷ 30 mL:
c(HF) = 0.09 M.
8 0
3 years ago
Read 2 more answers
The melting point of h2o is 0 degrees celsius. this is the same as its:
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This is the same as its freezing point

hope this helps
6 0
3 years ago
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