Answer:
both
Explanation:
id say that it could occur but also not as much. the moon would be smaller and further from the earth to where we would barely be able to see it. if the full moon is barely visible then im sure the total solar eclipse wouldn't be as noticeable as it is now. but thats just my opinion
You can stop the burning of methane with water or carbon dioxide extinguishers but problems arise when you try to use this to stop the burning of the magnesium.
Explanation:
To burn magnesium (Mg) and methane (CH₄) you need to react them with oxygen:
2 Mg (s) + O₂ (g) → 2 MgO + heat
CH₄ (g) + 2 O₂ (g) → CO₂ (g) + 2 H₂O (g) + heat
However at that temperatures magnesium (Mg) is able to react with water (H₂O) and carbon dioxide (CO₂).
Mg (s) + 2 H₂O (l) → Mg(OH)₂ (s) + H₂ (g)
2 Mg (s) + CO₂ (g) → 2 MgO (s) + C (s)
So the safe option to stop the burning of the magnesium is to limit the oxygen in the air.
we have used the following notations:
(s) - solid
(g) - gas
(l) - liquid
Learn more about:
combustion reactions
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False , it will make it lighter
Answer:
2.78 x 10²³
Explanation:
1 mole contains 6.02 x 10²³ hydrogen atoms => 0.46 mole contains 0.46(6.02 x 10²³) hydrogen atoms or 2.78 x 10²³ atoms.
Caution => When to use H vs H₂ => This problem is specific for 'hydrogen atoms' but some may simply say hydrogen. In such cases use H₂ or 'molecular hydrogen' is the focus. it's a matter of semantics, H vs H₂.
Step 1 : Write balanced chemical equation.
CaF₂ can be converted to F₂ in 2 steps. The reactions are mentioned below.
I] 
II] 
The final balanced equation for this reaction can be written as
Step 2: Find moles of CaF₂ Using balanced equation
We have 1.12 mol F₂
The mole ratio of CaF₂ and F₂ is 1:1
Step 3 : Calculate molar mass of CaF2.
Molar mass of CaF₂ can be calculated by adding atomic masses of Ca and F
Molar mass of CaF₂ = Ca + 2 (F)
Molar mass of CaF₂ = 40.08 + 18.998 = 78.08 g
Step 4 : Find grams of CaF₂
Grams of CaF₂ = 
Grams of CaF₂ = 87.45 g
87.45 grams of CaF2 would be needed to produce 1.12 moles of F2.
