Explanation:
Since the ratio is Cu/Fe, if some Fe were lost due to spillage, the Cu/Fe ratio would INCREASE because Fe would be lower.
Answer:
1.Very good electrical conductivity :<u> Metals</u> (Decreacing order of conductivity)
- <em>Silver > Copper > Gold > aluminium</em>
2. Amphoteric <u>: Metal elements</u>
- <em>Beryllium , Aluminium , Zinc </em>,
3.Gaseous at room temperature: mostly <u>Nobel gases elements</u> and some non - metal elements.
- <em>Helium ,neon , argon , krypton , fluorine , Oxygen , nitrogen</em>
4.Solid at room temperature:<u> Mostly Metals</u> (few non-metals, metalloid elements)
- <em>Metals (Sodium , potassium , calcium , gold are solid)</em>
<em>Non- metals(Carbon ,Boron )</em>
<em>Metalloids(antimony)</em>
<em>5.</em> Brittle <em>: </em><u>non - metals </u>(can't be rolled into wires)
<em>Hydrogen , carbon , sulfur , phosphorus</em><u> </u>
Explanation:
1 mol of Br = 79.9 g
15.7 g / 79.9 g = 0.196 moles of atoms
Answer:
6 days
Explanation:
The following data were obtained from the question:
Original amount (N₀) = 100 mg
Amount remaining (N) = 6. 25 mg
Time (t) = 24 days
Half life (t½) =?
Next, we shall determine the decay constant. This can be obtained as follow:
Original amount (N₀) = 100 mg
Amount remaining (N) = 6. 25 mg
Time (t) = 24 days
Decay constant (K) =?
Log (N₀/N) = kt / 2.303
Log (100/6.25) = k × 24 / 2.303
Log 16 = k × 24 / 2.303
1.2041 = k × 24 / 2.303
Cross multiply
k × 24 = 1.2041 × 2.303
Divide both side by 24
K = (1.2041 × 2.303) / 24
K = 0.1155 /day
Finally, we shall determine the half-life of the isotope as follow:
Decay constant (K) = 0.1155 /day
Half life (t½) =?
t½ = 0.693 / K
t½ = 0.693 / 0.1155
t½ = 6 days
Therefore, the half-life of the isotope is 6 days
