Answer:
Malleable
Explanation:
Because metals are malleable, they can be rolled into thin sheets or beaten into complex shapes.
kilo is 1,000 times the base unit <em>(tip for remembering: think money, 1k is 1,000)</em>
centi is 1/100 of the base unit <em>(tip for remembering: cent = 100)</em>
micro is 1/1,000,000 of the base unit
nano is 1/1,000,000,000 of the base unit <em>(tip for remembering: </em><em>n</em><em>ano will have </em><em>n</em><em>ine zeroes)</em>
milli is 1/1,000 of the base unit
mega is 1,000,000 times the base unit
Answer:
Explanation:
I'm assume that 3 Br2 + 6 OH¹-5 Br¹ + BrO3¹ + 3 H₂O is meant to read:
3 Br2 + 6 OH^-1 = 5 Br^-¹ + BrO3^-¹ + 3 H₂O
The balanced equation tells us that 3 moles of Br2 will react with 6 moles of OH^-1 to produce 5 moles of Br^-1.
The first step is to determine whether the Br2 or OH^-1 are limiting reagents. That is, is there enough of each to complete the reaction, with none left over.
We need 3 moles of Br2 for every 6 moles of OH^-1, a molar ratio of 1/2 (Br2/OH).
We are given 4.68 moles of Br2 and 8.12 moles of OH^-1. That is a ratio of 4.68/8.12 or 0.5764. This is higher than the ratio of 1/2 or 0.5 that is required. That means we have more than enough Br2. The limiting reagent is the OH^-1. Once it is consumed, the reaction stops and we are left with some unreacted Br2.
So we need the molar ratio of the OH^-1 to the Br from the balanced equation: We see that 6 moles of OH^-1 are required to produce 5 moles of Br, a 6/5 molar ratio.
Therefore, we may assume all 8.12 moles of the limiting reagent, OH^-1, will be consumed to produce *6/5) that amount of Br.
(8.12 moles OH^-1)*((6 moles Br)/(5 moles OH^-1)) = 9.75 moles of Br.
For curiosity's sake, we can determine the amount of unreacted Br2. 8.12 moles of OH^-1 would require (8.12 moles OH)*(1/2) = 4.06 moles of Br2.
4.68 moles starting Br2
4.06 moles consumed
0.62 moles remaining Br2
Answer:
89,4%
Explanation:
If you have a solutio of 2,5g of acetanilide in 50mL of water and you warm this solution to 100°C you will dissolve all acetanilide because the maximum solubility in 50mL will be:
5,5g / 100mL → 2,75g / 50mL.
Then, if you cold the water to 0°C the solubility in 50mL will be:
0,53g / 100mL → 0,265g / 50mL.
That means you will precipitate:
2,5g - 0,265g = <em>2,235g of acetanilide</em>
The theoretical percent recovery will be:
2,2365g / 2,5g ×100 = <em>89,4%</em>
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I hope it helps!
Answer:
Q = 44.9 j
Explanation:
Mass of iron = 5.0 g
Change in temperature = 20 °C
Specific heat of iron = 0.449 j/g.°C
Heat transferred = ?
Formula:
<em>Q = m.c. ΔT
</em>
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Solution:
<em>Q = m.c. ΔT
</em>
Q = 5 g × 0.449 j/g.°C × 20 °C
Q = 44.9 j
