C) 10 grams
Explanation:
Potassium chlorate (KCIO₃) have a solubility in water at 70 °C of 30 grams in 100 grams of water.
If our solution have already 20 grams of potassium chlorate salt dissolved that means we need another 10 grams of potassium chlorate to reach the saturation point.
Learn more:
Potassium chlorate solubility in water
brainly.com/question/10140327
Answer:
BRO WHAT THEY ANSWER TO UR BUT NOT MINE OK
Explanation: I THINK THE ASNSWER IS C BUT NOT SURE.... HOPE IT HELPS
I believe the answer is
At the moment it is the best way of explaining our scientific knowledge.
Answer:
MnO4⁻ (aq) + 8H⁺ (aq) + 5Fe³⁺ (aq) →Mn(aq)²⁺ + 4H2O (l) + 5Fe²⁺(aq)
Explanation:
a)
MnO4⁻ (aq) + 8H⁺ (aq) + 5e⁻ → Mn(aq)²⁺ + 4H2O (l)
b)
5Fe³⁺ (aq) +5e⁻ → 5Fe²⁺(aq)
c)
MnO4⁻ (aq) + 8H⁺ (aq) + 5Fe³⁺ (aq) →Mn(aq)²⁺ + 4H2O (l) + 5Fe²⁺(aq)
Answer:
1.25 gram of cesium-137 will remain.
Explanation:
Given data:
Half life of cesium-137 = 30 year
Mass of cesium-137 = 5.0 g
Mass remain after 60 years = ?
Solution:
Number of half lives passed = Time elapsed / half life
Number of half lives passed = 60 year / 30 year
Number of half lives passed = 2
At time zero = 5.0 g
At first half life = 5.0 g/2 = 2.5 g
At 2nd half life = 2.5 g/ 2 = 1.25 g
Thus. 1.25 gram of cesium-137 will remain.
