Answer: This can be quickly solved with "traintracks"
Explanation:
You start w/ grams of water and want to find moles of oxygen gas produced.
So you want to Convert:
Grams of water -> moles of water -> moles of oxygen gas.
The two things you need to know to set up the tracks are:
1)Molar mass of water- H2O
Hydrogen - 1.008(x2)
Oxygen - 16.00
Water - 18.016
Make a quick chart with each element represented, and count them up. HINT - leave the polyatomic anions together - in this case, PO4
Left Right
1 Ca 3
2 O 1
5 H 2
1 PO4 2
Begin by balancing like finding common denominators of fractions - apply to both sides:
I started by adding a 2 in front of H3PO4 on the left, them 6 in front of H2O on the right. Last, a 3 in front of Ca (OH)2. Then, re-count using the chart format to make sure you're right.
3Ca(OH)2 + 2H3PO4 = Ca3(PO4)2 + 6H2O
Answer:
This question is incomplete
Explanation:
This question is incomplete because the result of the described experiment would have better determined the type of scientific explanation to profer. However, the type of material that will preserve the relative hotness or temperature of the hot coffee for the longest time will be a material than can resist heat transfer. These materials tend to keep hot substances hot by not allowing the heat of the coffee to be conducted or pass through it. These materials are mostly insulators or made by placing an insulator between two heat conductors.
Generally, heat is usually transferred from a region of higher concentration to a region of lower concentration, hence when the heat is denied of this transfer, the heat will remain trapped in the "heat-donor" substance (in this case the hot coffee). Thus, the material chosen (A, B or C) will be the material that resists heat transfer the most based on the explanation above.
Metals are the most abundant
Answer:
800.0 mL.
Explanation:
- To solve this problem; we must mention the rule states the no. of millimoles of a substance before and after dilution is the same.
<em>(MV)before dilution of HCl = (MV)after dilution of HCl</em>
M before dilution = 12.0 M, V before dilution = 100.0 mL.
M after dilution = 1.5 M, V after dilution = ??? mL.
∵ (MV)before dilution of HCl = (MV)after dilution of HCl
∴ (12.0 M)(100.0 mL) = (1.5 M)(V after dilution of HCl)
<em>∴ V after dilution of HCl = (12.0 M)(100.0 mL)/(1`.5 M) = 800.0 mL.</em>
