Answer : 10 plague hearts are there in one map at a time in state of decay 2
Explanation : The State of Decay 2 consists of 10 Plague Hearts which one needs to destroy and they are placed on randomly dotted areas around the map inside a searchable location on it.
With each heart that one destroys, the remaining other ones become a little bit tougher and are more heavily guarded by the undead people.
Answer:
I think it's Trinitrogen Pentaseleniumide
Explanation:
Tri - three
Penta - five
Second element ends with -ide
O1Fl2
1. Assume an 100g sample, so the percentage will stay the same
2. Covert each element into their molar mass
29.6/16.00=1.8 mols of O
70.4/19.00=3.7 mols of Fl
3. Divide both by the smallest value of mol
1.8/1.8=1 O
3.7/1.8=2 Fl
4. Write the empirical formula:
O1Fl2
- <span>It is an odorless, colorless, and mostly inert gas, and continues to be colorless and odorless at a liquid state.
-</span><span>Nitrogen is believed to be the seventh most abundant element in the universe.
-At normal pressure, nitrogen liquefies at 77 K (-195.79 °C) and freezes at 63 K (-210.01 °C).
-Liquid nitrogen boils at -195.8 degrees Celsius.
-Due to its volatility, it is a fairly rare element on Earth, remaining as a gas in the atmosphere.
-Scientists in the eighteenth century recognized that there was a component of air that did not support combustion, thus discovering nitrogen.
-In 1772, Daniel Rutherford discovered what he called "noxious air," air that didn't contain oxygen.
-Large scale nitrogen manufacturing takes place through liquefaction of air and the fractional distillation of the resulting liquid air.All living things contain nitrogen, mostly in amino acids, DNA, and RNA.</span>
Answer:
The correct answer is - yes, 4.57 g of solute per 100 ml of solution
Explanation:
The correct answer is yes we can calculate the solubility of X in the water at 22.0°C. The salt will remain after the evaporate from the dissolved and cooled down at 26°C.
Then, the amount of solute dissolved in the 700 ml solution at 26°C is the weighed precipitate: 0.032 kg = 32 g.
Then solublity will be :
32. g solute / 700 ml solution = y / 100 ml solution
⇒ y = 32. g solute × 100 ml solution / 700 ml solution = 4.57 g.
Thus, the answer is 4.57 g of solute per 100 ml of solution.
