Answer:
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to calculate the required new volume by using the Charles' law as a directly proportional relationship between temperature and volume:
In such a way, we solve for V2 and plug in V1, T1 and T2 to obtain:
Regards!
Answer:Water contains almost one thousand times more matter than the same volume of air would, therefore it is denser. ... If 500 mL of a liquid has a density of 1.11 g/mL, what is its mass? ... Density is the mass per unit of volume of a substance. ... A block of platinum with a mass of 4,290 kilograms has a volume of .2 cubic meters.
Explanation:
Answer:
<u>~</u><u>Law of Conservation of </u><u>energy~</u>
The law of conservation of energy states that energy can neither be created nor destroyed, only energy can be converted from one form to another.
Answer:
Rate constant = 0.0237 M-1 s-1, Order = Second order
Explanation:
In this problem, it can be observed that as the concentration decreases, the half life increases. This means the concentration of the reactant is inversely proportional to the half life.
The order of reaction that exhibit this relationship is the second order of reaction.
In the second order of reaction, the relationship between rate constant and half life is given as;
t1/2 = 1 / k[A]o
Where;
k = rate constant
[A]o = Initial concentration
k = 1 / t1/2 [A]
Uisng the following values;
k = ?
t1/2 = 113
[A]o = 0.372M
k = 1 / (113)(0.372)
k = 1 / 42.036 = 0.0237 M-1 s-1
Since all of those percents add up to 100, you can just directly convert that to grams. So now you can use 2 grams H, 32.7 grams S, and 65.3 grams O. Use that info and convert that to moles for an answer of 2mol H, 1mol S, and 4mol O. In every empirical question you need to divide each quantity of moles by the lowest number. In this case, that number is one, so they stay the same, but it's important to remember that step. You're final chemical formula would be H2SO4 and the answer to your question would be that the subscript for oxygen is 4. Hope this helped!
