Answer : Option C) Early modern era
Explanation : The historical era which was marked by the beginnings of European explorations, conquests of the Americas, trans-Atlantic slave trade and the rise of a globally intertwined economy was Early modern era. It was later followed by the medieval period.
They are examples of elements.
Answer:
The correct answer is option D.
Explanation:
Rate of the reaction is a change in the concentration of any one of the reactant or product per unit time.

Rate of the reaction:
![R=-\frac{1}{1}\times \frac{d[NO_2]}{dt}=-\frac{1}{1}\times \frac{d[CO]}{dt}](https://tex.z-dn.net/?f=R%3D-%5Cfrac%7B1%7D%7B1%7D%5Ctimes%20%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7B1%7D%5Ctimes%20%5Cfrac%7Bd%5BCO%5D%7D%7Bdt%7D)
Rate of decrease in nitrogen dioxide concentration is equal to the rate of decrease in carbon monoxide.
Given rate expression of the reaction:
![R = k[NO2]^2[CO]](https://tex.z-dn.net/?f=R%20%3D%20k%5BNO2%5D%5E2%5BCO%5D)
Rate of the reaction on doubling concentration of nitrogen dioxide and carbon monoxide : R'
![R'=k(2\times [NO_2])^2(2\times [CO])=8\times k[NO2]^2[CO]=8R](https://tex.z-dn.net/?f=R%27%3Dk%282%5Ctimes%20%5BNO_2%5D%29%5E2%282%5Ctimes%20%5BCO%5D%29%3D8%5Ctimes%20k%5BNO2%5D%5E2%5BCO%5D%3D8R)
Doubling the concentrations of nitrogen dioxide and carbon monoxide simultaneously will increase the rate of the reaction by a factor of eight.
Hence, none of the given statements are true.
Answer:
Composition of the mixture:
%
%
Composition of the vapor mixture:
%
%
Explanation:
If the ideal solution model is assumed, and the vapor phase is modeled as an ideal gas, the vapor pressure of a binary mixture with
and
molar fractions can be calculated as:

Where
and
are the vapor pressures of the pure compounds. A substance boils when its vapor pressure is equal to the pressure under it is; so it boils when
. When the pressure is 0.60 atm, the vapor pressure has to be the same if the mixture is boiling, so:

With the same assumptions, the vapor mixture may obey to the equation:
, where P is the total pressure and y is the fraction in the vapor phase, so:
%
The fractions of B can be calculated according to the fact that the sum of the molar fractions is equal to 1.
Answer:
There are 20.8 g of fluorine in 55.5 g of copper (II) fluoride
Explanation:
x % by mass of a species in a specimen means there are x g of the species in total 100 g of a specimen
37.42 % F by mass means 100 g of copper (II) fluoride contains 37.42 g of F.
So, 100 g of copper (II) fluoride contains 37.42 g of F
55.5 g of copper (II) fluoride contains
g of F or 20.8 g of F
Hence there are 20.8 g of fluorine in 55.5 g of copper (II) fluoride.