Answer:
atomic mass of X is 48.0 amu
Explanation:
Let y be the atomic mass of X
Molar mass of O_2 is = 2×16 = 32 g / mol
X + O2 -----> XO_2
According to the equation ,
y g of X reacts with 32 g of O_2
24 g of X reacts with Z g of O_2
Z = ( 32×24) / y
But given that 24.0 g of X exactly reacts with 16.0 g of O_2
So Z = 16.0
⇒ (32×24) / y = 16.0
⇒ y = (32×24) / 16
y= 48.0
So atomic mass of X is 48.0 amu
Answer:
Q = 30284.88 j
Explanation:
Given data:
Mass of ethanol = 257 g
Cp = 2.4 j/g.°C
Chnage in temperature = ΔT = 49.1°C
Heat required = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Now we will put the values in formula.
Q = 257 g× 2.4 j/g.°C × 49.1 °C
Q = 30284.88 j
Answer:
Answer: a) 20g of H2O (18.02 g/mol) molecules=6.68x10^23
Explanation:
In order to find the amount of molecules of each of the options, we need to follow the following equation.

So, let´s get the number of molecules for each of the options.





the smalest number is in option a)
Best of luck.
Answer:
???
Explanation:
i would help answer but your post has no diagrams or at least there not showing up