Answer:C.people with perfect understanding of science.
70.33 L is the volume of 10 moles of a gas at 300 K held at a pressure of 3.5 atm.
<h3>What is volume?</h3>
Volume is the percentage of a liquid, solid, or gas's three-dimensional space that it occupies.
Liters, cubic metres, gallons, millilitres, teaspoons, and ounces are some of the more popular units used to express volume, though there are many others.
We will use ideal gas law to find the volume
PV = nRT
Can also be written as
V = (nRT)/P
Where,
P = pressure
V = volume
n = amount of substance
R = ideal gas constant
T = temperature
Here, we have given
P = 3.5 atm
V = to find
n = 10 moles
R = 0.08206 L⋅atm/K⋅mol
T = 300k
Lets substitute the values
V = (10 × 0.08206 × 300)/3.5
V = 70.33 L
Learn more about volume
brainly.com/question/463363
#SPJ10
chlorine has the atomic mass of 35.5 and is a non metal in the halogen family
Answer:
E) 2.38
Explanation:
The pH of any solution , helps to determine the acidic strength of the solution ,
i.e. ,
- Lower the value of pH , higher is its acidic strength
and ,
- Higher the value of pH , lower is its acidic strength .
pH is given as the negative log of the concentration of H⁺ ions ,
hence ,
pH = - log H⁺
From the question ,
the concentration of the solution is 0.0042 M , and being it a strong acid , dissociates completely to its respective ions ,
Therefore , the concentration of H⁺ = 0.0042 M .
Hence , using the above equation , the value of pH can be calculated as follows -
pH = - log H⁺
pH = - log ( 0.0042 M )
pH = 2.38 .
To
determine the empirical formula of the compound given, we need to determine the ratio of each element in the compound. To do that we assume to have 100 grams sample
of the compound with the given composition. Then, we calculate for the number
of moles of each element. We do as follows:<span>
mass moles
C 56.79 4.73
H 6.56 6.50
O 28.37 1.77
N 8.28 0.59
Dividing the number of moles of each element with
the smallest value, we will have the empirical formula:
</span> moles ratio
C 4.73 / 0.59 8
H 6.50 / 0.59 11
O 1.77 / 0.59 3
N 0.59 / 0.59 1<span>
</span><span>
The empirical formula would be C8H11O3N.</span>