An atom consists of three subatomic particles.
In the middle of atom it consists of nucleus which have two subatomic particles. They are protons and neutrons.
And there is a subatomic particle that is found orbiting around the nucleus in an atom, name of that subatomic particle is electron.
Hope this helps!
Magnesium. You can count the electrons in each level and because the number of electrons is the same with protons you have the atomic number based of which you can get the element in the periodic table
<u>Answer:</u> The volume of acid should be less than 100 mL for a solution to have acidic pH
<u>Explanation:</u>
To calculate the volume of acid needed to neutralize, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is HCl
are the n-factor, molarity and volume of base which is NaOH
We are given:
Putting values in above equation, we get:
For a solution to be acidic in nature, the pH should be less than the volume of acid needed to neutralize.
Hence, the volume of acid should be less than 100 mL for a solution to have acidic pH
Given:
<span>CS2 + 3O2 → CO2 + 2SO2
</span><span>114 grams of CS2 are burned in an excess of O2
</span>
moles CS2 = 114 g/76.143 g/mol → 114g * mol/76.143 g = 1.497 mol
<span>the ratio between CS2 and SO2 is 1 : 2 </span>
moles SO2 formed = 1.497 x 2 = 2.994 moles → 2nd option
The first order rate law has the form: -d[A]/dt = k[A] where, A refers to cyclopropane. We integrate this expression in order to arrive at an equation that expresses concentration as a function of time. After integration, the first order rate equation becomes:
ln [A] = -kt + ln [A]_o, where,
k is the rate constant
t is the time of the reaction
[A] is the concentration of the species at the given time
[A]_o is the initial concentration of the species
For this problem, we simply substitute the known values to the equation as in:
ln[A] = -(6.7 x 10⁻⁴ s⁻¹)(644 s) + ln (1.33 M)
We then determine that the final concentration of cyclopropane after 644 s is 0.86 M.
