Answer:
In liquids, particles are quite close together and move with random motion throughout the container. Particles move rapidly in all directions but collide with each other more frequently than in gases due to shorter distances between particles.
To find pH, use the following formula ---> pH= - log [H+]
so first we need to calculate the [H+] concentration using the OH concentration. to do this, we need to use this formula--> 1.0x10-14= [H+] X [OH-], so we solve for H+ and plug in
[H+]= 1.0X10-14/[OH-]---> 1.0 x 10-14/ 1.0 x 10-4= 1.0 x 10-10
now that we have the H+ concentration, we can solve of pH
pH= -log (1.0x10-10)= 10
answer is A
The value of log₂(x/4) is 22. Using the properties of the logarithm, the required value is calculated.
<h3>What are the required properties of the logarithm?</h3>
The required logarithm properties are
logₐx = n ⇒ aⁿ = x; and logₐ(xⁿ) = n logₐ(x);
Where a is the base of the logarithm.
<h3>Calculation:</h3>
It is given that,
log₄(x) = 12;
On applying the property logₐx = n ⇒ aⁿ = x; here a = 4;
So,
log₄(x) = 12 ⇒ 4¹² = x
⇒ x = (2²)¹² = 2²⁴
Then, calculating log₂(x/4):
log₂(x/4) = log₂(2²⁴/4)
= log₂(2²⁴/2²)
= log₂(2²⁴ ⁻ ²)
= log₂(2²²)
On applying the property logₐ(xⁿ) = n logₐ(x);
log₂(x/4) = 22 log₂2
We know that logₐa = 1;
So,
log₂(x/4) = 22(1)
∴ log₂(x/4) = 22.
Learn more about the properties of logarithm here:
brainly.com/question/12049968
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All Elements are obtained in the periodic table & water is Not an element it is a compound.
<span>A chemist adds 155.0ml of a 4.10 X 10^-5 mmol/L of a zinc oxalate (ZnC2O4)solution to a reaction flask. Calculate the mass in micrograms of zinc oxalate the chemist has added to the flask.
1mmol = 10^-3 mol
Therefore 4.10*10^-5mmol = 4.10*10^-8mol
molar mass ZnC2O4 = 65.39+(2*12.011)+(4*15.99) = 153.372g/mol
You have 4.10*10^-8 mol/litre =153.372 * 4.10*10^-8 = 6.29*10^-6 grams / litre (* see below)
But you have 155ml. Mass of ZnC2O4 = 155/1000*6.29*10^-6 g
Mass is = 9.75*10^-7 grams
1µg = 10^-6 g
You then have 9.75*10^-7/10^-6 = 0.975µg ZnC2O4
(*see below) at this point you could have said:
1µg = 10^-6 g therefore you have a solution of 6.29µg per litre,
155ml = 6.29*155/1000 = 0.975µg ZnC2O4</span>
