Answer:
work done = 750 J
Explanation:
Given data:
Force on object = 50 N
Distance covered = 15 m
Work done = ?
Solution:
W = F. d
W = work done
F = force
d = distance
Now we will put the values in formula.
W = 50 N × 15 m
W = 750 N. m = 750 J
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:
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This question provides us –
- Weight of
is = 47 g - Volume, V = 375 mL
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- Molar Mass of
–


<u>Using formula</u> –






- Henceforth, Molarity of the solution is = 1.7M
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40 g NaOH. You must use 40 g NaOH to prepare 10.0 L of a solution that has a pH of 13.
<em>Step 1</em>. Calculate the pOH of the solution
pOH = 14.00 – pH = 14.00 -13 = 1
<em>Step 2</em>. Calculate the concentration of NaOH
[NaOH] = [OH^(-)] = 10^(-pOH) mol/L = 10^(-1) mol/L = 0.1 mol/L
<em>Step 3</em>. Calculate the moles of NaOH
Moles of NaOH = 10.0 L solution × (0.1 mol NaOH/1 L solution) = 1 mol NaOH
<em>Step 4</em>. Calculate the mass of NaOH
Mass of NaOH = 1 mol NaOH × (40.00 g NaOH/1 mol NaOH) = 40 g NaOH
one mole of sulfuric acid will contain 2 moles of hydrogen atoms. The molar mass of sulfuric acid is 98.0795 g/mol. This means that every mole of sulfuric acid has a mass of 98.0795 g. Since you're dealing with one mole of sulfuric acid, it follows that you will also be dealing with two moles of hydrogen.