Answer:
2.81 × 10⁶ mm³
2.81 × 10⁻³ m³
Explanation:
Step 1: Given data
Length (l): 250 mm
Width (w): 225 mm
Thickness (t): 50 mm
Step 2: Calculate the volume of the textbook
The book is a cuboid so we can find its volume (V) using the following expression.
V = l × w × t = 250 mm × 225 mm × 50 mm = 2.81 × 10⁶ mm³
Step 3: Convert the volume to cubic meters
We will use the relationship 1 m³ = 10⁹ mm³.
2.81 × 10⁶ mm³ × 1 m³ / 10⁹ mm³ = 2.81 × 10⁻³ m³
None of them really sound like the answer, but my best guess would probably be the second one. It doesnt give much info.
To determine the pH of a solution which has 0.195 M hc2h3o2 and 0.125 M kc2h3o2, we use the ICE table and the acid dissociation constant of hc2h3o2 <span>to determine the concentration of the hydrogen ion present at equilibrium. We do as follows:
HC2H3OO = H+ + </span>C2H3OO-
KC2H3OO = K+ + C2H3OO-
Therefore, the only source of hydrogen ion would be the acid. We use the ICE table,
HC2H3OO H+ C2H3OO-
I 0.195 0 0.125
C -x +x +x
------------------------------------------------------------------
E 0.195-x x 0.125 + x
Ka = <span>1.8*10^-5 = (0.125 + x) (x) / 0.195 -x
x = 2.81x10^-5 M = [H+]
pH = - log [H+]
pH = -log 2.81x10^-5
pH = 4.55
Therefore, the pH of the resulting solution would be 4.55.</span>
Answer:
3.37kg of NaCl is the salt content of the dead sea
Explanation:
A concentration of 33.7% means there are 33.7g of NaCl in 100mL of solution.
To solve this question we need to convert 10.0L to mililiters, and then, solve for the mass of NaCl in grams and convert it to kg:
<em>Volume in milliliters:</em>
10.0L * (1000mL / 1L) = 10000mL
<em>Mass of NaCl:</em>
10000mL * (33.7g NaCl / 100mL) = 3370g NaCl
<em>In kg:</em>
3370g * (1kg / 1000g) =
<h3>3.37kg of NaCl is the salt content of the dead sea</h3>
