1. The <span>student with a measurement that might be in centimeters is A. Bill.
2. C</span><span>entimeters in 0.05 kilometers is </span>C. 5,000
solution:
0.05 km x (1000 meters/ 1 km)
= 50 meters x (100 cm/ 1 meter)
=5000 cm
Answer :
(a) 0.0152
(b) 0.0000000778
(c) 0.000001
(d) 1600.1
Explanation :
Scientific notation : It is the representation of expressing the numbers that are too big or too small and are represented in the decimal form with one digit before the decimal point times 10 raise to the power.
For example :
5000 is written as ![5.0\times 10^3](https://tex.z-dn.net/?f=5.0%5Ctimes%2010%5E3)
889.9 is written as ![8.899\times 10^{-2}](https://tex.z-dn.net/?f=8.899%5Ctimes%2010%5E%7B-2%7D)
In this examples, 5000 and 889.9 are written in the standard notation and
and
are written in the scientific notation.
(a) ![1.52\times 10^{-2}](https://tex.z-dn.net/?f=1.52%5Ctimes%2010%5E%7B-2%7D)
The standard notation is, 0.0152
(b) ![7.78\times 10^{-8}](https://tex.z-dn.net/?f=7.78%5Ctimes%2010%5E%7B-8%7D)
The standard notation is, 0.0000000778
(c) ![1\times 10^{-6}](https://tex.z-dn.net/?f=1%5Ctimes%2010%5E%7B-6%7D)
The standard notation is, 0.000001
(d) ![1.6001\times 10^{3}](https://tex.z-dn.net/?f=1.6001%5Ctimes%2010%5E%7B3%7D)
The standard notation is, 1600.1
Answer:
70.15 cm³
Solution:
Data Given;
Mass = 55 g
Density = 0.784 g.cm⁻³
Required:
Volume = ?
Formula Used:
Density = Mass ÷ Volume
Solving for Volume,
Volume = Mass ÷ Density
Putting values,
Volume = 55 g ÷ 0.784 g.cm⁻³
Volume = 70.15 cm³
Answer:
pH =3.8
Explanation:
Lets call the monoprotic weak acid HA, the dissociation equilibria in water will be:
HA + H₂O ⇄ H₃O⁺ + A⁻ with Ka = [ H₃O⁺] x [A⁻]/ [HA]
The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.
In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:
HA H₃O⁺ A⁻
Initial, M 0.40 0 0
Change , M -x +x +x
Equilibrium, M 0.40 - x x x
Lets express these concentrations in terms of the equilibrium constant:
Ka = x² / (0.40 - x )
Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,
7.3 x 10⁻⁶ = x² / 0.40 ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³
[H₃O⁺] = 1.71 x 10⁻³
Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.
pH = - log ( 1.71 x 10⁻³ ) = 3.8
Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.
22 is a
23 is d
24 is c
I believe the answers are correct.