Time t=270 minutes=270/60=9/2 hours
distance s=63 miles
average speed v=s/t=63 ÷

=63×

=14 miles/hour
Answer:
Latitude:
Longitude:
Explanation:
Lets begin by explaining the meaning of latitude and longitude as geogrephical coordinates:
Latitude is the angular distance between the equatorial line, and a specific point on the Earth. It is measured in degrees and is represented according to the hemisphere in which the point is located, which can be north or south latitude.
In this sense latitude
refers to the equatorial line that divides the Earth in two hemispheres (North and South).
Longitude represents the specific east–west position of a point on the Earth's surface, being longitude
the prime meridian or Greenwich meridian.
So, according to the figure, where the model of the Earth is divided by latitude lines separated by
and the longitude lines separated by
; we only have to count the lines from the equator to the line where the point A is, and count the lines fromo the Prime meridian to the line where point A is located.
Hence, point A location is:
Latitude:
Longitude:
There are many ways to test and identify metal. The easiest way is observing its color. Also how reflective it is. Other ways would be boiling point, melting point, density, or conductivity of the metal.
Hope This Helps and God Bless!
The balanced equation for the reaction is as follows;
Ca(OH)₂ + 2HBr --> CaBr₂ + 2H₂O
stoichiometry of Ca(OH)₂ to HBr is 1:2
number of Ca(OH)₂ moles reacted - 0.10 mol/L x 0.1000 L = 0.010 mol
Number of HBr moles added - 0.10 mol/L x 0.4000 = 0.040 mol
1 mol of Ca(OH)₂ needs 2 mol of HBr for neutralisation
therefore 0.010 mol of Ca(OH)₂ needs - 0.010 x 2 = 0.020 mol of HBr to be neutralised
but 0.040 mol of HBr has been added therefore number of moles of HBr in excess - 0.040 - 0.020 = 0.020 mol
then pH of the medium can be calculated using the excess H⁺ ions
HBr is a strong acid therefore complete ionization
[HBr] = [H⁺]
[H⁺] = 0.020 mol / (100.0 + 400.0 mL)
= 0.020 mol / 0.5 L
= 0.040 mol/L
pH = -log[H⁺]
pH = - log [0.040 M]
pH = 1.40
pH of the medium is 1.40