Answer:
The pH decreases very quickly, more quickly than earlier or later in the titration.
I think because SA-SB titration have the straight line where the equivalence point is where pH change is very steep and fast. Also pH is gonna decrease because after equivalence point you have more acid then base.
Explanation:
Answer is: <span>Sound waves travel faster in more dense material, and the more dense the material the louder the sound is.
</span>Sound travels approximately fifteen times faster through the metal railway tracks than through air. <span>The </span>speed of sound is faster in solid materials and slower in liquids or gases. The denser is material, t<span>he closer are the molecules to each other and the less time it takes to molecules to pass the sound to each other and sound is faster.</span>
Answer:
Number of moles = 0.01 mole
Explanation:
Given data:
Volume of gas = 0.1 L
Number of moles = ?
Temperature of gas = 25.0°C
Pressure of gas = 2.50 atm
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
Now we will convert the temperature.
25+273K = 298 K
by putting values,
2.50 atm × 0.1 L = n × 0.0821 atm.L/ mol.K × 298 K
0.25 atm.L = n × 24.47atm.L/ mol
n = 0.25 atm.L /24.47atm.L/ mol
n = 0.01 mol
Answer:
<h2>
Continue to move at 30 mph</h2>
Explanation:
Newton's first law of motion :
"Every object persists in its state of rest or uniform motion in a straight line unless it is compelled to change that state by forces exerted on it."
From the question we know that the net forces on the object were zero or that the there were no unbalanced forces on it.
Therefore we can assume that the object is moving along a straight line.
And the object was moving at a constant speed of 30 mph.
So it is clear from the Newton's first law that the object will remain in the state of motion as it was earlier.
That is the object will remain in motion at constant speed of 30 mph.
Given:
257J of heat
5500g of mercury
increase by 5.5
degrees Celsius
Required:
Specific heat of
mercury
Solution:
H
= mCpT
257J = (5500g of
mercury) Cp (5.5 degrees Celsius)
Cp = 8.5 x 10^-3
Joules per gram per degree Celsius