F = (mass)(acceleration) = ma
m = 55 kg
Vi = 20 m/s
t = 0.5 s
Vf = 0 m/s (since she was put to rest)
a=(Vf-Vi)/t
a=(0-20)/5
a = 40 m/s^2 (decelerating)
F = ma = (55 kg)(40 m/s^2)
F = 2200 N
Either covalent compounds are neutral, covalent compounds share electrons
Answer:
45.3°C
Explanation:
Step 1:
Data obtained from the question.
Initial pressure (P1) = 82KPa
Initial temperature (T1) = 26°C
Final pressure (P2) = 87.3KPa.
Final temperature (T2) =.?
Step 2:
Conversion of celsius temperature to Kelvin temperature.
This is illustrated below:
T(K) = T(°C) + 273
Initial temperature (T1) = 26°C
Initial temperature (T1) = 26°C + 273 = 299K.
Step 3:
Determination of the new temperature of the gas. This can be obtained as follow:
P1/T1 = P2/T2
82/299 = 87.3/T2
Cross multiply to express in linear form
82 x T2 = 299 x 87.3
Divide both side by 82
T2 = (299 x 87.3) /82
T2 = 318.3K
Step 4:
Conversion of 318.3K to celsius temperature. This is illustrated below:
T(°C) = T(K) – 273
T(K) = 318.3K
T(°C) = 318.3 – 273
T(°C) = 45.3°C.
Therefore, the new temperature of the gas in th tire is 45.3°C
Answer:
1. 0.45 mole
2. 49.95g
Explanation:
The following were obtained from the question:
Volume of solution = 300mL = 300/1000 = 0.3L
Molarity = 1.5 M
Mole of CaCl2 =?
1. We can obtain the mole of the solute as follow:
Molarity = mole of solute /Volume of solution
1.5 = mole of solute/0.3
Mole of solute = 1.5 x 0.3
Mole of solute = 0.45 mole
2. The grams in 0.45 mole of CaCl2 can be obtained as follow:
Molar Mass of CaCl2 = 40 + (35.5 x 2) = 40 + 71 = 111g/mol
Mole of CaCl2 = 0.45 mole
Mass of CaCl2 =?
Mass = number of mole x molar Mass
Mass of CaCl2 = 0.45 x 111
Mass of CaCl2 = 49.95g
