Answer:
Explanation:
The theory of natural selection was explored by 19th-century naturalist Charles Darwin. Natural selection explains how genetic traits of a species may change over time. This may lead to speciation, the formation of a distinct new species.
Answer:
238,485 Joules
Explanation:
The amount of energy required is a summation of heat of fusion, capacity and vaporization.
Q = mLf + mC∆T + mLv = m(Lf + C∆T + Lv)
m (mass of water) = 75 g
Lf (specific latent heat of fusion of water) = 336 J/g
C (specific heat capacity of water) = 4.2 J/g°C
∆T = T2 - T1 = 119 - (-20) = 119+20 = 139°C
Lv (specific latent heat of vaporization of water) = 2,260 J/g
Q = 75(336 + 4.2×139 + 2260) = 75(336 + 583.8 + 2260) = 75(3179.8) = 238,485 J
Answer:
C6H14O3F
Explanation:
The first step is to divide each compound by its molecular weight
Carbon
= 39.10/12
= 3.258
Hydrogen
= 7.67/1
= 7.67
Oxygen
= 26.11/16
= 1.63
Phosphorous
= 16.82/31
= 0.542
Flourine
= 10.30/19
= 0.542
The next step is to divide by the lowes value
3.258/0.542
= 6 mol of C
7.67/0.542
= 14 mol of H
1.63/0.542
= 3 mol of O
0.542/0.542
= 1 mol of P
0.542/0.542
= 1 mol of F
Hence the molecular formula is C6H14O3F
The action was him hitting the ball the reaction was the ball moving after being hit
Answer:
The new pressure will be 0.225 kPa.
Explanation:
Applying combined gas law:
![\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}](https://tex.z-dn.net/?f=%5Cfrac%7BP_1V_1%7D%7BT_1%7D%3D%5Cfrac%7BP_2V_2%7D%7BT_2%7D)
where,
are initial pressure and volume at initial temperature
.
are final pressure and volume at initial temperature
.
We are given:
![P_1=0.29 kPa\\V_1=V\\P_2=?\\V_2=V+50\% of V=1.5 V](https://tex.z-dn.net/?f=P_1%3D0.29%20kPa%5C%5CV_1%3DV%5C%5CP_2%3D%3F%5C%5CV_2%3DV%2B50%5C%25%20of%20V%3D1.5%20V)
![T_1=-25^oC=248.15 K](https://tex.z-dn.net/?f=T_1%3D-25%5EoC%3D248.15%20K)
![T _2 = 16^oC=289.15 K](https://tex.z-dn.net/?f=T%20_2%20%3D%2016%5EoC%3D289.15%20K)
Putting values in above equation, we get:
![\frac{0.29 kPa\times V}{248.15 K}=\frac{P_2\times 1.5V}{289.15 K}](https://tex.z-dn.net/?f=%5Cfrac%7B0.29%20kPa%5Ctimes%20V%7D%7B248.15%20K%7D%3D%5Cfrac%7BP_2%5Ctimes%201.5V%7D%7B289.15%20K%7D)
![P_1=0.225 kPa](https://tex.z-dn.net/?f=P_1%3D0.225%20kPa)
Hence, the new pressure will be 0.225 kPa.