Answer:
a) 72 °F= 22.22 °C
b) 213.8 °C= 416.84°F
c) 180 °C= 453.15 °K
d) 315 °K= 107.33 °F
e) 1750 °F= 1227.594 °K
f) 0 °K= -459.67 °F
Explanation:
Para realizar el intercambio de unidades debes tener en cuenta las siguientes conversiones:
- Fahrenheit a Celsius:

- Celsius a Fahrenheit: °F= °C*1.8 + 32
- Celsius a Kelvin: °K= °C + 273.15
- Kelvin a Fahrenheit: F= (K -273.15)*1.8 + 32
- Fahrenheit a Kelvin:

Entonces se obtiene:
a) 72 °F=
=22.22 °C
b) 213.8 °C= 213.8*1.8 + 32= 416.84°F
c) 180 °C= 180°C + 273.15= 453.15 °K
d) 315 °K= (315 -273.15)*1.8 + 32= 107.33 °F
e) 1750 °F=
= 1227.594 °K
f) 0 °K= (0 -273.15)*1.8 + 32= -459.67 °F
Answer:
Ethanol most easily forms hydrogen bonds.
Explanation:
The difference among the alcohols in this question is the size of carbonic chain and the position of the -OH group.
Ethanol has 2 carbons and the -OH group is terminal. The other alcohols have more carbons and the -OH group is not terminal. This means that the approximation of molecules will be facilitated for ethanol, and the interaction through hydrogen bons will be easier. However, for the other molecules, there will be steric hindrance, which will make it more difficult for the molecules to make hydrogen bonds.
The figure attached shows the alcohol structures.
X(OH)₂ + 2HCl = XCl₂ + 2H₂O
n{X(OH)₂}=m{X(OH)₂}/M{X(OH)₂}
n(HCl)=c(HCl)v(HCl)
n{X(OH)₂}=n(HCl)/2
m{X(OH)₂}/M{X(OH)₂}=c(HCl)v(HCl)/2
M{X(OH)₂}=2m{X(OH)₂}/{c(HCl)v(HCl)}
M{X(OH)₂}=2*8.65/{2.50*56.9*10⁻³}=121.6 g/mol
M(OH)=17.0 g/mol
M(X)=121.6-17.0*2=87.6 g/mol ⇒ X=Sr, strontium
Sr(OH)₂ + 2HCl = SrCl₂ + 2H₂O
The 8 hour shift is from 7.00 am to 3.00 pm.<span>
<span>At the beginning, from 7.00 am to 9.30am (2.5 hrs.),<span> the client has normal saline which </span></span>is
infusing at 125 ml/hr.
Hence amount of normal saline which was injected= 2.5 hrs. x 125 ml/hr.
= 312.5 ml
from 9.30am to 3.00 pm (5.5 hrs) lactated ringer solution, which is to infuse at 100 ml/hr was
given.
Hence amount of lactated ringer solution given = 5.5 hrs x 100 ml/hr
= 550 ml
Total amount of IV solution given = normal saline
+ lactated ringer solution
= 312.5 ml + 550
ml
= 862.5 ml
<span>
= 863 ml</span></span>