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Katyanochek1 [597]
4 years ago
12

What is the main intermolecular force in H2CO?

Chemistry
1 answer:
aivan3 [116]4 years ago
5 0
Dipole-dipole interactions, and London dispersion interactions
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Back titration experiment
iren2701 [21]

Answer:

Ouchie

Explanation:

4 0
3 years ago
Realice las siguientes conversiones: a) 72°F a °C, b) 213.8°C a °F, c)180°C a K, d) 315K a °F, e) 1750°F a K, f) 0K a °F.
bonufazy [111]

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: C=\frac{F-32}{1.8}
  • 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:K=\frac{F-32}{1.8} + 273.15

Entonces se obtiene:

a) 72 °F= \frac{72-32}{1.8}=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= \frac{1750-32}{1.8} + 273.15= 1227.594 °K

f) 0 °K= (0 -273.15)*1.8 + 32= -459.67 °F

7 0
3 years ago
Understand the Relationship Between the Structure of Alcohol and Their physical Properties Question Which of the following alcoh
IrinaK [193]

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.

6 0
4 years ago
An 8.65-g sample of an unknown group 2a metal hydroxide is dissolved in 85.0 ml of water. an acid-base indicator is added and th
garik1379 [7]
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

6 0
3 years ago
Read 2 more answers
At the beginning of the shift at 7:00 am, a client has 650 ml of normal saline solution left in the intravenous bag, which is in
madreJ [45]

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>

5 0
4 years ago
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