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3 years ago

Help please! I’ll mark brainliest to who ever answers this first!!

Chemistry

1 answer:

Nesterboy [21]3 years ago

6 0

Answer:

paleontologists is the answer to scientist who studies fossil

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What is the boiling point of a solution containing 203 g of ethylene glycol (C2H6O2) and 1035 g of water? (Kb for water is 0.52

Andru [333]

Answer:

101,37°C

Explanation:

Boiling point elevation is one of the colligative properties of matter. The formula is:

ΔT = kb×m (1)

Where:

ΔT is change in boiling point: (X-100°C) -X is the boiling point of the solution-

kb is ebulloscopic constant (0,52°C/m)

And m is molality of solution (mol of ethylene glycol / kg of solution). Moles of ethylene glycol (MW: 62,07g/mol):

203g × (1mol /62,07g) = 3,27moles of ethlyene glycol

Molality is: 3,27moles of ethlyene glycol / (1,035kg + 0,203kg) = 2,64m

Replacing these values in (1):

X - 100°C = 0,52°C/m×2,64m

X - 100°C = 1,37°C

X = 101,37°C

I hope it helps!

7 0

3 years ago

Any help with atomic radius, ionization & electronegativity? See picture please! I dont really understand any of the three.

enyata [817]

18)
a. Ra
d. Ag
19)
a. C
b. Br
c. Mg
20)
a. S
b. Br
c. O

Sorry wasn’t sure about some of them in #18
Hope the notes are helpful.

8 0

3 years ago

Us What is the percent of O in Cr2O3? (Cr = 52.00 amu, O = 16.00 amu) [? ]%

Gekata [30.6K]

The percent of O in Cr₂O₃ : 31.58%

<h3>Further explanation</h3>

Given

Cr = 52.00 amu, O = 16.00 amu

Required

The percent of O

Solution

MW Cr₂O₃ = 2 x Ar Cr + 3 x Ar O

MW Cr₂O₃ = 2.52+3.16

MW Cr₂O₃ =152 amu

$\tt \%O=\dfrac{3.Ar~O}{MW~Cr_2O_3}\times 100\%\\\\\%O=\dfrac{3.16}{152}\times 100\%\\\\\%O=31.58\%$

8 0

3 years ago

What are the basic functions and parts of an LED bulb?? Please Help!!

IgorLugansk [536]

Answer:

Lens

Heat Sink

Housing

Base

Circuit Board/Driver

LED Chips

Explanation:

Does this help you?

5 0

3 years ago

Convert the following temperatures to Kelvin:

maxonik [38]

Explanation:

A. 100°C to Kelvins

$T(K)=T(^oC)+273.15$

$T(K)=100(^oC)+273.15=373.15 K$

B 600°R to Kelvins

$(T)^oK=((T)^oR)\times 1.8$

$(T)^oK=600\times 1.8 K = 1080 K$

C. 98°F to Kelvins

$(T(K)-273.15)=(T(^oF)-32)\times \frac{5}{9}$

$(T(K))=(98(^oF)-32)\times \frac{5}{9}+273.15=309.81K$

D. 77.4°F to degree Celsius

$((T)^oC)=((T)^oF-32)\times \frac{5}{9}$

$(T)^oC =(77.4^oF-32)\times \frac{5}{9}=25.22^oC$

E. 77.4 K to degree Celsius

$T(^oC)=T(^K)-273.15$

$T(^oC)=77.4(K)-273.15=-195.75^oC$

F. 77.4°R to degree Celsius

$(T)^oC=((T)^oR-491.67)\times \frac{5}{9}$

$(T)^oC=((77.4)^oR-491.67)\times \frac{5}{9}=-230.15 ^oC$

7 0

3 years ago