An ion from a given element has 20 protons and 18 electrons.

if a volume of air occupying 12.0Lat 20celsius is heated to a new temperature of 100 celsius,what would be the new volume

emmasim [6.3K]

Answer:

V₂ = 15.3

Explanation:

Given data:

Initial volume = 12.0 L

Initial temperature = 20°C

Final temperature =100°C

Final volume = ?

Solution:

First of all we will convert the temperature into kelvin.

20°C + 273 = 293 K

100°C + 273 = 373 K

Formula:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 12.0 L × 373 K / 293 k

V₂ = 4476 L.K /293 k

V₂ = 15.3

V₂ = 1566 L.K / 298 K

V₂ = 5.3 L

6 0

3 years ago

Discribe how continents and oceans are carried on tectonic plates

Vanyuwa [196]

This question seems to be very basic .the ocean is above the plate, the plate material is heavier than water (it being rock). That's pretty much all of it. The plates grind a little and new land pushes up at plate boundaries but this does not seem to be related. The heavier material is below and the lighter above, those being rock and water respectively.
hope this helped u

7 0

2 years ago

Metabolic activity in the human body releases approximately 1.0 x 104kJ of heat per day. Assuming the body is 50 kg of water, ho

Zarrin [17]

Answer:

The body temperature would rise by 47.85 °C

The amount of water the body evaporates is 4.15 kg.

This makes sense because firstly the value obtained is positive then secondly it is a normal occurrence in the real world that in a place where the temperature is high the body usually produce sweat in order to balance its internal temperature

Explanation:

Considering the relationship (between the heat released and the mass of the object) as shown below

q = msΔT

where q is the heat released per day = $1.0 * 10^{4} kJ$

m is the mass of the body = 50 kg

ΔT is the temperature rise = ?

s is the specific heat of water = $4,18kJ/kg^{o} C$

substituting values we have

$1.0 * 10^{4} kJ$ = $(50kg) (4.18kJ/kg^{o}C )$ ΔT

ΔT = $\frac{1.0 * 10^{4}kg}{(50kg)(4.18kJ/kg^{o} C)}$

= 47.85°C

To maintain the normal body temperature (98.6F = 37°C) the amount of heat released by metabolism activity must be utilized for evaporation of some amount of water

Hence

$Amount of water that must be evaporated =\frac{heat released per day}{heat of vaporization of water}$

$= \frac{1.0* 10^{4}kJ}{2.41kJ/g} \\= 4149.38g\\=4.15kg$

Note (1 kg = 1000 g)

This makes sense because firstly the value obtained is positive then secondly it is a normal occurrence in the real world that in a place where the temperature is high the body usually produce sweat in order to balance its internal temperature

7 0

2 years ago

When silver tarnishes, ag atoms are oxidized to form ag+ ions. ag atoms are reduced to form ag+ i?

Tanya [424]

<h3><u>Answer;</u></h3>

Ag atoms are oxidized to form Ag+ ions

<h3><u>Explanation</u>;</h3>

<em><u>Silver tarnish is the result of the oxide on the silver surface reacting with hydrogen sulfide (H2S) in air. This leaves a black film of silver sulfide (Ag2S). </u></em> Silver atoms are oxidized to form Ag+ ions.
When a thin coating of silver sulfide forms on the surface of silver, it darkens the silver.
The offending silver tarnish can be removed by either striping it from the surface or reversing the chemical reaction and turn silver sulfide back into silver.

7 0

2 years ago

A 1.45 L buffer solution consists of 0.111 M butanoic acid and 0.281 M sodium butanoate. Calculate the pH of the solution follow

umka21 [38]

Answer: -5

Explanation:

5 0

2 years ago