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3 years ago

Which substance would retain the most heat for the longest period of time?

Chemistry

2 answers:

Sergio [31]3 years ago

6 0

Answer:

<h2>THIS IS ANSWER: <em> The different materials have different densities and capa city heat But capacity heats is how much unit of material , and the many materials or any metals is how measures of conductions of heat energy ...</em></h2>

Explanation:

<em>but capacity heats of material can be hold another unit and either materials (volume , mass) somethings like how ,capacity heat compare to the materials or can be molybdenum ,or but the better insulations or any heat energy and also take a long time to heats up of materials</em>

<h3><em> sana maka tulong?</em></h3>

Harrizon [31]3 years ago

4 0

<h3>Water</h3>

<em>Specific heat is defined as the amount of heat one gram of a substance must absorb or lose to change its temperature by one degree Celsius.</em>

For water, this amount is one calorie, or 4.184 Joules.

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Ascorbic acid (H2C6H6O6; H2Asc for this problem), known as vitamin C, is a diprotic acid (Ka1= 1.0x10–5 and Ka2= 5x10–12) found

Igoryamba

Answer:

The concentrations are :

$[HAsc^-]=0.000702 M$

$[Asc^{2-}]=5.92\times 10^{-8} M$

The pH of the solution is 3.15.

Explanation:

$H_2Asc\rightleftharpoons HAs^-+H^+$ $K_{a1}=1.0\times 10^{-5}$

Initial

c 0 0

Equilibrium

c-x x x

$K_{a1}=\frac{[HAs^-][H^+]}{[H_2Asc]}$

$1.0\times 10^{-5}=\frac{x\times x}{(c-x)}$

$1.0\times 10^{-5}=\frac{x^2}{(0.050-x)}$

Solving for x:

x = 0.000702 M

$[HAsc^-]=0.000702 M$

$HAsc^-\rightleftharpoons As^{2-}+H^+$ $K_{a2}=5\times 10^{-12}$

Initially

x 0 0

At equilibrium ;

(x - y) y y

$K_{a2}=\frac{[As^{2-}][H^+]}{[HAsc^-]}$

$5\times 10^{-12}=\frac{y\times y}{(x-y)}$

$5\times 10^{-12}=\frac{y^2}{(x-y)}$

Putting value of x = 0.000702 M

$5\times 10^{-12}=\frac{y^2}{(0.000702 -y)}$

$y=5.92\times 10^{-8} M$

$[Asc^{2-}]=5.92\times 10^{-8} M$

Total concentration of $[H^+]=x+y=0.000702 M+5.92\times 10^{-8} M=7.0206\times 10^{-4} M$

The pH of the solution :

$pH=-\log[H^+]$

$pH=-\log[7.0206\times 10^{-4} M}=3.15$

7 0

3 years ago

predict where an unknown element that has these properties could fit in the periodic table 62 protons and electrons

Vesnalui [34]

Position of element in periodic table is depend on the electronic configuration of element.

Element with 62 electrons has following electronic configuration:
<span>1s2 2s2 </span>2p6 <span>3s2 </span>3p6 4s2 3d10 4p6 <span>5s2 </span>4d10 5p6 4f6 <span>6s<span>2
</span></span>
From above electronic configuration, it can be seen that highest value of principal quantum number, where electron is present, is 6. Hence, element belongs to 6th period.

Further, last electron has entered f-orbital, hence it is a f-block element. Position of f-block element is the bottom of periodic table.

Further, there are 6 electrons in f-orbital. Hence, it is the 6th f-block element in 6th period of periodic table.

5 0

3 years ago

Write the structure of the following compound:<br>=> 3-(4-chlorophenyl)-2-methylpropane

Rudiy27

Answer:

See explanation and image attached

Explanation:

The IUPAC system of nomenclature enables the structure of molecules to be written seamlessly from the name of the compound. Hence it is commonly called the systematic nomenclature.

The parent chain here is propane. It is substituted at the 2- position by a methyl group and at the 3-position by 4-chlorophenyl group as we can see in the image attached to this answer.

7 0

3 years ago

Soda pop is carbonated with CO2. Mark puts one bottle of soda pop in the refrigerator and leaves the other out in the hot sunlig

fomenos

Answer:

The one left in the hot sunlight.

Explanation:

The solubility of gases decreases when temperature increases. The gas in the soda pop (CO2) left in the sun will not stay dissolved as much as the on left in the refrigerator.

3 0

3 years ago

Complete combustion of a 0.600-g sample of a compound in a bomb calorimeter releases 24.0 kJ of heat. The bomb calorimeter has a

coldgirl [10]

The final temperature, t₂ = 30.9 °C

<h3>Further explanation</h3>

Given

24.0 kJ of heat = 24,000 J

Mass of calorimeter = 1.3 kg = 1300 g

Cs = 3.41 J/g°C

t₁= 25.5 °C

Required

The final temperature, t₂

Solution

Q = m.Cs.Δt

Q out (combustion of compound) = Q in (calorimeter)

24,000 = 1300 x 3.41 x (t₂-25.5)

t₂ = 30.9 °C

3 0

3 years ago

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