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3 years ago

What is the specific heat of copper when heated to 221.32

Chemistry

1 answer:

Dafna1 [17]3 years ago

5 0

Answer:

The specific heat of copper when heated to 221.32 (not listed form of heat measurement) is 221.32 (not listed form of heat measurement).

Explanation:

uh not really sure what else there is here, I may be missing something

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The question is in the attached word document.

Alekssandra [29.7K]

Answer:i read it Explanation:

3 0

3 years ago

Which equation represents a double replacement reaction?

AveGali [126]

Answer:

C) LiOH + HCl → LiCl + H₂O

General Formulas and Concepts:

Chemistry - Reactions

Synthesis Reactions: A + B → AB
Decomposition Reactions: AB → A + B
Single-Replacement Reactions: A + BC → AB + C
Double-Replacement Reactions: AB + CD → AD + BC

Explanation:

Step 1: Define

RxN A: 2Na + 2H₂O → 2NaOH + H₂

RxN B: CaCO₃ → CaO + CO₂

RxN C: LiOH + HCl → LiCl + H₂O

RxN D: CH₄ + 2O₂ → CO₂ + 2H₂O

Step 2: Identify

RxN A: Single Replacement Reaction

RxN B: Decomposition Reaction

RxN C: Double Replacement Reaction

RxN D: Combustion Reaction

7 0

3 years ago

How does an atom become an atom of another element is

ivann1987 [24]

Can you give me the anwsers

3 0

3 years ago

Which is a cause of polarity in water molecules? partial negative charge on the hydrogen atoms partial positive charge on the ox

Tamiku [17]

Explanation:

Water is a polar solvent as the hydrogen and oxygen atom has large difference in their electronegativities.

Oxygen atom is highly electronegative as compared to hydrogen atom therefore, it pulls the electrons of hydrogen atom closer towards itself.

As a result, two poles will create forming a partial positive charge on the hydrogen atom and partial negative charge on the oxygen atom.

Thus, we can conclude that high electronegativity difference between oxygen and hydrogen is the cause of polarity in water molecules.

7 0

3 years ago

The raw water supply for a community contains 18 mg/L total particulate matter. It is to be treated by addition of 60 mg alum (A

s344n2d4d5 [400]

Solution :

Given :

The steady state flow = 8000 $m^3 /d$

$= 80 \times 10^5 \ I/d$

The concentration of the particulate matter = 18 mg/L

Therefore, the total quantity of a particulate matter in fluid $= 80 \times 10^5 \ I/d \times 18 \ mg/L$

$= 144 \times 10^6 \ mg/g$

$= 144 \ kg/d$

If 60 mg of alum $[Al_2(SO_4)_3.14 H_2O]$ required for one litre of the water treatment.

So Alum required for $80 \times 10^5 \ I/d$

$= 80 \times 15^5 \ I/d \times 60 \ mg \ alum /L$

$= 480 \times 10^6 \ mg/d$

or 480 kg/d

Therefore the alum required is 480 kg/d

1 mg of the alum gives 0.234 mg alum precipitation, so 60 mg of alum will give $= 60 \times 0.234 \text{ of alum ppt. per litre}$

$= 14.04$ mg of alum ppt. per litre

480 kg of alum will give = 480 x 0.234 kg/d

= 112.32 kg/d ppt of alum

Daily total solid load is $= 144 \ kg/d + 112.32 \ kg/d$

= 256.32 kg/d

So, the total concentration of the suspended solid after alum addition $= 18 \ mg/L + 60 \times 0.234$

= 32.04 mg/L

Therefore total alum requirement = 480 kg/d

b). Initial pH = 7.4

The dissociation reaction of aluminium hydroxide as follows :

$Al(OH)_3 \rightleftharpoons Al^{3+} + 3OH^{-}$

After addition, the aluminium hydroxide pH of water will increase due to increase in $OH^-$ ions.

Therefore, the pH of water will be acceptable range after the addition of aluminium hydroxide.

c). The reaction of $CO_2$ and water as follows :

$CO_2 (g) + H_2O (l) \rightarrow H_2CO_3$

For the atmospheric pressure :

$p_{CO_2} = 3.5 \times 10^{-4} \ atm$

And the pH is reduced into the range of 5.9 to 6.4

6 0

3 years ago