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kati45 [8]
2 years ago
10

A 100g piece of iron at 150oC is emersed in 268.5 g of water at 20oC. the temperature of both iron and water became 25oC. If the

specific heat of iron is 0.449 J/g.K, find the specific heat of water?​
Chemistry
1 answer:
borishaifa [10]2 years ago
4 0

Answer:

cW = 4.2J/g.k

Explanation:

Heat gained = Heat lost

iron(mc∆€) = water(mc∆€)

mi= 100g

ci = 0.449

∆€i= 150-25 = 125

...

mw= 268.5g

cw = ?

∆€w=25-20 = 5

...

100×0.449×125= 268.5×cw×5

5612.5= 1342.5cw

cw = 5612.5/1342.5

cw= 4.18 ~ 4.2J/g.K

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Explanation:

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Acetone, a common solvent, has density 0.79g/cm3 at 20 degrees celcius. What is the volume of 85.1g of acetone at 20 degrees cel
RoseWind [281]
1)

a)

D = 0.79 g/cm³           m = 85.1 g

D = m / V

0.79 = 85.1 / V

V = 85.1 / 0.79 => 107.72 cm³
____________________________________________
b) D = m / V

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hope this helps!

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2 years ago
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What is the bronsred base of NO2- +H2O HNO2 +OH-
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Bronsted lowry bases

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3 years ago
.950 L of .420 M H2SO4 is mixed with .900 L of .260 M KOH. What concentration of sulfuric acid remains after neutralization?
kipiarov [429]
H₂SO₄:

V=0,95L
Cm=0,420mol/L

n = CmV = 0,42mol/L * 0,95L = 0,399mol

KOH:

V=0,9L
Cm=0,26mol/L

n = CmV = 0,26mol/L * 0,9L = 0,234mol

H₂SO₄            +           2KOH ⇒ K₂SO₄ + 2H₂O
1mol                :           2mol
0,399mol         :           0,234mol
                                    limiting reagent
reamins: 0,399mol - 0,117mol = 0,282mol

n = 0,282mol
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3 0
2 years ago
Suppose Gabor, a scuba diver, is at a depth of 15m15m. Assume that: The air pressure in his air tract is the same as the net wat
alexandr1967 [171]

Answer:

The ration of molar concentration is "2.5".

Explanation:

The given values are:

Average density of salt water,

= 1.03 \ g/cm^3

Net pressure,

= 2.00 \ atm

Increase in pressure,

= 1.00 \ atm

Now,

The under water pressure will be:

=  \frac{15 \ m}{10}\times 1 \ atm +1 \ atm

=  1.5\times 1+1

=  1.5+1

=  2.5 \ atm

hence,

The ratio will be:

=  \frac{(\frac{n}{V})_{15m} }{(\frac{n}{V})_{surface} }

or,

=  \frac{P}{P_s}

=  \frac{2.5}{1}

=  2.5

7 0
3 years ago
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