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ladessa [460]
3 years ago
8

If the heat of fusion of ice is 335 j/g, how much heat is needed to melt 4.36 g of ice at 0.0 °c?

Chemistry
1 answer:
Vadim26 [7]3 years ago
7 0
Heat=335*4.36
=1460.6 joules
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By what factor would a scuba diver’s lungs expand if she ascended rapidly to the surface from a depth of 125 ft without inhaling
vova2212 [387]

It is found that scuba diver’s can safely ascend up to 52.53 ft without Breathing out.

By what factor would a scuba diver’s lungs expand if she ascended rapidly to the surface from a depth of 125 ft without inhaling or exhaling then

p_{1} =patm+pH_{2} O

p_{1} = 101325+ρgh_{1}

It is given that height is 125ft. Put the value of h in above formula:

h1  =125ft=38.1m

ρ=1.04g/mL=1040kg/m^{3}

g=9.81                                  

p_{1} =101325Pa+388711.44

p_{1}   =490036.44‬Pa                

p_{2} =p atm  =101325Pa

It is known that volume and pressure can be expressed as:

V*P=const.

where, V is volume and P is pressure.

Now,              

V_{1} *p_{1} =V_{2}  *p_{2}

V_{2} /V_{1}=p_{2} /p_{1}

V_{2} /V_{1} =490036.44/101325

V_{2} /V_{1}=4.84

Assume constant temperature

d of seawater = 1.04 g/mL; d of Hg = 13.5 g/mL.

now p_{1} =p atm​+pH_{2} O =490036.44Pa

V*p=const                              

where, V is volume and P is pressure.

Now,              

V_{1} *p_{1} =V_{2}  *p_{2}

V_{2} /V_{1}=p_{2} /p_{1}

V_{2} /V_{1} =490036.44/X

p_{2} =490036.44pa/(V2/V1) =326690.96Pa

p_{2} =patm +pH_{2} O

p_{2} =101325Pa+ρgh_{2}

326690.96Pa=101325Pa+ρgh_{2}

ρgh1  =151987.5-101325=225365.96‬‬Pa

ρ=1,04g/mL=1040kg/m3

g=9.81

h_{2} =225365.96/‬ρ∗g

​h_{2}  =225365.96  / ‬1040∗9.81

h_{2}  =22.09m= 72.47ft

ΔH=H_{1} -H_{2}

=125-72.47

=52.53ft

So she can safely ascend up to 52.53 ft without Breathing out

To know more about Scuba diver here

brainly.com/question/15430942

#SPJ4

6 0
1 year ago
2 Points
levacccp [35]
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