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Nutka1998 [239]
2 years ago
5

A. How many Joules of energy are required to raise 3.50kg of water from 38.5 to 75.0C ? (specific heat of water is 4.184J/g C)(1

000g=1kg)
b. How many grams of carbon are present if it requires 37.5 kJoules of energy to heat a sample from 30.0 to 55.0C ? (specific heat of carbon is 0.71 J/g C)

c. If 480.0 Joules is applies to a 50.0g sample of Hg in a thermometer that reads 25.0C, what will be the final temperature of the sample? (SH mercury is 0.14)

d. A 25.0kg sample of an unknown metal X requires 875kJ of energy to heat it from 55.0 to 125.0C. What is the specific heat of the unknown metal?


5.690x103 miles to meters.

900.0kg to grams

Please also give sig figs to all answers.

Answer fast please.
Chemistry
2 answers:
DochEvi [55]2 years ago
7 0

a.

A substance's specific heat tells you how much heat is required to increase the temperature of 1 g of that substance by 1°C.

The equation that establishes a relationshop between heat and change in temperature is

q = m • c • ∆T, where

q - heat absorbed

c - the specific heat of the substance, in your case of water

ΔT - the change in temperature, defined as the difference between the final temperature and the initial temperature

so:

q = 1.00 g • 4.18 J/g×°C • (75.0 - 38.5)°C

q = 152,57 J

just apply this formula for all exercises

Marta_Voda [28]2 years ago
5 0

A. How many Joules of energy are required to raise 3.50kg of water from 38.5 to 75.0C ? (specific heat of water is 4.184J/g C)(1000g=1kg)

Answer:

q = 534,506 Joules

q = 534.506 KiloJoules

534.506 KiloJoules or 534,506 Joules energy required to raise 3.50kg of water from 38.5 to 75.0C

<h3>What is sp.heat ?</h3>

Sp. Heat is the heat required to increase the temperature of the 1 mass of a given substance by a  1^{0} C temperature.

The formula of specific heat Cp = \frac{q}{m (∆T)}

Where,

q = energy of substance (Joules / KiloJoules),

Cp = Specific heat capacity of the substance (J/Kg.C),

m = mass of the substance

∆T = Change in temp.

Explanation:

Given data from que: -

Mss of the water (m) = 3.50 Kg = 3.50 × 1000 gm = 3500 gm

specific heat capacity of water (Cp) = 4.184J/g C

Change in temp (∆T) = 75.0 - 38.5 = 36.5

now, put all above given data in formula

we get

Cp = \frac{q}{m (∆T)}

q = Cp ×m×∆T

q = 4.184×3500×36.5

q = 534,506 Joules

q = 534.506 KJ

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Consider the reaction below for which K = 78.2 atm-1. A(g) + B(g) ↔ C(g) Assume that 0.386 mol C(g) is placed in the cylinder re
borishaifa [10]

Answer:

1.65 L

Explanation:

The equation for the reaction is given as:

                        A            +            B           ⇄        C

where;

numbers of moles = 0.386 mol C  (g)

Volume =  7.29 L

Molar concentration of C = \frac{0.386}{7.29}

= 0.053 M

                        A            +            B           ⇄        C

Initial               0                           0                      0.530    

Change          +x                          +x                       - x

Equilibrium      x                           x                      (0.0530 - x)

K = \frac{[C]}{[A][B]}

where

K is given as ; 78.2 atm-1.

So, we have:

78.2=\frac{[0.0530-x]}{[x][x]}

78.2= \frac{(0.0530-x)}{(x^2)}

78.2x^2= 0.0530-x

78.2x^2+x-0.0530=0  

Using quadratic formula;

\frac{-b+/-\sqrt{b^2-4ac} }{2a}

where; a = 78.2 ; b = 1 ; c= - 0.0530

= \frac{-b+\sqrt{b^2-4ac} }{2a}   or \frac{-b-\sqrt{b^2-4ac} }{2a}

= \frac{-(1)+\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}  or \frac{-(1)-\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}

= 0.0204  or -0.0332

Going by the positive value; we have:

x = 0.0204

[A] = 0.0204

[B] = 0.0204

[C] = 0.0530 - x

     = 0.0530 - 0.0204

     = 0.0326

Total number of moles at equilibrium = 0.0204 +  0.0204 + 0.0326

= 0.0734

Finally, we can calculate the volume of the cylinder at equilibrium using the ideal gas; PV =nRT

if we make V the subject of the formula; we have:

V = \frac{nRT}{P}

where;

P (pressure) = 1 atm

n (number of moles) = 0.0734 mole

R (rate constant) = 0.0821 L-atm/mol-K

T = 273.15 K  (fixed constant temperature )

V (volume) = ???

V=\frac{(0.0734*0.0821*273.15)}{(1.00)}

V = 1.64604

V ≅ 1.65 L

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ehidna [41]

Answer:

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Explanation:

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58,5 g---1 mol NaCl

64 g--------x= (64 g x1 mol NaCl)/58,5 g= 1, 09 mol NaCl

A solution molar--> moles of solute in 1 L of solution:

4 L-----1,09 mol NaCl

1L----x0( 1L x1,09 mol NaCl)/4L =0,27moles NaCl--->0,27M

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