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3 years ago

A piece of material with a mass of 755 g is heated to 84.5 c and added to 50 g of water at 5

1 answer:

6 0

Alright sorry you're getting the answer hours later, but i can help with this.
so you're looking for specific heat, the equation for it is <span>macaΔTa = - mbcbΔTb with object a and object b. that's mass of a times specific heat of a times final minus initial temperature of a equals -(mass of b times specific heat of b times final minus initial temperature of b)
</span>so putting in your values is, 755g * ca * (75 celsius - 84.5 celsius) = -(50g * cb * (75 celsius - 5 celsius))
well we know the specific heat of water is always 4180J/kg celsius, so put that in for cb
with a bit of simplification to the equation by doing everything on each side first you have, -7172.5 * ca = -14630000
divide both sides by -7172.5 so you can single out ca and you get, ca= 2039.74
add units for specific heat which are J/kg celsius and the specific heat of the material is 2039.74 J/kg celsius

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3 years ago

A cylinder with a moving piston expands from an initial volume of 0.350 L against an external pressure of 1.90 atm. The expansio

myrzilka [38]

Answer:

1.84 L

Explanation:

Using the equation for reversible work:

$W = -P*(V_{2} - V_{1})$

Where:

W is the work done (J) = -287 J.

Since the gas did work, therefore W is negative.

P is the pressure in atm = 1.90 atm.

However, work done is in joules and pressure is in atm. We can use the values of universal gas constant as a convenient conversion unit. R = 8.314 J/(mol*K); R = 0.0821 (L*atm)/(mol*K)

Therefore, the conversion unit is 0.0821/8.314 = 0.00987 (L*atm)/J

$V_{1}$ is the initial volume = 0.350 L

$V_{2}$ is the final volume = ?

Thus:

(-287 J)*0.00987 (L*atm)/J = -1.9 atm*( $V_{2}$ - 0.350) L

$V_{2}$ = [(287*0.00987)+(1.9*0.350)]/1.9 = (2.833+0.665)/1.9 =1.84 L

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3 years ago

What are some chemical properties of Gallium?

Natali [406]

Answer:

Gallium is silvery white and soft enough to be cut with a knife. It takes on a bluish tinge because of superficial oxidation. Unusual for its low melting point (about 30 °C [86 °F]), gallium also expands upon solidification and supercools readily, remaining a liquid at temperatures as low as 0 °C (32 °F).

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3 years ago

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irinina [24]

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2 years ago

If a certain gas occupies a volume of 19 L when the applied pressure is 9.5 atm, find the pressure when the gas occupies a volum

kakasveta [241]

Answer: 37.6 atm

Explanation:

Given that,

Initial volume of gas (V1) = 19L

Initial pressure of gas (P1) = 9.5 atm

Final volume of gas (V2) = 4.8L

Final pressure of gas (P2) = ?

Since pressure and volume are given while temperature remains the same, apply the formula for Boyle's law

P1V1 = P2V2

9.5 atm x 19L = P2 x 4.8L

180.5 atm•L = 4.8L•P2

Divide both sides by 4.8L

180.5 atm•L/4.8L = 4.8L•P2/4.8L

37.6 atm = P2

Thus, the final pressure is 37.6 atmospheres.

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3 years ago