13. Convert 125.0 calories to Joules. A. 523J B. 210J C. 41.84J D. 764.1 J How m

Question

13. Convert 125.0 calories to Joules. A. 523J B. 210J C. 41.84J D. 764.1 J How much energy (calories) is needed to raise 12.0 grams ethanol at 14. 24.0 to 42.0 ? specific heat of ethanol-0.58 cal/g C A. 18.5 cal B. 99.0 cal C. 125.2 cal D. 341.4 cal 15. Convert 2280 mm Hg to atmospheres A. 350 atm B. 1140 atm C. 89.5 atm D. 3.00 atm A sample of sulfur dioxide occupies a volume of 652 mL at 40.0 °C and 720 mm Hg. What volume will the sulfur dioxide occupy at 0 °C, and 1 atm of pressure? 16. A. 140 mL B. 0.540L C. 825 mL D. 0.375 L A balloon with a volume of 2.0 L is filled with a gas at 3 atmospheres. If the pressure is reduced to 0.5 atmospheres without a change in temperature, what would be the volume of the balloon? 17. A. 4.0L B. 120. L C. 8.0L D. 12.0L 18. How many moles of gas are contained in 890.0 mL at 21.0 °C and 750.0 mm Hg pressure? A. 0.037 B. 1.24 C. 8.91 D. 2.31x102 19. At what pressure would 0.150 mole of nitrogen gas at 23.0 °C occupy 8.90 L? A. 0.200 atm B. 1250 mm Hg C, 0.724 atm D. 311 mm Hg

Explanation / Answer

Answers:-

13) Answer:- 523 J

Thermochemical calories to Joules

1 cal = 4.184 J

So, therefore

125.0 cal × 4.184 J = 523 J

14)Answer:- 125.2 cal

Use the equation,

q = m × c × T

                                        where,  

q = the heat absorbed / given off

                                       m = the mass of the sample

c = the specific heat of the substance

                                    T = the change in temperature

                                              Final temperature – initial temperature

In your case, the sample goes from 24 to 42, which means that,

T = 42 – 24 = 18

Plug your values into the above equation, to find the amount of heat absorbed by the water

q(water) = 12g × 0.58cal/g × 18

= 125.2 cal

15)Answwer:- 3.00 atm

1 atm = 760 mmHg

So, therefore

2280 mmHg × (1atm / 760 mmHg) = 3.00 atm

16)Answer:- 0.540 L

Therefore, the formula is,

P1V1 / T1 = P2V2 / T2

Where,

P1, V1 and T1 is the initial pressure, volume and temperature

And

P2, V2 and T2 is the final pressure, volume and temperature

You must notice that,

1 atm = 760 mmHg

We have given that,

P1 = 720 mmHg   ,   V1 = 652 ml   and T1 = 40+273 = 313 K

P2 = 760 mmHg    ,   V2 = ?           and T2 = 0 + 273 = 273 K

SO rearrange the above equation to solve for V2

V2 = (P1V1 / T1) × (T2 / P2)

Put all the value we have

V2 =( (720 mmHg × 652 ml) / 313k ) × (273 k/ 760mmHg)

V2 = (720mmHg × 652ml × 273 k )/ (313k × 760 mmHg)

V2 = 540 ml or 0.540 L

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