Answer the ones that are circled. (don't put decimals for the division put remainder)if u can't see any tell me.

Lesson 2.7:
1) / divide 195 by 18 to get 10.833.
/ to get the whole number, round up to get '11'.
11 cartons

Lesson 2.8:
3)  ÷  = 7
5) 6,681 ÷ 53 = 126 with a remainder of 3

Lesson 2.9:
1) / let 'x' equal the inches of snow that reidville gets
/ '12x' equal the inches of snow that greenboro gets
/ you should set up the problem like: 
/ add 'x + 12x' to get '13x'. 
/ divide both sides by '13'.
/ 5 goes into 13 to get 65. simplify the fraction to '5'.
/ x = 5
Reidville gets 5 inches of snow

2) / let 'x' equal the miles ansley ran in one month
/ you should set up the problem like: 
/ add '15x + x' to get '16x'
/ divide both sides by '16'
/ 12 goes into 16 to get 192. simplify the fraction to 12.
/ x = 12
Ansley runs 180 miles (Zack only runs 12)

Question:

A piston–cylinder device contains 0.85 kg of refrigerant- 134a at -10°C. The piston that is free to move has a mass of 12 kg and a diameter of 25 cm. The local atmospheric pressure is 88 kPa. Now, heat is transferred to refrigerant-134a until the temperature is 15°C. Determine (a) the final pressure, (b) the change in the volume of the cylinder, and (c) the change in the enthalpy of the refrigerant-134a.

a) 90.4 kPa

b) 0.0205 m³

c) 17.4 kJ/kg

Explanation:

Given:

Mass, m = 0.85 kg

a) The final pressure here is equal to the initial pressure. Let's use the formula:

= 90398 Pa

≈ 90.4 KPa

Final pressure = 90.4 kPa

b) Change in volume of the cylinder:

To find the initial and final volume, let's use the values from the A-13 table for refrigerant-134a, at initial values of 90.4 kPa and -10°C and final values of 90.4 kPa and 15°C

v1 = 0.2302m³/kg

h1 = 247.76 kJ/kg

v2 = 0.2544 m³/kg

h2 = 268.2 kJ/kg

Change in volume is calculated as:

Δv = m(v2 - v1)

Δv = 0.85(0.2544 - 0.2302)

= 0.0205 m³

Change in volume = 0.0205 m³

c) Change in enthalpy

Let's use the formula:

Δh = m(h2 - h1)

= 0.85(268.2 - 247.76)

= 17.4 kJ/kg

Change in enthalpy = 17.4 kJ/kg