Tap water used for drinking is considered in the DIL calculations; true
Neutrons don’t carry an electrical charge, meaning that adding or subtracting them from the nucleus will not change the electrical charge of the nucleus of an atom. But, adding/removing neutrons changes the mass of the nucleus. This is how isotopes are formed.
The 7160 cal energy is required to melt 10. 0 g of ice at 0. 0°C, warm it to 100. 0°C and completely vaporize the sample.
Calculation,
Given data,
Mass of the ice = 10 g
Temperature of ice = 0. 0°C
- The ice at 0. 0°C is to be converted into water at 0. 0°C
Heat required at this stage = mas of the ice ×latent heat of fusion of ice
Heat required at this stage = 10 g×80 = 800 cal
- The temperature of the water is to be increased from 0. 0°C to 100. 0°C
Heat required for this = mass of the ice×rise in temperature×specific heat of water
Heat required for this = 10 g×100× 1 = 1000 cal
- This water at 100. 0°C is to be converted into vapor.
Heat required for this = Mass of water× latent heat
Heat required for this = 10g ×536 =5360 cal
Total energy or heat required = sum of all heat = 800 +1000+ 5360 = 7160 cal
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Answer:
∇T = 51.68°C
Explanation:
Mass = 150g
Heat Energy (Q) = 1.0*10³J
Change in temperature ∇T = ?
Q = mc∇T
Q = heat energy
M = mass
C = specific heat capacity of the gold = 0.129j/g°C
∇T = change in temperature
Q = Mc∇T
1.0*10³ = 150 * 0.129 * ∇T
1000 = 19.35∇T
Solve for ∇T
∇T = 1000 / 19.35
∇T = 51.679°C = 51.68°C
The change in temperature of gold was 51.68°C
Answer:
The correct answer is A. 140 atm
Explanation:
We use the gas formula, which results from the combination of the Boyle, Charles and Gay-Lussac laws. According to which at a constant mass, temperature, pressure and volume vary, keeping constant PV / T. We convert the unit Celsius into Kelvin:
0 ° C = 273K, 67 ° C = 273 + 67 = 340K; 94 ° C = 273 + 94 = 367K
P1xV1 /T1= P2x V2/T2
P2= ((P1xV1 /T1)xT2)/V2
P2=((88,89atm x 17L/340K)x367K)/12L= <em>135,927625 atm</em>
