Answer:
because it will help u
because it will make u to understand the question very well
heyyy Powerpuff girls! what advantage does globe maps have over a flat map?
Answer:
The advantage of the globe is that it promotes visual accuracy. Students need to use a globe frequently if they are to form accurate mental maps. The advantage of the world map is that you can see the entire world at one time. The disadvantage is that world maps distort shape, size, distance, and direction.
Explanation:
miss girl- you could've just looked this up
(just make it look like you typed it)
Answer:
The advantage of the globe is that it promotes visual accuracy. Students need to use a globe frequently if they are to form accurate mental maps. The advantage of the world map is that you can see the entire world at one time. The disadvantage is that world maps distort shape, size, distance, and direction.
Friend?
The diagram represents a plant leaf cell and two different molecules used in the process of photosynthesis. Molecule 1 and 2 are most likely
A. Carbon dioxide and water
B. Nitrogen and water
C. carbon dioxide and nitrogen
D. Nitrogen and oxygen
you take 295.5 g of a solid at 30.0 c and let it melt in 425 g of water. the water temperature decreases from 85.1 c to 30.0 c. calculate the heat of fusion of this solid
Answer:
[tex]\Delta _{fus}H=332 \frac{J}{g}[/tex]
Explanation:
Hello!
In this case, since the heat of fusion of a solid substance stands for the energy required to melt it, which is a phase transition from solid to liquid, we can see that the heat lost by water is that gained by the solid, so we can write:
[tex]Q_{solid}=-Q_{w}[/tex]
Thus, by using the water data and its specific heat (4.184), we obtain:
[tex]Q_{solid}=-m_{w}C_{w}(T_f-T_i)\\\\Q_{solid}=-425g*4.184\frac{J}{g\°C}*(30.0-85.1)\°C\\\\ Q_{solid}=97,978.82J=98.0kJ[/tex]
Next, since the heat of fusion of a substance is usually represented in terms of energy per amount of substance, we use the mass of solid to obtain:
[tex]\Delta _{fus}H=\frac{98.0kJ}{295.5g}\\\\ \Delta _{fus}H=0.332\frac{kJ}{g}=332 \frac{J}{g}[/tex]
Best regards!
The heat of fusion of the solid is 0.332 KJ/g or 332 J/g
We'll begin by calculating the heat loss by the water. This can be obtained as follow:
Mass of water (M) = 425 g
Initial temperature of water (T₁) = 85.1 °C
Final temperature (T₂) = 30 °C
Change in temperature (ΔT) = T₂ – T₁ = 30 – 85.1 = –55.1 °C
Specific heat capacity of water (C) = 4.184 J/gºC
Heat loss (Q) =?Q = MCΔT
Q = 425 × 4.184 × –55.1
Q = –97978.82 J
Divide by 1000 to express in KJ
Q = –97978.82 / 1000
Q = –97.98 KJ Finally, we shall determine the heat fusion of the solid.Heat loss by water = –97.98 KJ
Heat gained by the solid (Q) = 97.98 KJ
Mass of solid (m) = 295.5 g
Heat of fusion (Hf) =?Q = m•Hf
97.98 = 295.5 × Hf
Divide both side by 295.5
Hf = 97.98 / 295.5
Hf = 0.332 KJ/g or 332 J/gLearn more: https://brainly.com/question/15563205