Answer:
Fig 3 is a mixture
Explanation:
Fig 1 has just one kind of atoms.
fig 2 shows 2 atoms of the same kind bonded together (compound)
fig 4 shows atoms of 2 different kinds bonded together (compound)
only in fig 3 we see uneven distribution of two kinds of atoms. By definition a mixture contains two or more kind of substances randomly arranged. So fig 3 shows arrangement of mixture
Answer:
14. 13.2cg = 1.32dg
15. 3.8m = 0.0038km
16. 24.8L = 24800mL
17. 0.87kL = 870L
18. 26.01cm = 0.0002601km
19. 0.001hm = 10cm
Explanation:
14. 13.2/10 = 1.32
15. 38/1000 = 0.0038
16. 24.8(1000) = 24,800
17. 0.87(1000) = 870
18. 26.01/100000 = 0.0002601
19. 0.001hm(10000) = 10
An easy way to do these by yourself is to familiarize yourself with what each prefix means. Once you do this, you can multiply the value of the prefix when converting from a smaller unit of measurement to a larger one and divide the value of the prefix when converting from a large unit of measurement to a smaller one.
Answer : The heat required is, 1904 calories.
Explanation :
The process involved in this problem are :
The expression used will be:
![\Delta H=m\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]](https://tex.z-dn.net/?f=%5CDelta%20H%3Dm%5Ctimes%20%5CDelta%20H_%7Bfusion%7D%2B%5Bm%5Ctimes%20c_%7Bp%2Cl%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D)
where,
m = mass of ice = 17 g
= specific heat of liquid water = 
= enthalpy change for fusion = 
Now put all the given values in the above expression, we get:
![\Delta H=17g\times 80.0cal/g+[17g\times 1cal/g^oC\times (32.0-0)^oC]](https://tex.z-dn.net/?f=%5CDelta%20H%3D17g%5Ctimes%2080.0cal%2Fg%2B%5B17g%5Ctimes%201cal%2Fg%5EoC%5Ctimes%20%2832.0-0%29%5EoC%5D)
Therefore, the heat required is, 1904 calories.
<span>Measurements by traditional analog tools are less prone to experimental error.</span>
