A crushed garlic will have a lot of flavor when placed in food due to the surface area that is in contact with the food. When we have a large piece of garlic, only the external part touches the food and its full capacity is not used. When we reduce the size of the year by crushing the internal parts that were not in contact with the food, now they will be, in addition, liquids are also released due to the pressure exerted on the garlic and these liquids mix more easily with the food and they give it more flavor. For better understanding we can see the following figure:
Simply to understand it, in the figure, there is a clove of whole garlic represented by the rectangle that will have a height of 3 and a width of 1, the units do not matter in this case. The area that is in contact will be equal to 8, but if we divide the garlic into three equal parts, it will have a contact area greater than 12. Therefore, the more we divide the garlic, the more area it will be in contact with the food and will give it more flavor.
Answer:
it's the lighting
Explanation:
if you were to use a prism you will see the rainbow. the color black sucks up all colors white sucks up none. that why in the summer black it the hottest to ware and white is the best to wear and in the winter vise versa
Spent fuel that can no longer be used to create energy is waste.
20600Cal
Explanation:
Given parameters:
Mass of water = 319.5g
Initial temperature = 35.7°C
Final temperature = 100°C
Unknown:
Calories needed to heat the water = ?
Solution:
The calories is the amount of heat added to the water. This can be determined using;
H = m c Ф
c = specific heat capacity of water = 4.186J/g°C
H is the amount of heat
Ф is the change in temperature
H = m c (Ф₂ - Ф₁)
H = 319.5 x 4.186 x (100 - 35.7) = 85996.56J
Now;
1kilocalorie = 4184J
85996.56J to kCal; = 20.6kCal = 20600Cal
learn more:
Specific heat brainly.com/question/3032746
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Answer:
Explanation:
Initial burette reading = 1.81 mL
final burette reading = 39.7 mL
volume of NaOH used = 39.7 - 1.81 = 37.89 mL .
37.89 mL of .1029 M NaOH is used to neutralise triprotic acid
No of moles contained by 37.89 mL of .1029 M NaOH
= .03789 x .1029 moles
= 3.89 x 10⁻³ moles
Since acid is triprotic , its equivalent weight = molecular weight / 3
No of moles of triprotic acid = 3.89 x 10⁻³ / 3
= 1.30 x 10⁻³ moles .