Answer:
200 liters of fluoridated drinking water would have to be consumed by as person to reach the toxic level.
Explanation:
Toxic level of fluoride ions = 0.2 g per 70 kg body weight
Concentration of fluoride ions in drinking water = 1 mg/ L = 0.001 g/L
1 mg = 0.001 g
Let the volume of drinking water consumed by person to reach the toxic level of fluoride concentration be V.
For 70 kg body weight 0.2 grams of fluoride ions will be toxic.
200 liters of fluoridated drinking water would have to be consumed by as person to reach the toxic level.
Heat = mass * heat capacity of water * change in temperature
mass = 5.25 g
heat capacity of water = 4.186 joule/gram °C
Change in temperature = 62.8°C - 5.3°C = 57.5 °C
Plug in the values
heat = 5.25 g * 4.186 joule/gram °C * 57.5 °C = 1263.6 J
Rounded to two three significant figures, it is 1260 J of energy needed.
In terms of calories, the heat capacity of water is 1 calorie/gram °C. So do the plugging in all over again.
mass = 5.25 g
heat capacity of water = 1 calorie/gram °C
Change in temperature = 62.8°C - 5.3°C = 57.5 °C
heat = 5.25 g * 1 calorie/gram °C * 57.5 °C = 301.9 calories
Rounded to 3 significant figures, it is 302 calories
Q=SM∆T=4.18*5.25*(62.8-4.3)=1280 J
1280 J * (1 cal/4.18 J) = 307 cal
Answer:
Endothermic: ice melting into water, and an instant ice pack turning cold Explanation:
Answer: 2 The Proximity toward the ocean
Explanation:
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Answer:
The crystal field stabilization energy for the octahedral ion hexacyanomanganate(III) , if the wavelength of maximum absorption for the ion is 600 nm is -<u> 1987.59kJ/mole</u>
Explanation:
Lets calculate -:
Crystal field stabilization energy -
where h = planks constant =
c= velocity of light =
=
=
= =
=
Thus , the crystal field stabilization energy for the octahedral ion hexacyanomanganate(III) is 1987.59kJ/mole