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Lina20 [59]
3 years ago
5

3. The distance from the thumb to the little finger on Mrs. McCord's hand is 98.0mm. Convert this to

Chemistry
1 answer:
lianna [129]3 years ago
4 0

Answer:

0.098m

Hope it helps

I am sure about the answer

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3 years ago
Determine the number of significant figures in 478 cm​
Grace [21]

Answer:

There are 3 significant figures on this one.

3 0
3 years ago
True<br> or False: A force has to touch an object in order to affect it.<br> Hellppp
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3 years ago
One isotope of Br has a half-life of 16.5 hours. How much of a 2.00 gram sample remains at the end of 1.00 day?
valina [46]

Answer:

half life=16.5hrs

sample=2g

total time=1day=24hrs

no. of half lives=total time/half life

no. of half lives=24hrs/16.5hrs=1.45approx1.5

sample left=1/2n=1/2[1.5]=0.707gapprox

Explanation:

4 0
3 years ago
Air at 25°C with a dew point of 10 °C enters a humidifier. The air leaving the unit has an absolute humidity of 0.07 kg of water
madreJ [45]

Explanation:

The given data is as follows.

         Temperature of dry bulb of air = 25^{o}C

          Dew point = 10^{o}C = (10 + 273) K = 283 K

At the dew point temperature, the first drop of water condenses out of air and then,

        Partial pressure of water vapor (P_{a}) = vapor pressure of water at a given temperature (P^{s}_{a})

Using Antoine's equation we get the following.

            ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{T(K) - 46.854}

            ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{283 - 46.854}

                               = 0.17079

                   P^{s}_{a} = 1.18624 kPa

As total pressure (P_{t}) = atmospheric pressure = 760 mm Hg

                                   = 101..325 kPa

The absolute humidity of inlet air = \frac{P^{s}_{a}}{P_{t} - P^{s}_{a}} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}

                  \frac{1.18624}{101.325 - 1.18624} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}

                 = 0.00735 kg H_{2}O/ kg dry air

Hence, air leaving the humidifier has a has an absolute humidity (%) of 0.07 kg H_{2}O/ kg dry air.

Therefore, amount of water evaporated for every 1 kg dry air entering the humidifier is as follows.

                 0.07 kg - 0.00735 kg

              = 0.06265 kg H_{2}O for every 1 kg dry air

Hence, calculate the amount of water evaporated for every 100 kg of dry air as follows.

                0.06265 kg \times 100

                  = 6.265 kg

Thus, we can conclude that kg of water the must be evaporated into the air for every 100 kg of dry air entering the unit is 6.265 kg.

3 0
3 years ago
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