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ddd [48]
3 years ago
6

Warm air _________, while cool air ____________.

Chemistry
2 answers:
d1i1m1o1n [39]3 years ago
8 0

Answer: Warm air rises, cool air drops

Explanation:

bagirrra123 [75]3 years ago
4 0
The person who answer fist is right
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Which of the following is equal to 2.0 liters? 200 mL 2,000 cm3 20 m3 20,000 mm3
ololo11 [35]
It equals 2000  but in this case it would be 20,000mm3
4 0
4 years ago
What volume (mL) of the sweetened tea described in Example 1 contains the same amount of sugar (mol) as 10 mL of the soft drink
S_A_V [24]

The question is incomplete, the complete question is:

What volume (mL) of the sweetened tea described in Example 3.14 contains the same amount of sugar (mol) as 10 mL of the soft drink in this example. The example is attached below.

<u>Answer:</u> 75 mL of sweetened tea will contain the same amount of sugar as in 10 mL of soft drink

<u>Explanation:</u>

We first calculate the number of moles of soft drink in a volume of 10 mL

The formula used to calculate molarity:

\text{Molarity of solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (mL)}} .....(1)

Taking the concentration of soft drink from the example be = 0.375 M

Volume of solution = 10 mL

Putting values in equation 1, we get:

0.375=\frac{\text{Moles of sugar in soft drink}\times 1000}{10}\\\\\text{Moles of sugar in soft drink}=\frac{0.375\times 10}{1000}=0.00375mol

<u>Calculating volume of sweetened tea:</u>

Moles of sugar = 0.00375 mol

Molarity of sweetened tea = 0.05 M

Putting values in equation 1, we get:

0.05=\frac{0.00375\times 1000}{\text{Volume of sweetened tea}}\\\\\text{Volume of sweetened tea}=\frac{0.00375\times 1000}{0.05}=75mL

Hence, 75 mL of sweetened tea will contain the same amount of sugar as in 10 mL of soft drink

5 0
3 years ago
Perform unit conversions and determine Re for the case when a fluid with density of 92.8 lbm/ft3 and viscosity of 4.1 cP (centip
erastovalidia [21]

Answer:

Re=309926.13

Explanation:

density=92.8lbm/ft3*(0.45kg/1lbm)*(1ft3/0.028m3)=1491.43kg/m3

viscosity=4.1cP*((1*10-3kg/m*s)/1cP)=0.0041kg/m*s

velocity=237ft/min*(1min/60s)*(0.3048m/1ft)=1.2m/s

diameter=28inch*(0.0254m/1inch)=0.71m

Re=(density*velocity*diameter)/viscosity=(1491.43kg/m3*1.2m/s*0.71m)/0.0041kg/m*s

Re=309926.13

4 0
3 years ago
What are the limitations of Bronsted-Lowery concept???
HACTEHA [7]

Answer:

There is no limitation at all because that is the basic theory of solution electrolysis

3 0
4 years ago
Read 2 more answers
A certain drug has a half-life in the body of 3.5h. Suppose a patient takes one 200.Mg pill at :500PM and another identical pill
tekilochka [14]

Answer:

The amount of drug left in his body at 7:00 pm is 315.7 mg.

Explanation:

First, we need to find the amount of drug in the body at 90 min by using the exponential decay equation:

N_{t} = N_{0}e^{-\lambda t}

Where:

λ: is the decay constant = ln(2)/t_{1/2}

t_{1/2}: is the half-life of the drug = 3.5 h

N(t): is the quantity of the drug at time t

N₀: is the initial quantity

After 90 min and before he takes the other 200 mg pill, we have:

N_{t} = 200e^{-\frac{ln(2)}{3.5 h}*90 min*\frac{1 h}{60 min}} = 148.6 mg

Now, at 7:00 pm we have:

t = 7:00 pm - (5:00 pm + 90 min) = 30 min

N_{t} = (200 + 148.6)e^{-\frac{ln(2)}{3.5 h}*30 min*\frac{1 h}{60 min}} = 315.7 mg    

Therefore, the amount of drug left in his body at 7:00 pm is 315.7 mg (from an initial amount of 400 mg).

I hope it helps you!

3 0
3 years ago
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