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RSB [31]
3 years ago
7

Living and nonliving things cheetahs interact with

Chemistry
2 answers:
skelet666 [1.2K]3 years ago
7 0
Or other big cats like some lions
arsen [322]3 years ago
4 0
Cheetahs interact with their own kind

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Calculate the osmotic pressure of a solution containing 24.6g of glycerin in 250ml of solution at 298K
gavmur [86]
The formula to calculate osmotic pressure is 
Osmotic Pressure = M R T
M = Molarity
R = Ideal Gas Constant
T = Temperature in Kelvin

So, 
24.6/.2254kg=109.139g /kg  >>>>> Molarity
109.139 x mols/92 g = 1.186 mols kg^-1
1.186 x 0.08134 x 298 K = 28.755 atm
<span>1.06852 x 0.08134 x 298K= 26.5 atm

The answer is 26.5</span>
6 0
3 years ago
One difference between mixtures and pure substances is that
defon

Answer:

The mixture is made up of different atoms and pure substance is made up of same type of atom.

The main difference is that mixture can be separated into its component by physical mean while pure substances can not be separated by physical process

Explanation:

Mixture:

  • The properties of the mixture are not same and contains the properties of all those component present in it.
  • it is a combination of one or more Pure substances and can be separated by simple physical methods.
  • it have varying boiling and melting point

Examples are:

  • mixture of salt and sand
  • Salt water is mixture of water and NaCl and can be separated by physical mean.
  • Alloys: its a mixture of different metal
  • Air: mixture of gases

Pure Substance:

Pure substances are those made of same type of atoms all elements and compounds are pure substances.

  • it can not be separated by simple physical mean
  • it have very constant and consistent melting and boiling point

Examples are:

  • Water : contain only water molecule
  • All elements: all elements are pure substance made of same atoms
  • All compounds: can not be separated by physical mean.

8 0
3 years ago
Read 2 more answers
What is the maximum concentration of Ag⁺ that can be added to a 0.00750 M solution of Na₂CO₃ before a precipitate will form? (Ks
Firlakuza [10]

Answer:

\large \boxed{1.64\times 10^{-5}\text{ mol/L }}

Explanation:

Ag₂CO₃(s) ⇌2Ag⁺(aq) + CO₃²⁻(aq); Ksp = 8.10 × 10⁻¹²

                           2x      0.007 50 + x

K_{sp} =\text{[Ag$^{+}$]$^{2}$[CO$_{3}^{2-}$]} = (2x)^{2}\times 0.00750 = 8.10 \times 10^{-12}\\0.0300x^{2} = 8.10 \times 10^{-12}\\x^{2} = 2.70 \times 10^{-10}\\x = \sqrt{2.70 \times 10^{-10}} = \mathbf{1.64\times 10^{5}} \textbf{ mol/L}\\\text{The maximum concentration of Ag$^{+}$ is $\large \boxed{\mathbf{1.64\times 10^{-5}}\textbf{ mol/L }}$}

 

3 0
3 years ago
Express 0.044 km in meters.
SVEN [57.7K]

Answer:

44

Explanation:

You multiply the length value by 1000

6 0
3 years ago
Read 2 more answers
In an electrically heated boiler, water is boiled at 140°C by a 90 cm long, 8 mm diameter horizontal heating element immersed in
RideAnS [48]

Explanation:

The given data is as follows.

Volume of water = 0.25 m^{3}

Density of water = 1000 kg/m^{3}

Therefore,  mass of water = Density × Volume

                       = 1000 kg/m^{3} \times 0.25 m^{3}

                       = 250 kg  

Initial Temperature of water (T_{1}) = 20^{o}C

Final temperature of water = 140^{o}C

Heat of vaporization of water (dH_{v}) at 140^{o}C  is 2133 kJ/kg

Specific heat capacity of water = 4.184 kJ/kg/K

As 25% of water got evaporated at its boiling point (140^{o}C) in 60 min.

Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg

Heat required to evaporate = Amount of water evapotaed × Heat of vaporization

                           = 62.5 (kg) × 2133 (kJ/kg)

                           = 133.3 \times 10^{3} kJ

All this heat was supplied in 60 min = 60(min)  × 60(sec/min) = 3600 sec

Therefore, heat supplied per unit time = Heat required/time = \frac{133.3 \times 10^{3}kJ}{3600 s} = 37 kJ/s or kW

The power rating of electric heating element is 37 kW.

Hence, heat required to raise the temperature from 20^{o}C to 140^{o}C of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)

                      = 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)

                     = 125520 kJ  

Time required = Heat required / Power rating

                       = \frac{125520}{37}

                       = 3392 sec

Time required to raise the temperature from 20^{o}C to 140^{o}C of 0.25 m^{3} water is calculated as follows.

                    \frac{3392 sec}{60 sec/min}

                     = 56 min

Thus, we can conclude that the time required to raise the temperature is 56 min.

4 0
3 years ago
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