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kvv77 [185]
3 years ago
6

A chemist prepares a solution of silver(I) nitrate by measuring out of silver(I) nitrate into a volumetric flask and filling the

flask to the mark with water.
Chemistry
1 answer:
leva [86]3 years ago
4 0

Answer:

3.33 M

Explanation:

It seems your question is incomplete, however, that same fragment has been found somewhere else in the web:

" <em>A chemist prepares a solution of silver nitrate (AgNO3) by measuring out 85.g of silver nitrate into a 150.mL volumetric flask and filling the flask to the mark with water.</em>

<em>Calculate the concentration in mol/L of the chemist's silver nitrate solution. Be sure your answer has the correct number of significant digits.</em> "

In this case, first we <u>calculate the moles of AgNO₃</u>, using its molecular weight:

  • 85.0 g AgNO₃ ÷ 169.87 g/mol = 0.500 mol AgNO₃

Then we<u> convert the 150 mL of the volumetric flask into L</u>:

  • 150 / 1000 = 0.150 L

Finally we <u>divide the moles by the volume</u>:

  • 0.500 mol AgNO₃ / 0.150 L = 3.33 M
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Its B. Increasing Atomic Number.


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4 Al + 3O2 → 2Al2O3 If 14.6 grams Al are reacted, how many liters of O2 at STP would be required?
melomori [17]

Answer: 9.08 L

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}    

\text{Moles of} Al=\frac{14.6g}{27g/mol}=0.54moles

4Al+3O_2\rightarrow 2Al_2O_3

According to stoichiometry :

4 moles of Al require  = 3 moles of O_2

Thus 0.54 moles of Al will require=\frac{3}{4}\times 0.54=0.405moles  of O_2

Standard condition of temperature (STP)  is 273 K and atmospheric pressure is 1 atm respectively.

According to the ideal gas equation:

PV=nRT

P = Pressure of the gas = 1 atm

V= Volume of the gas = ?

T= Temperature of the gas = 273 K      

R= Gas constant = 0.0821 atmL/K mol

n=  moles of gas= 0.405

V=\frac{nRT}{P}=\frac{0.405\times 0.0821\times 273}{1}=9.08L

Thus 9.08 L of O_2 at STP would be required

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Which force keeps the planets from floating off into space?
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Be sure to answer all parts. A person drinks four glasses of cold water (3.2 degree C) every day. The volume of each glass is 2.
Lelechka [254]

Explanation:

(a). The given data is as follows.

Volume of one glass of water = 2.2 \times 10^{2} ml = 220 ml

Volume of 4 glass of water = 220 \times 4 = 880 ml

We known that density of water is 1 g/ml.  Therefore, calculate the mass of water as follows.

       Mass of water = 880 ml \times 1 g/ml

                               = 880 gm

                               = 0.88 Kg               (as 1 kg = 1000 g)

The relation between heat energy, mass and temperature change is as follows.

            Q = mC \Delta T

\Delta T = (37 - 3.2)^{o}C = 33.8^{o}C

Putting the given values into the above formula as follows.

         Q = mC \Delta T

             = 0.88 \times 4.186 J/g^{o}C \times 33.8^{o}C

             = 124.5 kJ

Hence, the body have to supply 124.5 kJ to raise the temperature of the water to 37 degree C.

(b).      As we know that the heat of fusion of ice is 333 J/g.

So, energy required for 8.4 \times 10^{2} g or 840 g is as follows.

          333 \times 840 = 279.72 kJ

Heat capacity of water= 4.184 J/g^{o}C

Now, heat energy will be as follows.

           Q = 4.184 \times 840 g \times 37^{o}C

               = 130.03 kJ

Therefore, total heat required = (279.72 + 130.03) kJ

                                                  = 409.75 kJ

Hence, for the given situation your body should lose 409.75 kJ  of heat.

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