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gtnhenbr [62]
3 years ago
12

What volume does .0685 mol of gas occupy at STP? 1mole =22.4

Chemistry
2 answers:
SashulF [63]3 years ago
6 0

Answer : The volume of gas at STP is, 1.53 liter

Solution : Given,

Moles of a gas = 0.0685 mole

As we know that at STP,

1 mole of gas contains 22.4 liter volume of gas

As, 1 mole of gas contains 22.4 liter volume of gas

So, 0.0685 mole of gas contains 0.0685\times 22.4=1.53L volume of gas

Therefore, the volume of gas at STP is, 1.53 liter

GuDViN [60]3 years ago
4 0
V = nRT/P
V = 0.685 mol*(.0821 L*atm/K*mol)*273 K/1 atm
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If 20.0 grams of ice takes about 6680j...then 40.0 grams of ice would take ___ .
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Two cars travel next to each other. One speedometer reads
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Answer:

The pair of arrows which represents the relationship of speeds of the two cars is;

The second option as shown in the attached drawing

Explanation:

The given parameters are;

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The blue arrow = 20 m/s

Therefore, given that the speeds of both cars are equal (20 m/s = 72 km/h = 20 m/s),  the pair of arrows that represent the relationship of speeds of the two cars is two equal length blue arrows which is the second option

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3 years ago
Predict how many grams of KCI is produced from 40 grams of K?
Nutka1998 [239]

Explanation:

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5 0
3 years ago
Define and then describe the difference between a polar covalent bond, nonpolar covalent bond and ionic bond.
Romashka [77]
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3 years ago
The true absorbance for a 1.0 x 10 −5 M solution is 0.7526. If the percentage stray light for a spectrophotometer is 0.56%, calc
Korvikt [17]

Answer:

The percentage deviation is  \Delta M = 1.87%

Explanation:

From the question we are told that  

     The concentration is of the solution is C = 1.0*10^{-5} M

     The true absorbance A = 0.7526

      The percentage of transmittance due to stray light z = 0.56% =\frac{0.56}{100}  = 0.0056

Generally Absorbance is mathematically represented as

           A = -log T

Where T is  the percentage of true transmittance

    Substituting value  

          0.7526 = - log T

              T = 10^{-0.7526}

                  = 0.177

                  = 17.7%

The Apparent absorbance is mathematically represented

           A_p = -log (T +z)

Substituting values

           A_p = -log(0.177 + 0.0056)

                = -log(0.1826)

               = 0.7385

The percentage by which apparent absorbance deviates from known absorbance is mathematically evaluated as

       \Delta A = \frac{A -A_p}{A} * \frac{100}{1}

              = \frac{0.7526 - 0.7385}{0.7526} * \frac{100}{1}

             \Delta A = 1.87%  

Since Absorbance varies directly with concentration the percentage deviation of the apparent concentration from know concentration  is

              \Delta M = 1.87%

           

6 0
3 years ago
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