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Nadya [2.5K]
3 years ago
11

Chemical properties of potassium

Chemistry
1 answer:
tester [92]3 years ago
6 0
<span><span>Atomic number19 
</span><span>Atomic mass<span>39.0983g.mol -1
</span></span><span>Electronegativity according to Pauling0.8
</span><span>Density<span>0.86
 g.cm -3 at 0 °C
</span></span><span>Melting point <span>63.2 °C</span></span></span>
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For ethanol, propanol, and n-butanol the boiling points, surface tensions, and viscosities all increase. what is the reason for
Contact [7]
Moving from Ethanol through Propanol to Butanol the physical properties like boiling points, surface tension and viscosity increases because of the increases in intermolecular interactions between the molecules of given compounds.

Explanation:
                   Ethanol, propanol and butanol all have hydroxyl groups in common, means all have hydrogen bond intractions between their molecules. So, taking the hydrogen bonding interaction constant we are left with only the difference in the number of carbon atoms.
                    Butanol has the greatest physical properties than other two because it has four carbon atom chain. So, as we know the London Dispersion forces or Van der Waal forces increases with increase in molecular size and chain length of hydrocarbon.
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3 years ago
Balance the following reaction. As2S3 + 9O2 → 2As2O3 + SO2
kumpel [21]

Answer:

2As2S3 + 9O2 = 2As2O3 + 6SO2

Explanation:

8 0
3 years ago
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how many mL of a 3.5 M sodium hydroxide will be needed to neutralize 15mL of a 4.3 M hydrochloric acid solution?
kobusy [5.1K]
Step 1 : write a valanced equation..
NaOH + HCl 》NaCl + H2O

Step 2 : find the number of mole of HCl..
1000 ml ..contains 4.3 mole
15ml... (4.3÷1000)×15 =...

Stem 3 : use mole ratio....
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So mole is same as calculated above...

Step 4 :
3.5 mole of NaOH is in 1000ml
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3 years ago
How many milliliters of a 3.4 M NaCl solution would be needed to prepare each solution?
Ksenya-84 [330]

Answer:

a. Approximately 1.3\; \rm mL.

b. Approximately 7.2\; \rm mL.

Explanation:

The unit of concentration "\rm M" is equivalent to "\rm mol \cdot L^{-1}", which means "moles per liter."

However, the volume of both solutions were given in mililiters \rm mL. Convert these volumes to liters:

\displaystyle 45\; \rm mL = 45\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.045\; \rm L.

\displaystyle 330\; \rm mL = 330\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.330\; \rm L.

In a solution of volume V where the concentration of a solute is c, there would be c \cdot V (moles of) formula units of this solute.

Calculate the number of moles of \rm NaCl formula units in each of the two solutions:

Solution in a.:

n = c \cdot V = 0.045\; \rm L \times 0.10\; \rm mol \cdot L^{-1} = 0.0045\; \rm mol.

Solution in b.:

n = c \cdot V = 0.330\; \rm L \times 0.074\; \rm mol \cdot L^{-1} = 0.02442\; \rm mol.

What volume of that 3.4\; \rm M (same as 3.4 \; \rm mol \cdot L^{-1}) \rm NaCl solution would contain that many

For the solution in a.:

\displaystyle V = \frac{n}{c} = \frac{0.0045\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0013\; \rm L.

Convert the unit of that volume to milliliters:

\displaystyle 0.0013\; \rm L = 0.0013\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 1.3\; \rm mL.

Similarly, for the solution in b.:

\displaystyle V = \frac{n}{c} = \frac{0.02442\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0072\; \rm L.

Convert the unit of that volume to milliliters:

\displaystyle 0.0072\; \rm L = 0.0072\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 7.2\; \rm mL.

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