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andrew-mc [135]
3 years ago
8

How many moles are in 13.0 grams of water​

Chemistry
1 answer:
ira [324]3 years ago
6 0

13.0/18=.07222222

sig figs: 0.722 mole of water

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Unabbreviated electron configuration of magnesium
fenix001 [56]

Answer:

Mg₁₂ = 1s² 2s² 2p⁶ 3s²

Explanation:

Abbreviated and unabbreviated electronic configuration:

The abbreviated electronic configuration uses the noble gas configuration i.e complete electronic shells. For example, the atomic number of neon is ten and magnesium is twelve. The abbreviated electronic configuration of magnesium is written by using the neon abbreviation in following way:

The electronic configuration of neon is given below:

Ne₁₀ = 1s² 2s² 2p⁶

The abbreviated electronic configuration of magnesium:

Mg₁₂ = [Ne] 3s²

While the unabbreviated electronic configuration is written without using noble gas electronic configuration.

Unabbreviated electronic configuration of magnesium:

Mg₁₂ = 1s² 2s² 2p⁶ 3s²

5 0
3 years ago
PLEASE HELP!!!!! The table below shows the volume of two samples, X and Y, when placed in three containers of different volumes.
Dahasolnce [82]

Answer:

B m8

Explanation:

7 0
3 years ago
Determine the amount of heat(in Joules) needed to boil 5.25 grams of ice. (Assume standard conditions - the ice exists at zero d
Yuki888 [10]
Following are important constant that used in present calculations
Heat of fusion of H2O = 334 J/g 
<span>Heat of vaporization of H2O = 2257 J/g </span>
<span>Heat capacity of H2O = 4.18 J/gK 
</span>
Now, energy required for melting of ICE = <span>  334 X 5.25 = 1753.5 J .......(1)
Energy required for raising </span><span>the temperature water from 0 oC to 100 oC =  4.18 X 5.25 X 100 = 2195.18 J .............. (2)
</span>Lastly, energy required for boiling water = <span>  2257X 5.25 = 11849.25 J ......(3)
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Thus, total heat energy required for entire process = (1) + (2)  + (3)
                                                                        = 1753.5 + 2195.18 + 11849.25
                                                                        = </span><span>15797.93 J 
</span><span>                                                                        = 15.8 kJ
</span><span>Thus, 15797.93 J of energy is needed to boil 5.25 grams of ice.</span>
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3 years ago
3. 78 mL of 2.5 M phosphoric acid is neutralized with 500 mL of potassium hydroxide. What is the
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Hence, concentration of base is 1.17 M
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radioisotope actinium 225 has half life of 10 days, I begin with 16 kg of this isotope, how much will remain after 40 days?
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N(t)=N_{0}(\frac{1}{2})^\frac{t}{\tau_{_\frac{1}{2}}}}\\\\&#10;N_{0}=16kg\\&#10;t=40days\\&#10;\tau_{_\frac{1}{2}}}=10days\\\\\\&#10;N(t)=16kg*(\frac{1}{2})^{\frac{40}{10}}=16kg*\frac{1}{16}=1kg
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3 years ago
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