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lina2011 [118]
3 years ago
7

What's the percentage of ZnF2

Chemistry
1 answer:
Elza [17]3 years ago
6 0
I don’t understand the question
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Which statement would support a merit of the Bronsted-Lowry base theory has over the Arrhenius base theory?
butalik [34]

Answer:

Explanation:

Bronsted Base is an H+ acceptor

No good answer Bronstead base does not accept hydroxide or electrons

5 0
3 years ago
Read 2 more answers
HELPPPP ANSWERSS!!!<br><br> A. <br> 13<br><br> B. <br> 8<br><br> C. <br> 15<br><br> D. <br> 5
KATRIN_1 [288]

Answer:

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Explnation:

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4 0
2 years ago
The total volume of seawater is 1.5 x 1021L .Assume that seawater contains 3.1 percent sodium chloride by mass and that its dens
son4ous [18]

Answer:

Mass in kg = 4.7*10^19 kg

Mass in tons = 5.2*10^16 tons

Explanation:

<u>Given:</u>

Total volume of sea water = 1.5*10^21 L

Mass % NaCl in seawater = 3.1%

Density of seawater = 1.03 g/ml

<u>To determine:</u>

Total mass of NaCl in kg and in tons

<u>Calculation:</u>

Unit conversion:

1 L = 1000 ml

The volume of seawater in ml is:

=\frac{1.5*10^{21}L*1000ml }{1L} =1.5*10^{24} ml

Mass\ seawater = Density*volume = 1.03g/ml*1.5*10^{24} ml=1.5*10^{24}g

Mass\ NaCl\ in\ seawater = \frac{3.1}{100}*1.5*10^{24}  g=4.7*10^{22} g

To convert mass from g to Kg:

1000 g = 1 kg

Mass\ seawater(kg) = \frac{4.7*10^{22}g*1kg }{1000g} =4.7*10^{19} kg

To convert mass from g to tons:

1 ton = 9.072*10^6 g

Mass\ seawater(tons) = \frac{4.7*10^{22}g*1ton }{9.072*10^{6}g } =5.2*10^{16} tons

5 0
3 years ago
Is electrical conductivity an extensive property or an intensive property?
Thepotemich [5.8K]
Your Answer Will Be Intensive Property
8 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)
</span><span>
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>
7 0
3 years ago
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