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omeli [17]
2 years ago
15

A container filled with gas is connected to an open-end manometer that is filled with mineral oil. the pressure in the gas conta

iner is 753 mm hg and atmospheric pressure is 724 mm. how high will the level rise in the manometer if the densities of hg and mineral oil are 13.6 g/ml and 0.822 g/ml respectively?
Chemistry
1 answer:
Vesnalui [34]2 years ago
3 0
<span>The difference in pressure is equal to the pressure exerted by the displaced fluid. Since the Pressure difference is equal, density of fluid is inversely proportional to the height of the column of displaced fluid => (Ď1)(h1) = (Ď2)(h2) (13.6 g/ml)(753 - 724 mm Hg) = (0.822 g/ml)(h2) h2 = (13.6 g/ml)(753 - 724 mm Hg) / (0.822 g/ml) h2 = 480 mm</span>
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This information is taken directly from Exercise 3 in the CHEM 111/112 Laboratory Manual. Another example of excellence in the p
kolbaska11 [484]

Answer:

<h2>No</h2>

the information was not cited correctly....

Explanation:

I hope the following explanation will help you a lot.

5 0
2 years ago
If the mass percentage composition of a compound is 72.1% Mn and 27.9% O, its empirical formula is
mojhsa [17]

Answer:

MnO- Manganese Oxide

Explanation:

Empirical formula: This is the formula that shows the ratio of elements

present in a  

compound.

   

How to determine Empirical formula

1. First arrange the symbols of the elements present in the compound

alphabetically to  determine the real empirical formula. Although, there

are exceptions to this rule, E.g H2So4

2. Divide the percentage composition by the mass number.

3. Then divide through by the smallest number.

4. The resulting answer is the ratio attached to the elements present in

a compound.

           

                                                                              Mn                         O    

                         

% composition                                                      72.1                      27.9    

                       

Divide by mass number                                       54.94                     16  

                                 

                                                                               1.31                      1.74    

                       

Divide by the smallest number                         1.31                      1.31                          

                                                                               1                    1.3

                                                 

The resulting ratio is 1:1

 

Hence the Empirical formula is MnO, Manganese oxide

8 0
2 years ago
. Calculate the mass of O2 produced if 3.450 g potassium chlorate is completely decomposed by heating in presence of a catalyst
Vesnalui [34]
 <span>2 KClO3(s) → 3 O2(g) + 2 KCl(s) 

</span><span>Note: MnO2 (Manganese Dioxide) is not part of the reaction. A catalyst lowers the activation energy and increases both forward and reverse reactions at equal rates. 
</span>
molar mass of KClO3 = 122.5
Moles of KClO3 =  3.45 / 122.55 = 0.028

Moles of O2 produce = \frac{3}{2} \times 0.028

= 0.042 moles

molar mass of O2 = 32

so, mass of O2 = 32 x 0.042  = 1.35 g



5 0
2 years ago
What is the mass in grams of 7.5 x 10^15 atoms of nickel?
satela [25.4K]
Atomic mass Ni = 58.69 a.m.u

58.69 g ----------------- 6.02x10²³ atoms
?? g --------------------- 7.5x10¹⁵ atoms

58.69x (7.5x10¹⁵) / 6.02x10²³

=> 7.31x10⁻⁷ g
6 0
2 years ago
Read 2 more answers
How many grams o an ointment base must be added to 45 g o clobetasol (T EMOVAT E) ointment, 0.05% w/w, to change its strength to
saw5 [17]

Answer:

Drug calculation

If we have 45g of clobetasol  = 0.05%w/w

Then what mass in g of clobetasol is in 0.03%w/w = 45 x 0.03/0.05 =27g

It means that 27g of clobetasol must be added to change the drug strength to 0.03% w/w

3 0
2 years ago
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