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sladkih [1.3K]
3 years ago
14

A mixture of helium and hydrogen gases, at a total pressure of 751 mm Hg, contains 0.447 grams of helium and 0.193 grams of hydr

ogen. What is the partial pressure of each gas in the mixture
Chemistry
1 answer:
Margaret [11]3 years ago
4 0

Answer:

D0wnload Phot0Math........................

Explanation:

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The teacher in item 10 also needs to order plastic tubing. If each of the 60 students need 750 mm of tubing, what length of tubi
Alex777 [14]

45 m. If each student needs 750 mm of tubing, the teacher should order 45 m of tubing.

a) Find the <em>length in millimetres</em>

Length = 60 students x (750 mm tubing/1 student) = 45 000 mm tubing

b) Convert <em>millimetres to metres </em>

Length = 45 000 mm tubing x (1 m tubing/1000 mm tubing) = 45 m tubing

6 0
2 years ago
A little aluminum boat (mass of 14.50
Valentin [98]
When the volume is 450 cm3 so the boat will displace 450 g.

so we have  450 g - 14.5g = 435.5 left 

so to get how many pennies can be added to the boat before it sinks can be determined by:

divided the mass for the one penny:

no.of Penny = 435.5 / 2.5g

                     = 174.2 

∴ the boat will float with 174 pennies and will start to sink with 175 

5 0
3 years ago
Marcus measured the masses and volumes of samples of four different substances, and he calculated their densities. The table sho
Zina [86]

<u>Given:</u>

Calculated density values-

Aluminum = 2.7 g/cm3

Copper = 9.0 g/cm3

Iron = 7.9 g/cm3

Titanium = 4.8 g/cm3

Unknown sample mass = 9.5 g

Sample volume = 2.1 cm3

<u>To determine:</u>

The identity of the unknown sample

<u>Explanation:</u>

'Density' is a physical parameter which can be used to identify the nature of the unknown substance.

Density = Mass/Volume

For the unknown sample

Density = 9.5 g/2.1 cm3 = 4.52 g/cm3

This matches closely with the calculated density of titanium

Ans: The unknown substance is made of titanium

4 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
3 years ago
36
LiRa [457]
Omg that is that ( but I do not know this
4 0
2 years ago
Read 2 more answers
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