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ziro4ka [17]
3 years ago
10

Chlorine dioxide is used as a disinfectant and bleaching agent. In water, it reacts to form chloric acid (HClO3),

Chemistry
1 answer:
Sliva [168]3 years ago
5 0
All done no worries

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. A 5.00 – g sample of lead is bluish-white in color and is very soft and malleable. It has a melting point of 327.4°C and a boi
Alexus [3.1K]

Answer:

The mass of the lead

Explanation:

The 5g mass of the lead is an extensive property.

An extensive property is a physical property of matter which depends on the amount of matter that is present there in. Mass, volume e.t.c are all extensive properties. The more the quantity of the lead, the more its mass and the volume it occupies.

Melting point, boiling point, density are all intensive propeties. These properties do not rely on the amount of matter present. Any amount of lead will have the same density.

5 0
3 years ago
In a thin layer chromatography experiment, a plate of length 9.3 cm was used and a horizontal line was made at 1.45 cm above the
dezoksy [38]

Answer: The R_f value is 0.664

Explanation:

Distance travelled by solvent front = (7.7-1.45)cm = 6.25 cm

Distance travelled by unknown = (5.6-1.45) cm = 4.15 cm

 The retention factor or the R_f value is defined as the ratio of distance traveled by the unknown to the distance traveled by the solvent front.

R_f=\frac{\text {distance travelled by unknown}}{\text {distance travelled by solvent}}

R_f=\frac{4.15}{6.25}=0.664

Thus the R_f value is 0.664

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What is the limiting reagent when a 2.00 g sample of ammonia is mixed with 4.00 g of oxygen?​
UNO [17]

Answer:

Ammonia is limiting reactant

Amount of oxygen left  = 0.035 mol

Explanation:

Masa of ammonia = 2.00 g

Mass of oxygen = 4.00 g

Which is limiting reactant = ?

Balance chemical equation:

4NH₃ + 3O₂     →     2N₂ + 6H₂O

Number of moles of ammonia:

Number of moles = mass/molar mass

Number of moles = 2.00 g/ 17 g/mol

Number of moles = 0.12 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 4.00 g/ 32 g/mol

Number of moles = 0.125 mol

Now we will compare the moles of ammonia and oxygen with water and nitrogen.

                      NH₃          :            N₂

                        4             :             2

                      0.12           :           2/4×0.12 = 0.06

                      NH₃         :            H₂O

                        4            :             6

                        0.12       :           6/4×0.12 = 0.18

                       

                       O₂            :            N₂

                        3             :             2

                      0.125        :           2/3×0.125 = 0.08

                        O₂           :            H₂O

                        3              :             6

                        0.125       :           6/3×0.125 = 0.25

The number of moles of water and nitrogen formed by ammonia are less thus ammonia will be limiting reactant.

Amount of oxygen left:

                        NH₃          :             O₂

                           4            :              3

                           0.12       :          3/4×0.12= 0.09

Amount of oxygen react = 0.09 mol

Amount of oxygen left  = 0.125 - 0.09 = 0.035 mol

3 0
3 years ago
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