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Tresset [83]
1 year ago
6

Franklin was performing an experiment by combining hydrochloric acid and sodium hydroxide. He measured the mass of his reactant

materials to be 35g. The reaction resulted in the production of salt and water. He reported that his products weighed 32g. Which best describes the results of Franklin’s experiment? No error occurred, some of the products are always lost as heat. An error occurred, the mass of the reactants should equal the mass of the products. An error occurred, the products should weigh more than the reactants. No error occurred, water is not weighed when determining the weight of the products.
Chemistry
2 answers:
ArbitrLikvidat [17]1 year ago
8 0

Answer:

B

Explanation:

VARVARA [1.3K]1 year ago
4 0

Answer:

B

Explanation:

took the test

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Draw a sodium formate molecule. The structure has been supplied here for you to copy. To add formal charges, click the button be
Karo-lina-s [1.5K]

The Molecule of Sodium Formate along with Formal Charges (in blue) and lone pair electrons (in red) is attached below.

Sodium Formate is an ionic compound made up of a positive part (Sodium Ion) and a polyatomic anion (Formate).

Nomenclature:

                       In ionic compounds the positive part is named first. As sodium ion is the positive part hence, it is named first followed by the negative part i.e. formate.

Name of Formate:

                             Formate ion has been derived from formic acid ( the simplest carboxylic acid). When carboxylic acids looses the acidic proton of -COOH, they are converted into Carboxylate ions.

E.g.

                    HCOOH (formic acid)    →     HCOO⁻ (formate)  +  H⁺

                H₃CCOOH (acetic acid)     →      H₃CCOO⁻ (acetate)  +  H⁺

Formal Charges:

                           Formal charges are calculated using following formula,

          F.C  =  [# of Valence e⁻] - [e⁻ in lone pairs + 1/2 # of bonding electrons]

For Oxygen:

                    F.C  =  [6] - [6 + 2/2]

                    F.C  =  [6] - [6 + 1]

                    F.C  =  6 - 7

                    F.C  =  -1

For Sodium:

                    F.C  =  [1] - [0 + 0/2]

                    F.C  =  [1] - [0]

                    F.C  =  1 - 0

                    F.C  =  +1

5 0
2 years ago
How could installing new technology, such as scrubber machines, affect the factories required to install them? Name a positive a
Sindrei [870]

Answer:

Installing new technology, such as scrubbers, in factories will decrease their harmful emissions. This helps improve the safety of the surrounding community and the workers. But this technology is expensive and requires time and effort to install.

Explanation:

Hope this helped :)

5 0
2 years ago
For each pair below, select the sample that contains the largest number of moles. Pair A 2.50 g O2 2.50 g N2
raketka [301]

Answer:

Explanation:

Pair  2.50g of O₂ and 2.50g of  N₂

The atoms sample with the largest number of moles since the masses are the same would be the one with lowest molar mass according the the equation below:

Number of moles = \frac{mass }{molarmass}

Atomic mass of O = 16g and N = 14g

Molar mass of O₂ = 16 x 2 = 32gmol⁻¹

Molar mass of N₂ = 14 x 2 = 28gmol⁻¹

Number of moles of O₂ = \frac{2.5}{32} = 0.078mole

Number of moles of N₂ = \frac{2.5}{28} =  0.089mole

We see that N₂ has the largest number of moles

4 0
2 years ago
Read 2 more answers
Which gas tank will empty first? acetylene (C2H2) oxygen (O2)
Monica [59]
The answer is oxygen. (02)
4 0
2 years ago
Read 2 more answers
Air in a 0.3 m3 cylinder is initially at a pressure of 10 bar and a temperature of 330K. The cylinder is to be emptied by openin
sergiy2304 [10]

Answer:

(a) Temperature = 330 K, and mass = 0.321 kg

(b) T₂ = 171.56 K, mass = 0.32223 kg

Explanation:

For a constant temperature process we have

p₁v₁ = p₂v₂

Where p₁ = initial pressure = 10 bar = 1000000 Pa

p₂ = final pressure = 1 atm = 101325 Pa

v₁ initial volume = 0.3 m³

v₂ = final volume = unknown

From the relation we have v₂ = 2.96 m³

Therefore at constant temperature 2.93 m³ - 0.3 m³ or 2.66 m³ will be expelled from the container

Temperature = 330 K, and mass =

Also from the relation p1v1 = mRT1

We have, (1000000×0.3)/(8314×330) = 109..337 mole

For air mass

Mass = 3.171 kg

After opening we have

p2v2/(RT1) = n2 = 11.07 mol or 0.321 kg

or

(b) This is said to be adiabatic condition hence

Here

But cp = 29 (J/mol K).

and p₁v₁ = RT₁ therefore R = 1000000*0.3/330 = 909.1 J/mol·K

And For perfect gas γ = 1.4

Hence T₂ = 171.56 K

γ =cp/cv therefore cv=cp/γ = 29/1.4 = 20.714 (J/mol K). and R =cp-cv = 8.29 J/mol·K

Therefore p1v1/(RT1) = 109.66 moles and we have

p2v2/(R×T2) = 11.11 mole left

For air that is 0.32223 kg

5 0
2 years ago
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