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denpristay [2]
3 years ago
5

Which of the following is not an example of a polymer? (2 points)

Chemistry
1 answer:
ahrayia [7]3 years ago
3 0
Answer is "sucrose".

Polymer is a large molecule which is made from repeating units. The smallest repeating unit is called as monomer. <span>Polystyrene, nylon and PVC are examples for polymers. But sucrose is a disaccharide which is made from glucose and fructose. Hence, sucrose is not an example of polymer. </span>
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Why are cells the building blocks of an organism.​
ladessa [460]

Answer:A cell is the smallest unit of life, also called the 'building blocks of life' because cells multiply and differentiate to form a multicellular organism as well as give rise to new organism by forming gametes or reproductive spores.

Explanation:

8 0
3 years ago
Under standard-state conditions, which of the following species is the best reducing agent? a. Ag+ b. Pb c. H2 d. Ag e. Mg2+
eimsori [14]

<u>Answer:</u> The correct answer is Option b.

<u>Explanation:</u>

Reducing agents are defined as the agents which help the other substance to get reduced and itself gets oxidized. They undergo oxidation reaction.

X\rightarrow X^{n+}+ne^-

For determination of reducing agents, we will look at the oxidation potentials of the substance. Oxidation potentials can be determined by reversing the standard reduction potentials.

For the given options:

  • <u>Option a:</u>  Ag^+

This ion cannot be further oxidized because +1 is the most stable oxidation state of silver.

  • <u>Option b:</u>  Pb

This metal can easily get oxidized to Pb^{2+} ion and the standard oxidation potential for this is 0.13 V

Pb\rightarrow Pb^{2+}+2e^-;E^o_{(Pb/Pb^{2+})}=+0.13V

  • <u>Option c:</u>  H_2

This metal can easily get oxidized to H^{+} ion and the standard oxidation potential for this is 0.0 V

H_2\rightarrow 2H^++2e^-;E^o_{(H_2/H^{+})}=0.0V

  • <u>Option d:</u>  Ag

This metal can easily get oxidized to Ag^{+} ion and the standard oxidation potential for this is -0.80 V

Ag\rightarrow Ag^{+}+e^-;E^o_{(Ag/Ag^{+})}=-0.80V

  • <u>Option e:</u>  Mg^{2+}

This ion cannot be further oxidized because +2 is the most stable oxidation state of magnesium.

By looking at the standard oxidation potential of the substances, the substance having highest positive E^o potential will always get oxidized and will undergo oxidation reaction. Thus, considered as strong reducing agent.

From the above values, the correct answer is Option b.

8 0
3 years ago
(ill give brainliest)
gizmo_the_mogwai [7]

Answer:

It is because the object that is charged attracts the uncharged because it is giving electrons and passing it on to the object. As shown in the image the balloon is attracting the paper that is not charged. Actually telling,the charged object will cause the uncharged object to become charged, with a positive charge on one side and a negative charge on the other side. This process is called induction.

Explanation: Pls Mark Brainliest

4 0
2 years ago
Which element above has 8 protons in its nucleus?
Nataliya [291]
A. Oxygen. Oxygen has an atomic mass of 16. The atomic mass of an atom is the combined weight o the protons and neutrons. Since Oxygen's atomic mass is 16, it has 8 protons and 8 neutrons.
6 0
3 years ago
Read 2 more answers
S8 + 24 F2 ⟶ 8 SF6
Arturiano [62]

Answer:

Theoretical Yield of SF₆ = 2.01 moles

Explanation: If you understand and can apply the methodology below, you will find it applies to ALL chemical reaction stoichiometry problems based on the balanced standard equation; i.e., balanced to smallest whole number coefficients.

Solution 1:

Rule => Convert given mass values to moles, solve problem using coefficient ratios. Finish by converting moles to the objective dimensions.

Given      S₈            +          24F₂            =>    8SF₆

             425g                    229g                      ?

= 425g/256g/mol.      = 226g/38g/mol.

= 1.66 moles S₈          = 6.03 moles F₂ <= Limiting Reactant

<em>Determining Limiting Reactant => Divide moles each reactant by their respective coefficient; the smaller value will always be the limiting reactant. </em>

S₈ = 1.66/1 = 1.66

F₂ = 6.03/24 = 0.25 => F₂ is the limiting reactant

<em>Determining Theoretical Yield:</em>

Note: When working problem do not use the division ratio results for determining limiting reactant. Use the moles F₂ calculated from 229 grams F₂ => 6.03 moles F₂. The division procedure to define the smaller value and limiting reactant is just a quick way to find which reactant controls the extent of reaction.  

Given      S₈            +          24F₂            =>    8SF₆

             425g                    229g                      ?

   = 425g/256g/mol. = 226g/38g/mol.

= 1.66 moles S₈          = 6.03 moles F₂ <= Limiting Reactant

<em>Max #moles SF₆ produced from 6.03 moles F₂ and an excess S₈ </em>

Since coefficient values represent moles, the reaction ratio for the above reaction is 24 moles F₂ to 8 moles SF₆. Such implies that the moles of SF₆ (theoretical) calculated from 6.03 moles of F₂ must be a number less than the 6.03 moles F₂ given. This can be calculated by using a ratio of equation coefficients between 24F₂ and 8SF₆  to make the outcome smaller than 6.03. That is,

moles SF₆ = 8/24 x 6.03 moles = 2.01 moles SF₆ (=> theoretical yield)  

S₈ + 24F₂ => 8SF₆

moles SF₆ = 8/24(6.03) moles = 2.01 moles

You would NOT want to use 24/8(6.03) = 18.1 moles which is a value >> 6.03.        

This analysis works for all reaction stoichiometry problems.

Convert to moles => divide by coefficients for LR => solve by mole mole ratios from balanced reaction and moles of given.    

____________________

Here's another example just for grins ...

             C₂H₆O   +   3O₂     =>     2CO₂    + 3H₂O

Given:    253g          307g               ?               ?

a. Determine Limiting Reactant

b. Determine mass in grams of CO₂ & H₂O produced        

Limiting Reactant

moles  C₂H₆O = 253g/46g/mol = 5.5 moles  => 5.5/1 = 5.5

moles  O₂ = 307g/32g/mol = 9.6 moles         =><em>  9.6/24 = 0.4 ∴ O₂ is L.R.</em>

But the problem is worked using the mole values; NOT the number results used to ID the limiting reactant.  

 C₂H₆O   +       3O₂          =>     2CO₂    + 3H₂O

------------ 9.6 mole (L.R.)              ?               ?

mole yield CO₂ = 2/3(9.6)mole = 6.4 mole  (CO₂ coefficient < O₂ coefficient)

mole yield H₂O = 9.6mole  = 9.6mole (coefficients O₂ & CO₂ are same.)

mole used C₂H₆O = 1/3(9.6)mole = 3.2 mole (coefficient  C₂H₆O < coefficient O₂)

For grams => moles x formula weight (g/mole)

7 0
3 years ago
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