I need help:((( like now pls

Chemistry

2 answers:

makvit [3.9K]3 years ago

7 0

Answer: The first is Genes and the second one is the structure.

Explanation:

Korvikt [17]3 years ago

6 0

Answer:

I think #5 is its structure

Explanation:

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2. (Exercise 4.1.6) A liquid adhesive consists of a polymer dissolved in a solvent. The amount of polymer in the solution is imp

Lilit [14]

Answer:

800 lb of pure solvent , 1700 lb of 20% solution and 500 lb of 10% solution will be mixed to form 3000 lb of 13 % solution .

Explanation:

3000 lb of 13% solution is required .

Total adhesive in weight = 3000 x .13 = 390 lb of adhesive

Available = 500 lb of 10% solution = 50 lb of adhesive

Rest = 390 - 50 = 340 lb required .

rest mass of solution = 3000 - 500 = 2500 lb

mass of adhesive required = 340 lb

Let the mass of 20% required be V

mass of adhesive = .20 V

.20 V = 340

V = 1700

rest of the volume = 2500 - 1700 = 800 lb which will be of pure solvent

So 800 lb of pure solvent , 1700 lb of 20% solution and 500 lb of 10% solution will be mixed to form 3000 lb of 13 % solution .

6 0

4 years ago

The acetylene tank contains 35.0 mol C2H2, and the oxygen tank contains 84.0 mol O2.

harkovskaia [24]

Answer:- As per the question is asked, 35.0 moles of acetylene gives 70 moles of carbon dioxide but if we solve the problem using the limiting reactant which is oxygen then 67.2 moles of carbon dioxide will form.

Solution:- The balanced equation for the combustion of acetylene is:

$2C_2H_2(g)+5O_2(g)\rightarrow 4CO_2(g)+2H_2O(g)$

From the balanced equation, two moles of acetylene gives four moles of carbon dioxide. Using dimensional analysis we could show the calculations for the formation of carbon dioxide by the combustion of 35.0 moles of acetylene.

$35.0molC_2H_2(\frac{4molCO_2}{2molC_2H_2})$

= $70molCO_2$

The next part is, how we choose 35.0 moles of acetylene and not 84.0 moles of oxygen.

From balanced equation, there is 2:5 mol ratio between acetylene and oxygen. Let's calculate the moles of oxygen required to react completely with 35.0 moles of acetylene.

$35.0molC_2H_2(\frac{5molO_2}{2molC_2H_2})$

= $87.5molO_2$

Calculations shows that 87.5 moles of oxygen are required to react completely with 35.0 moles of acetylene. Since only 84.0 moles of oxygen are available, the limiting reactant is oxygen, so 35.0 moles of acetylene will not react completely as it is excess reactant.

So, the theoretical yield should be calculated using 84.0 moles of oxygen as:

$84.0molO_2(\frac{4molO_2}{5molO_2})$