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3 years ago

8

Do you think a mutation (created during reproduction) in one gene is likely to result in a new trait in an offspring? Why or why

not?

1 answer:

Mashutka [201]3 years ago

8 0

Answer:

Yes and no

Explanation:

It can be silent or neutral mutation that has no effect on the structure or functioning of the organism

OR

It can be a useful (advantageous to the organism)

or a harmful mutation (causes genetic disorders)

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Hoochie [10]

a. volume of NO : 41.785 L

b. mass of H2O : 18 g

c. volume of O2 : 9.52 L

<h3>Further explanation</h3>

Given

Reaction

4 NH₃ (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l)

Required

a. volume of NO

b. mass of H2O

c. volume of O2

Solution

Assume reactants at STP(0 C, 1 atm)

Products at 1000 C (1273 K)and 1 atm

a. mol ratio NO : O2 from equation : 4 : 5, so mo NO :

$\tt \dfrac{4}{5}\times 0.5=0.4$

volume NO at 1273 K and 1 atm

$\tt V=\dfrac{nRT}{P}=\dfrac{0.4\times 0.08206\times 1273}{1}=41.785~L$

b. 15 L NH3 at STP ( 1mol = 22.4 L)

$\tt \dfrac{15}{22.4}=0.67~mol$

mol ratio NH3 : H2O from equation : 4 : 6, so mol H2O :

$\tt \dfrac{6}{4}\times 0.67=1$

mass H2O(MW = 18 g/mol) :

$\tt mass=mol\times MW=1\times 18=18~g$

c. mol NO at 1273 K and 1 atm :

$\tt n=\dfrac{PV}{RT}=\dfrac{1\times 35.5}{0.08206\times 1273}=0.34$

mol ratio of NO : O2 = 4 : 5, so mol O2 :

$\tt \dfrac{5}{4}\times 0.34=0.425$

Volume O2 at STP :

$\tt 0.425\times 22.4=9.52~L$

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Which answer best describes the potential energy diagram for a reaction that takes place in water, if the temperature of the sur

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A 34.0 gram sample of an un-known hydrocarbon is burned in excess oxygen to form 93.5 grams of carbon dioxide and 76.5 grams of

natka813 [3]

Answer:

The answer to your question is CH₄

Explanation:

Process

1.- Write the possible chemical reaction

CxHy + O₂ ⇒ CO₂ + H₂O

2.- Calculate the amount of carbon in the reaction

Molecular mass of CO₂ = 12 + 32 = 44 g

44 g of CO₂ ------------------- 12 g of carbon

93.5 g of CO₂ ---------------- x

x = (93.5 x 12) / 44

x = 25.5 g of carbon

3.- Calculate the mass of hydrogen in the sample

Molecular mass of water = 2 + 16 = 18 g

18 g of water ------------- 2 g of hydrogen

76.5 g of water -------- x g of hydrogen

x = (76.5 x 2) / 18

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4.- Calculate the moles of carbon and hydrogen in the sample

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25.5g of carbon ---------- x

x = (25.5 x 1) / 12

x = 2.12 moles of carbon

1 g of hydrogen ----------- 1 mol

8.5 g of hydrogen -------- x

x = (8.5 x 1) / 1

x = 8.5 moles of hydrogen

5.- Divide both number of moles by the lowest number

carbon = 2.12 / 2.12 = 1

hydrogen = 2.12 = 8,5 / 2.12 = 4

6.- Write the molecular formula

CH₄

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