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FromTheMoon [43]
3 years ago
9

WILL GIVE BRAINLIEST ANSWER ASAP!!

Chemistry
1 answer:
Margarita [4]3 years ago
4 0

Morphology

Explanation:

The shape and pigment of the hairs are the morphological elements of the two hair strands.

Morphology deals with the study of the form and structure of organisms, plants and their components.

  • The shape, size, color, textures, orientation are all morphological elements that can be used to draw biological comparison between things.
  • It also entails the relationship between the structures with each other.
  • Morphology is an expression of the genotypic make up of an organism.

Learn more:

Traits     brainly.com/question/2908634

#learnwithBrainly

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A student titrates 10.00 milliliters of hydrochloric acid of unknown molarity with 1.000 m naoh. it takes 21.17 milliliters of b
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Mole ratio for the reaction is 1:1
no of moles in NaOH that reacted= 1*21.17/1000=0.02117mols
 molarity of HCl=0.02117*10/1000
                         =2.117M
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Explain why the following medicines should not be thrown down the drains
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Because medications contains chemicals, it would dissolve into the water and not only would it pollute the water, it could flow out into areas where people are exposed to these waters.
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What mass of copper sulphate dissolves in 100g of water at 70 °C​
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3 years ago
Which reactant will be used up first if 78.1g of o2 is reacted with 62.4g of c4h10?
dlinn [17]

Answer:

Reagent O₂ will be consumed first.

Explanation:

The balanced reaction between O₂ and C₄H₁₀ is:

2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O

Then, by reaction stoichiometry, the following amounts of reactants and products participate in the reaction:

  • C₄H₁₀: 2 moles
  • O₂: 13 moles
  • CO₂: 8 moles
  • H₂O: 10 moles

Being:

  • C: 12 g/mole
  • H: 1 g/mole
  • O: 16 g/mole

The molar mass of the compounds that participate in the reaction is:

  • C₄H₁₀: 4*12 g/mole + 10*1 g/mole= 58 g/mole
  • O₂: 2*16 g/mole= 32 g/mole
  • CO₂: 12 g/mole + 2*16 g/mole= 44 g/mole
  • H₂O: 2*1 g/mole + 16 g/mole= 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of reactants and products participate in the reaction:

  • C₄H₁₀: 2 moles* 58 g/mole= 116 g
  • O₂: 13 moles* 32 g/mole= 416 g
  • CO₂: 8 moles* 44 g/mole= 352 g
  • H₂O: 10 moles* 18 g/mole= 180 g

If 78.1 g of O₂ react, it is possible to apply the following rule of three: if by stoichiometry 416 g of O₂ react with 116 g of C₄H₁₀, 62.4 g of C₄H₁₀ with how much mass of O₂ do they react?

mass of O_{2} =\frac{416grams of O_{2}*62.4 grams ofC_{4}H_{10}   }{116 grams of C_{4}H_{10}}

mass of O₂= 223.78 grams

But 21.78 grams of O₂ are not available, 78.1 grams are available. Since you have less mass than you need to react with 62.4 g of C₄H₁₀, <u><em>reagent O₂ will be consumed first.</em></u>

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