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Harlamova29_29 [7]
3 years ago
9

Please help due right now 10 POINTS

Chemistry
1 answer:
Darya [45]3 years ago
3 0

Answer:

F=ma

Explanation:

F=m×a

according to that F÷m=a and also F ÷a=m

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Which one is correct
anyanavicka [17]

Answer:

number 3.

Explanation:

4 0
3 years ago
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using the equation you wrote determine how many moles of butane c4h10 are needed to react with 5.5 moles of oxygen
GaryK [48]

Answer:

0.846 moles.

Explanation:

  • This is a stichiometric problem.
  • The balanced equation of complete combustion of butane is:

C₄H₁₀ + 6.5 O₂ → 4 CO₂ + 5 H₂O

  • It is clear from the stichiometry of the balanced equation that complete combustion of 1.0 mole of butane needs 6.5 moles of O₂ to produce 4 moles of CO₂ and 5 moles of H₂O.

<u><em>Using cross multiplication:</em></u>

  • 1.0 mole of C₄H₁₀ reacts with → 6.5 moles of O₂
  • ??? moles of C₄H₁₀ are needed to react with → 5.5 moles of O₂
  • The number of moles of C₄H₁₀ that are needed to react with 5.5 moles of O₂ = (1.0 x 5.5 moles of O₂) / (6.5 moles of O₂) = 0.846 moles.
3 0
3 years ago
C + 2H2 -&gt; CH4
ankoles [38]

Answer:

<u>C) 4</u>

Explanation:

<u>The reaction</u> :

  • C (s) + 2H₂ (g) ⇒ CH₄ (g)

       12g      4g             16g

Hence, based on this we can say that : <u>2 moles of hydrogen gas are needed to produce 16g of methane.</u>

<u />

<u>For 32g of methane</u>

  • Number of moles of H₂ = 32/16 × 2
  • Number of moles of H₂ = <u>4</u>
4 0
2 years ago
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Whose main job is to help a plant cell make proteins
hoa [83]

Answer:

Water

Explanation:

Because water is very important

7 0
3 years ago
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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>

3 0
3 years ago
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