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enot [183]
3 years ago
10

How do we know the car has moved

Chemistry
1 answer:
Anuta_ua [19.1K]3 years ago
8 0

Answer:

We need more information to solve this question. Take a picture of the problem and post it and I will try to help.

Explanation:

You might be interested in
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
What is the hydronium ion concentration of a solution whose pH is 4.12?
Degger [83]
A is may be the correct answer
3 0
3 years ago
PLEASE HURRY!!!
bekas [8.4K]

Answer:

high ph -basic,low pH -acidic

Explanation:

hi I'm from India nice to meet you

5 0
3 years ago
(LO 3N, 4G, 4O) Aluminum sulfate is also involved in dying fabrics. The gelatinous precipitate formed in the reaction with dilut
brilliants [131]

9.4 × 10⁻³ mg (0.73 mmoles) of Al(OH)₃ is formed

Explanation:

We have the following chemical reaction:

Al₂(SO₄)₃(aq) + 6 NaOH(aq) → 2 Al(OH)₃(s) + 3 Na₂SO₄(aq)

The precipitate mentioned by the problem is aluminium hydroxide Al(OH)₃.

Now to determine the number of moles of sodium hydroxide NaOH we use the following formula:

molar concentration =  number of moles / volume

number of moles = molar concentration × volume

number of moles of NaOH = 0.088 M × 25 mL = 2.2 mmoles

number of moles of Al₂(SO₄)₃ = 5.6 × 10⁻³ moles = 5.6 mmoles (found in the  problem text)

We see from the chemical reaction that 1 mole of Al₂(SO₄)₃ requires 6 moles of NaOH so 5.6 mmoles of Al₂(SO₄)₃ would require 6 times more NaOH which is 33.6 mmoles and we have only 2.2 mmoles. The limiting reactant will be NaOH.

Now we devise the following reasoning:

if        6 mmoles of NaOH produces 2 mmoles of Al(OH)₃

then  2.2 mmoles of NaOH produces X mmoles of Al(OH)₃

X = (2.2 × 2) / 6 = 0.73 mmoles of Al(OH)₃

mass of Al(OH)₃ = number of moles / molecular weight

mass of Al(OH)₃ = 0.73 / 78

mass of Al(OH)₃ =  9.4 × 10⁻³ mg

Learn more:

precipitation reaction

brainly.com/question/10400269

7 0
3 years ago
A 3.50 g sample of an unknown compound containing only C , H , and O combusts in an oxygen‑rich environment. When the products h
statuscvo [17]

Explanation:

First, calculate the moles of CO_{2} using ideal gas equation as follows.

                PV = nRT

or,          n = \frac{PV}{RT}

                = \frac{1 atm \times 4.41 ml}{0.0821 Latm/mol K \times 293 K}      (as 1 bar = 1 atm (approx))

                = 0.183 mol

As,   Density = \frac{mass}{volume}

Hence, mass of water will be as follows.

                Density = \frac{mass}{volume}

             0.998 g/ml = \frac{mass}{3.26 ml}    

                 mass = 3.25 g

Similarly, calculate the moles of water as follows.

        No. of moles = \frac{mass}{\text{molar mass}}

                              =  \frac{3.25 g}{18.02 g/mol}            

                              = 0.180 mol

Moles of hydrogen = 0.180 \times 2 = 0.36 mol

Now, mass of carbon will be as follows.

       No. of moles = \frac{mass}{\text{molar mass}}

          0.183 mol =  \frac{mass}{12 g/mol}            

                              = 2.19 g

Therefore, mass of oxygen will be as follows.

              Mass of O = mass of sample - (mass of C + mass of H)

                                = 3.50 g - (2.19 g + 0.36 g)

                                = 0.95 g

Therefore, moles of oxygen will be as follows.

          No. of moles = \frac{mass}{\text{molar mass}}

                               =  \frac{0.95 g}{16 g/mol}            

                              = 0.059 mol

Now, diving number of moles of each element of the compound by smallest no. of moles as follows.

                         C              H           O

No. of moles:  0.183        0.36       0.059

On dividing:      3.1           6.1            1

Therefore, empirical formula of the given compound is C_{3}H_{6}O.

Thus, we can conclude that empirical formula of the given compound is C_{3}H_{6}O.            

6 0
3 years ago
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