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bulgar [2K]
3 years ago
9

Even if two cars perform the same amount of work, one car can still have more power.

Chemistry
1 answer:
nikdorinn [45]3 years ago
4 0
True because one car can have more power and just be able to to the easy stuff and the hard stuff
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What genotype will appear in box 4?
Pani-rosa [81]
The genotype will be (tt)
Hope this helps!
7 0
3 years ago
Read 2 more answers
Suppose that 0.410 mol of methane, CH4(g), is reacted with 0.560 mol of fluorine, F2(g), forming CF4(g) and HF(g) as sole produc
Ber [7]

Answer:

The balanced equation for this reaction will be

                            CH4 + 4F2    →  CF4 + 4HF

We can see that 1 mole of methane requires 4 moles of fluorine but we have 0.41 moles of CH4 and 0.56mole of F2

So using the unitary method we will get that

  • 1 mole of CH4 → 4 mole of 4 mole of fluorine
  • 0.41 mole of methane  →  4*0.41 = 1.64 mole of fluorine for complete reaction

but we have only 0.56 mole of fluorine that means fluorine is the limiting reagent and the product will only be formed by only this amount of fluorine.

  • 4 moles of  fluorine →  1 mole of CF4
  • 0.56 mole →  \frac{1}{4} * 0.56 = 0.14mole of CF4
  • 4 moles of fluorine →  4 moles of HF
  • 0.56 mole of fluorine →  0.56 mole of HF

now to find the heat released we have the formula as

DELTA H = n * Delta H of product - n *delta H of reactant

where n is the moles of the reactant and product.

note: since no information is given about the enthalpies of the species we leave it on general equation also you need to add the product side enthalpy of the species present and similarly on the product side.

8 0
3 years ago
Does anyone know the answer?
valentina_108 [34]

An artery carries blood away from the heart, a vein carries blood to the heart.

3 0
3 years ago
Solid aluminum and gaseous oxygen read in a combination reaction to produce aluminum oxide. 4Al(s) + 3O_2(g) rightarrow 2Al_2O_3
Kryger [21]

Answer:

4.7 g. Option 5 is the right one.

Explanation:

4Al(s) + 3O₂ (g) ⇄ 2Al₂O₃ (s)

We convert the mass of reactants to moles, in order to determine the limiting.

2.5 g Al / 26.98 g/mol → 0.092 moles of Al

2.5 g O₂ / 32g/mol → 0.078 moles of O₂

Ratio is 4:3. 4 moles of Al react with 3 moles of O₂

Then, 0.092 moles of O₂ would react with (0.092 . 3)/ 4 = 0.069 moles O₂

We have 0.078 moles of O₂ and we need 0.069 moles, the oxygen is the limiting in excess. Therefore the Al is the limiting reactant.

Ratio is 4:2. 4 moles of Al, can produce 2 moles of Al₂O₃

Then, 0.092 moles of Al would produce (0.092 .2) / 4 = 0.046 moles

If we convert the moles to mass, we find the anwer:

0.046 mol . 101.96 g/mol = 4.69 g

5 0
3 years ago
A 0.580 g sample of a compound containing only carbon and hydrogen contains 0.480 g of carbon and 0.100 g of hydrogen. At STP, 3
Sati [7]

Answer:

Molecular formula for the gas is: C₄H₁₀

Explanation:

Let's propose the Ideal Gases Law to determine the moles of gas, that contains 0.087 g

At STP → 1 atm and 273.15K

1 atm . 0.0336 L = n . 0.082 . 273.15 K

n = (1 atm . 0.0336 L) / (0.082 . 273.15 K)

n = 1.500 × 10⁻³ moles

Molar mass of gas = 0.087 g / 1.500 × 10⁻³ moles = 58 g/m

Now we propose rules of three:

If 0.580 g of gas has ____ 0.480 g of C _____ 0.100 g of C

58 g of gas (1mol) would have:

(58 g . 0.480) / 0.580 = 48 g of C  

(58 g . 0.100) / 0.580 = 10 g of H

 48 g of C / 12 g/mol = 4 mol

 10 g of H / 1g/mol = 10 moles

7 0
3 years ago
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