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1 year ago

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To win in a qualifying round in a duck race, a duck's rate of change needs to be 35.4 meters/second. Kate was training her duck

at a rate of change of 33.9 meters/second. What was Kate's percent error? Does she need to train her duck at a higher or lower rate of change?

1 answer:

Setler [38]1 year ago

3 0

Answer:

i believe that 14 at a higher rate is the answer

Explanation:

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Based on the bond energies for the reaction below, what is the enthalpy of the reaction?HC≡CH (g) + 5/2 O₂ (g) → 2 CO₂ (g) + H₂O

Anuta_ua [19.1K]

Answer:

1219.5 kj/mol

Explanation:

To reach this result, you must use the formula:

ΔHºrxn = Σn * (BE reactant) - Σn * (BE product)

ΔHºrxn = [1 * (BE C = C) + 2 * (BE C-H) + 5/2 * (BE O = O)] - [4 * (BE C = O) + 2 * (BE O-H).

The BE values are:

BE C = C: 839 kj / mol

BE C-H: 413 Kj / mol

BE O = O: 495 kj / mol

BE C = O = 799 Kj / mol

BE O-H = 463 kj / mol

Now you must replace the values in the above equation, the result of which will be:

ΔHºrxn = [1 * 839 + 2 * (413) + 5/2 * (495)] - [4 * (799) + 2 * (463) = 1219.5 kj/mol

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1 year ago

A solid metal oxide crystallizes in a cubic unit cell. In the unit cell there are metal (M) ions on every corner and in the cent

kotegsom [21]

Answer:

The empirical formula of the solid metal oxide is : $MO$

Explanation:

M atom is present in the every corner and in the center of every face.

Number of m atoms :

$\frac{1}{8}\times 8+\frac{1}{2}\times 6=4$

Number of tetrahedral voids in F.C.C = 2n = 2 × 4 = 8

Oxide ion is present in the half of the tetrahedral void

Number oxide ions = $\frac{8}{2}=4$

The molecular formula of the solid metal oxide is : $M_4O_4=MO$

The empirical formula represent the lowest number of atoms present in a compound.

The empirical formula of the solid metal oxide is : $MO$

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1 year ago

The initial temperature of the water in a constant-pressure calorimeter is 24°C. A reaction takes place in the calorimeter, and

kupik [55]

Answer:

Explanation:

For a chemical reaction, the enthalpy of reaction (ΔHrxn) is … ... to increase the temperature of 1 g of a substance by 1°C; its units are thus J/(g•°C). ... Both Equations 12.3.7 and 12.3.8 are under constant pressure (which ... The specific heat of water is 4.184 J/g °C (Table 12.3.1), so to heat 1 g of water by 1 ..

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2 years ago

Read 2 more answers

Compare and contrast the way waves transmit energy and the way particles transmit energy

Maksim231197 [3]

<span>Waves transfer energy through vibration. just like electromagnetic waves, energy is transferred through vibrations of electric and magnetic fields. In sound waves, energy is transferred through vibration of air particles or particles of a solid through which the sound travels. In water waves, energy is transferred through the vibration of the water particles. While particles transfer energy through conduction and convection.</span>

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2 years ago

One of the emission spectral lines for Be31 has a wavelength of 253.4 nm for an electronic transition that begins in the state w

max2010maxim [7]

Explanation:

The given data is as follows.

$\lambda$ = 253.4 nm = $253.4 \times 10^{-9}m$ (as 1 nm = $10^{-9}$ )

$n_{1}$ = 5, $n_{2}$ = ?

Relation between energy and wavelength is as follows.

E = $\frac{hc}{\lambda}$

= $\frac{6.626 \times 10^{-34} Js \times 3 \times 10^{8} m/s}{253.4 \times 10^{-9}}$

= $0.0784 \times 10^{-17}$ J

= $7.84 \times 10^{-19} J$

Hence, energy released is $7.84 \times 10^{-19} J$ .

Also, we known that change in energy will be as follows.

$\Delta E = -2.178 \times (Z)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{n^{2}_{1}}$

where, Z = atomic number of the given element

$7.84 \times 10^{-19} J = -2.178 \times (4)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{(5)^{2}}$

$\frac{7.84 \times 10^{-19} J}{34.848} = \frac{1}{n^{2}_{2}} - \frac{1}{(5)^{2}}$

0.02 + 0.04 = $\frac{1}{n^{2}_{1}}$

$n_{1}$ = $\sqrt{\frac{1}{0.06}}$

= 4

Thus, we can conclude that the principal quantum number of the lower-energy state corresponding to this emission is n = 4.

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1 year ago