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2 years ago

1) The hydrocarbon C15 H32 burns to form carbon dioxide and water. Write the equation for the reaction.

Chemistry

1 answer:

Dahasolnce [82]2 years ago

8 0

Explanation:

Ch4+heat=CO2.

Ans

It is not balanced equation if you want it say in comment

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4/5(2x 5)−4=1

uysha [10]

5/8=0.625 4/5=.8

5/8 because 5/8=0.625 and because its 2x you multiply 0.625 by 2=1.25 then you do 1.25+5=6.25 4/5=.8 6.25 times .8=5 5-4=1

X=5/8 or 0.625

5 0

3 years ago

Calculate the wavelength, in nanometers, of the light emitted by a hydrogen atom when its electron falls from the n = 7 to the n

Gre4nikov [31]

Answer:

The wavelength of the light emitted by a hydrogen atom for the given transition is 2166 nm.

Explanation:

The energy of nth energy levels of the H atom is given as:

$E_n = -2.18 \times 10^{-18} \times \frac{1}{n^2} J$

Energy of the seventh energy level = $E_7$

$E_7=-2.18 \times 10^{-18} \times \frac{1}{7^2} J$

$E_7=-2.18 \times 10^{-18} \times \frac{1}{7^2} J=-4.4490\times 10^{-20} J$

Energy of the seventh energy level = $E_4$

$E_4=-2.18 \times 10^{-18} \times \frac{1}{4^2} J$

$E_4=-2.18 \times 10^{-18} \times \frac{1}{16} J=-1.3625\times 10^{-19} J$

Energy of the light emitted will be equal to the energy difference of the both levels.

$E=E_7-E_4=-4.4490\times 10^{-20} J-(-1.3625\times 10^{-19} J)$

$E=9.176\times 10^{-20} J$

Wavelength corresponding to energy E can be calculated by using Planck's equation:

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

$\lambda =\frac{hc}{E}=\frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{9.176\times 10^{-20} J}=2.166\times 10^{-6} m=2166 nm$

The wavelength of the light emitted by a hydrogen atom for the given transition is 2166 nm.

8 0

2 years ago

Which type of electromagnetic wave has the greatest wavelength?(1 point)

Kruka [31]

The answer is A ……..

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2 years ago

25.0 mL of a hydrofluoric acid solution of unknown concentration is titrated with 0.200 M NaOH. After 20.0 mL of the base soluti

lesantik [10]

Answer:

[HF]₀ = 0.125M

Explanation:

NaOH + HF => NaF + H₂O

Adding 20ml of 0.200M NaOH into 25ml of HF solution neutralizes 0.004 mole of HF leaving 0.004 mole NaF in 0.045L with 0.001M H⁺ at pH = 3. This is 0.089M NaF and 0.001M HF remaining.

=> 45ml of solution with pH = 3 and contains 0.089M NaF from titration becomes a common ion problem.

HF ⇄ H⁺ + F⁻

C(eq) [HF] 10⁻³M 0.089M (<= soln after adding 20ml 0.200M NaOH)

Ka = [H⁺][F⁻]/[HF]₀ => [HF]₀ = [H⁺][F⁻]/Ka

[HF]₀ = (0.001)(0.089)/(7.1 x 10⁻⁴) M = 0.125M

6 0

2 years ago

4. Traditional bomb calorimetry can be used to find the energy content of food. Why is it

VladimirAG [237]

Answer:

B. it accounts for all the energy in the good even if some of its largely excreted by th body

3 0

2 years ago