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faust18 [17]
3 years ago
15

Which of the following molecules has a double bond? A. butane B. 2-butyne C. cyclopentane D. 1-methylcyclohexene

Chemistry
1 answer:
Tanzania [10]3 years ago
3 0
D. would be my guess im almost positive that its right
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Calculate the period when the frequency is 50Hz.​
liraira [26]

Answer:

For example, a wave with a time period of 2 seconds has a frequency of 1 ÷ 2 = 0.5 Hz. A radio wave has a time period of 0.0000003333333 seconds.

8 0
3 years ago
What is the energy of a photon that emits a light of frequency 6.42 x 1014 Hz?<br><br>​
svetlana [45]

Answer:

Option B. 4.25×10¯¹⁹ J

Explanation:

From the question given above, the following data were obtained:

Frequency (f) = 6.42×10¹⁴ Hz

Energy (E) =?

Energy and frequency are related by the following equation:

Energy (E) = Planck's constant (h) × frequency (f)

E = hf

With the above formula, we can obtain the energy of the photon as follow:

Frequency (f) = 6.42×10¹⁴ Hz

Planck's constant (h) = 6.63×10¯³⁴ Js

Energy (E) =?

E = hf

E = 6.63×10¯³⁴ × 6.42×10¹⁴

E = 4.25×10¯¹⁹ J

Thus, the energy of the photon is 4.25×10¯¹⁹ J

3 0
3 years ago
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Question 7: *
Otrada [13]

Answer:

2

Explanation:

Thermal, chemical and electromagnetic is the right answer

6 0
3 years ago
Acidic<br> Basic<br> Neutral
lana66690 [7]
The answer is acidic
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3 years ago
Combustion of 9.511 grams of c4h10 will yield ____ grams of CO2
Flauer [41]

Answer:

\boxed{28.81}

Explanation:

We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.  

M_r:      58.12                   44.01

           2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O

m/g:     9.511

1. Moles of C₄H₁₀

\text{Moles of C$_{4}$H$_{10} $} = \text{ 9.511 g C$_{4}$H$_{10} $} \times \dfrac{\text{1 mol C$_{4}$H$_{10} $}}{\text{ 58.12 g C$_{4}$H$_{10} $}} = \text{0.1636 mol C$_{4}$H$_{10}$}

2. Moles of CO₂

The molar ratio is 8 mol CO₂:2 mol C₄H₁₀

\text{Moles of CO}_{2} =\text{0.1636 mol C$_{4}$H$_{10} $} \times \dfrac{\text{8 mol CO}_{2}}{\text{2 mol C$_{4}$H$_{10}$}} = \text{0.6546 mol CO}_{2}

3. Mass of CO₂

\text{Mass of CO}_{2} = \text{0.6546 mol CO}_{2} \times \dfrac{\text{44.01 g CO}_{2}}{\text{1 mol CO}_{2}} = \textbf{28.81 g CO}_{2}\\\\\text{The combustion will form $\boxed{\textbf{28.81 g CO}_{2}}$}

8 0
3 years ago
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