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Triss [41]
3 years ago
8

1. a.) Calculate the wavelength of light which has a frequency of 5.25 x 10 14 Hz.

Chemistry
1 answer:
Scilla [17]3 years ago
3 0
<h3>Answer:</h3>

5.71 × 10² nm

<h3>Explanation:</h3>

The product of wavelength and frequency of a wave gives the speed of the wave.

Therefore;

Velocity of wave = Wavelength × Frequency

c = f ×λ

In our case;

Frequency = 5.25 × 10^14 Hz

Speed of light = 2.998 × 10^8m/s

But;

λ = c ÷ f

  = 2.998 × 10^8m/s ÷  5.25 × 10^14 Hz

  = 5.71 × 10^-7 m

But; 1 M = 10^9 nm

Therefore;

wavelength = 5.71 × 10^-7 × 10^9

                  = 5.71 × 10² nm

The wavelength of light wave 5.71 × 10² nm

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Explanation:

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If a sample of HF gas at 694.9 mmHg has a volume of 3.463 Land the volume is changed to 5.887 L, then what will be the new press
VARVARA [1.3K]

Answer:

\large \boxed{\text{381.7 mmHg}}

Explanation:

Data:

p₁ = 694.9 mmHg; V₁ = 3.463 L

p₂ = ?;                     V₂ = 5.887 L

Calculation:

\begin{array}{rcl}p_{1}V_{1} & = & p_{2}V_{2}\\\text{648.9 mmHg} \times \text{3.463 L} & = & p_{2} \times\text{5.887 L}\\\text{2247.1 mmHg} & = & 5.887p_{2}\\p_{2} & = & \dfrac{\text{2247.1 mmHg}}{5.887}\\\\& = &\textbf{381.7 mmHg}\\\end{array}\\\text{The new pressure of the gas is $\large \boxed{\textbf{381.7 mmHg}}$}

4 0
3 years ago
What were the results on Rutherford's goldfoil experiment
diamong [38]

Answer:

Explanation:

The relative massive alpha particles could go through the gold foil without being deviated of their trajectory or only small deviations due to the uniformity distribution positive charge of the protons.

4 0
3 years ago
HC2H3O2 (aq) + H2O (l) ⇔ C2H3O2- (aq) + H3O+ (aq) Ka = 1.8 x 10-5
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The concentration of [H3O⁺]=2.86 x 10⁻⁶ M

<h3>Further explanation</h3>

In general, the weak acid ionization reaction  

HA (aq) ---> H⁺ (aq) + A⁻ (aq)  

Ka's value  

\large {\boxed {\bold {Ka \: = \: \frac {[H ^ +] [A ^ -]} {[HA]}}}}

Reaction

HC₂H₃O₂ (aq) + H₂O (l) ⇔  (aq) + H₃O⁺ (aq) Ka = 1.8 x 10⁻⁵

\tt Ka=\dfrac{[C_2H_3O^{2-}[H_3O^+]]}{[HC_2H_3O_2]}}\\\\1.8\times 10^{-5}=\dfrac{0.22\times [H_3O^+]}{0.035}

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5 0
3 years ago
State whether the error introduced by each of the following problems would result in a high or a low value for the Cu recovery o
vampirchik [111]

Answer:

Low value for copper recovery

Explanation:

The percentage recovery is obtained from;

Percent recovery = amount of substance you actually collected / amount of substance you were supposed to collect × 100

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