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astra-53 [7]
3 years ago
15

In one or two sentences, make and justify a claim about the frequency of a wave if the wavelength increases and the velocity sta

ys the same.
Chemistry
1 answer:
oksian1 [2.3K]3 years ago
4 0

Answer:

The relationship between the frequency of a wave and the wavelength is an inverse relationship, therefore, if the wavelength of a wave increases and the velocity of the wave stays the same, the frequency of the wave decreases.

An example is the ocean wave, if the distance between successive wave crests, which is the wavelength, 'λ', of the wave, increases, while the velocity, 'v', of the wave coming to shore remains the same, the number of wave crest arriving at the beach in a given period of time, which is the frequency of the wave, 'f',  decreases according to the following equation;

f = v/λ

Explanation:

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There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with
grandymaker [24]

\text{Hg} \text{O} and \text{Hg}_{2} \text{O}.

Assuming complete decomposition of both samples,

  • m(\text{Hg}) = m(\text{residure})
  • m(\text{O}) = m(\text{loss})

First compound:

  • m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g
  • m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g

n = m/M; 0.6498 \; g of the first compound would contain

  • n(\text{O atoms}) = 0.048 \; g  / 16 \; g \cdot mol^{-1}= 0.003 \; mol
  • n(\text{Hg atoms}) = 0.6018 \; g  / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol

Oxygen and mercury atoms seemingly exist in the first compound at a 1:1 ratio; thus the empirical formula for this compound would be \text{Hg} \text{O} where the subscript "1" is omitted.

Similarly, for the second compound

  • m(\text{O}) = m(\text{loss}) = 0.016 \; g
  • m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012  \; g

n = m/M; 0.4172 \; g of the first compound would contain

  • n(\text{O atoms}) = 0.016 \; g  / 16 \; g \cdot mol^{-1}= 0.001 \; mol
  • n(\text{Hg atoms}) = 0.4012 \; g  / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol

n(\text{Hg}) : n(\text{O}) \approx  2:1 and therefore the empirical formula

\text{Hg}_{2} \text{O}.

8 0
3 years ago
How many moles of oxygen are required to produce 37.15 g CO2? 37.15 g CO2 = mol O2
NNADVOKAT [17]

Answer:

0.84 moles of oxygen are required.

Explanation:

Given data:

Mass of CO₂ produced = 37.15 g

Number of moles of oxygen = ?

Solution:

Chemical equation:

C + O₂     →     CO₂

Number of moles of  CO₂:

Number of moles = mass/molar mass

Number of moles = 37.15 g/ 44 g/mol

Number of moles = 0.84 mol

Now we will compare the moles of oxygen and carbon dioxide.

                          CO₂         :       O₂  

                              1           :         1

                            0.84       :       0.84

0.84 moles of oxygen are required.

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Which information about a chemical reaction is provided by a potential energy diagram?(1) the oxidation states of the reactants
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D. The energy released or absorbed during the reaction 

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Burning gasoline is an _____________ reaction.
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Answer:

chemical.............

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An emerald gemstone is an impure form of the mineral beryl, Be3Al2Si16O18. What is the percentage of silicon in the mineral?
solong [7]

Answer:

b. 54.9%

Explanation:

An emerald gemstone has the formula Be₃Al₂Si₁₆O₁₈. We can find the mass of each element in 1 mole of Be₃Al₂Si₁₆O₁₈ by multiplying the molar mass of the element by its atomicity.

Be: 3 × 9.01 g = 27.03 g

Al: 2 × 26.98 g = 53.96 g

Si: 16 × 28.09 g = 449.4 g

O: 18 × 16.00 g = 288.0 g

Total mass = 818.4 g

The mass percentage of silicon is:

(449.4 g / 818.4 g) × 100% = 54.91%

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3 years ago
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