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Morgarella [4.7K]
3 years ago
8

When a horn is moving away from an observer the sound waves from it will pass by the observer____ frequently, causing the pitch

to____
Chemistry
1 answer:
xxMikexx [17]3 years ago
8 0
The answer: less frequency, causing the pitch to be lowered. 

This is known as the Doppler effect. When you have an object that is producing sound while moving, the side that it is moving towards to will have waves that will become a bit more squished together, while the other side that it is moving away from will have waves that are spaced a bit more apart from each other. The only determining factor in this is the magnitude of the speed, so the faster the object is moving towards or away from you, then the more severe pitch change the object will have.
In this case, it's just asking in general what would happen if the horn was moving away from you while it was giving off sound waves. Therefore, since it is moving away from you, its sound waves are a bit further apart, resulting in a lowered frequency and pitch.
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Given the balanced equation:
Vikki [24]

Answer: Option (3) is the correct answer.

Explanation:

When a more reactive element or atom replaces a less reactive atom then this type of reaction is known as single displacement reaction.

In the given reaction, potassium iodide reacts with fluorine and results in the formation of potassium fluoride and iodine.

Here, fluorine being more reactive displaces iodine from potassium iodide.

Therefore, it is a single replacement or displacement reaction.

8 0
3 years ago
A mixture of gases contains 4.46 moles of neon (Ne), 0.74 mole of argon (Ar), and 2.15 moles of xenon (Xe). Calculate the partia
viva [34]

Explanation:

The partial pressure of an individual gas is equal to the total pressure of the mixture multiplied by the mole fraction of the gas.

Total pressure = 2atm

Mole Fraction = number of moles / total number of moles

Neon

Mole Fraction = 4.46 / 7.35 = 0.607

Partial Pressure = 0.607 * 2 = 1.214 atm

Argon

Mole Fraction = 0.74 / 7.35 = 0.101

Partial Pressure = 0.101 * 2 = 0.202 atm

Xenon

Mole Fraction = 2.15 / 7.35 = 0.293

Partial Pressure = 0.293 * 2 = 0.586 atm

5 0
3 years ago
When 136g of glycine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezin
loris [4]

The given question is incomplete. The complete question is:

When 136 g of glycine are dissolved in 950 g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 136 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for sodium chloride in X.

Answer: The vant hoff factor for sodium chloride in X is 1.9

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=T_f^0-T_f=8.2^0C = Depression in freezing point

K_f = freezing point constant

i = vant hoff factor = 1 ( for non electrolyte)

m= molality =\frac{136g\times 1000}{950g\times 75.07g/mol}=1.9

8.2^0C=1\times K_f\times 1.9

K_f=4.32^0C/m

Now Depression in freezing point for sodium chloride is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=20.0^0C = Depression in freezing point

K_f = freezing point constant  

m= molality = \frac{136g\times 1000}{950g\times 58.5g/mol}=2.45

20.0^0C=i\times 4.32^0C\times 2.45

i=1.9

Thus vant hoff factor for sodium chloride in X is 1.9

3 0
2 years ago
I need help FAST ASAP
kiruha [24]
In this item, we are simply to find the ions that may bond and are able to form a formula unit. We are also instructed to give out their name. There are numerous possible combinations of ions to form a compound. Some answers are given in the list below.

1.  Na⁺     ,    Cl⁻    , NaCl   ---> sodium chloride (this is most commonly known as table salt)

2. C⁴⁺       , O²⁻     , CO₂  ---> carbon dioxide

3. Al³+     , Cl⁻       , AlCl₃   ----> aluminum chloride

4. Ca²⁺     , Cl⁻     , CaCl₂    ---> calcium chloride

5. Li⁺        , Br⁻      , LiBr       ---> lithium bromide

6. Mg³⁺     , O²⁻      , Mg₂O₃   ----> magnesium oxide

7. K⁺        , I⁻          , KI   ---> potassium iodide

8. H⁺        , Cl⁻        , HCl  --> hydrogen chloride

9. H⁺        , Br⁻         , HBr ----> hydrogen bromide

10. Na⁺    , Br⁻         , NaBr   ---> sodium bromide
6 0
3 years ago
Using the balanced equation for the combustion of ethane: 2C2H6 + 7O2 → 4CO2 + 6H2O, how many moles of O2 needed to produce 12 m
pickupchik [31]

Answer:

14 moles of oxygen needed to produce 12 moles of H2O.

Explanation:

We are given that balance eqaution

2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O

We have to find number of moles of O2 needed  to produce 12 moles of H2O.

From given equation

We can see that

6 moles of   H2O produced by Oxygen =7 moles

1 mole of   H2O produced by Oxygen=\frac{7}{6}moles

12 moles of H2O produced by Oxygen=\frac{7}{6}\times 12moles

12 moles of H2O produced by Oxygen=7\times 2moles

12 moles of H2O produced by Oxygen=14 moles

Hence, 14 moles of oxygen needed to produce 12 moles of H2O.

3 0
2 years ago
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