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2 years ago

If you stretch back a rubber band and release it, it shoots across the room. What type of energy conversion has occurred?

2 answers:

5 0

<span>Elastic to mechanical is the energy conversion that occurs. Elastic, the pulling back and stretching of the rubber band, mechanical is the release!</span>

Crank2 years ago

4 0

Answer: elastic to mechanical

Explanation: According to the law of conservation of energy, energy can neither be created nor be destroyed. It can only be transformed from one form to another.

Elastic energy is the energy stored as a result of applying a force to deform an elastic object. The energy is stored until the force is removed. When the force is removed, the object gets back to its original shape.The deformation can involve compressing, stretching or twisting of an object.

Mechanical energy is the energy possessed by a body by virtue of its position and motion.

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A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the

Romashka [77]

This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, $C_6H_4N_2O_4$

Explanation :

First we have to calculate the mass of benzene.

$\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}$

$\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g$

Now we have to calculate the molar mass of unknown compound.

Given:

Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g = 0.04395 kg

Formula used :

$\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}$

where,

$\Delta T_f$ = change in freezing point = $5.53-1.37=4.16^oC$

$\Delta T_s$ = freezing point of solution

$\Delta T^o$ = freezing point of benzene

Molal-freezing-point-depression constant $(K_f)$ for benzene = $5.12^oC/m$

m = molality

Now put all the given values in this formula, we get

$4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}$

$\text{Molar mass of unknown compound}=180.6g/mol$

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.575}{1.193}=2.99\approx 3$

For H = $\frac{2.4}{1.193}=2.01\approx 2$

For N = $\frac{1.193}{1.193}=1$

For O = $\frac{2.381}{1.193}=1.99\approx 2$

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_3H_2N_1O_2$

The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{180.6}{84}=2$

Molecular formula = $(C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4$

Therefore, the molecular of the compound is, $C_6H_4N_2O_4$

3 0

2 years ago

A student with an unknown code of Q3A conducts a qualitative analysis of her unknown. Her results indicated the formation of a y

insens350 [35]

Answer:

Grey precipitate implies the presence of silver ions

Yellow precipitate implies the presence of lead II ions

Explanation:

Qualitative analysis provides us a quick method of identifying ions present in a sample by chemical reactions involving simple reagents. Precipitates having a unique colour is formed. The identity of ions in the sample is deduced from the colour of precipitate obtained when particular reagents are added.

In the question, a precipitate containing silver ions upon standing turn into grey colour. Similarly, lead II ions give a yellow precipitate.

6 0

2 years ago

Please help I'm not good at science

Citrus2011 [14]

Answer:

i'm not able to see images but if you type it all i can answer:)

Explanation:

8 0

2 years ago

An unknown gas diffuses 1.25 times faster than CO2 gas. What is the molar mass of the unknown gas?

hodyreva [135]

Answer:

28.16 g/mol

Explanation:

From Graham's law;

Let the rate of diffusion of gas X be 1.25

Let the rate of diffusion of CO2 be 1

Molecular mass of gas X= M

Molecular mass of CO2 = 44g/mol

1.25/1=√44/M

(1.25/1)^2 = 44/M

1.5625 = 44/M

M= 44/1.5625

M= 28.16 g/mol

4 0

2 years ago

Write the net cell equation for this electrochemical cell. Phases are optional. Do not include the concentrations. Co ( s ) ∣ ∣

Tatiana [17]

Answer: The overall equation will be $Co(s)+2Ag^+(aq)\rightarrow Co^{2+}(aq)+Ag(s)$

Explanation:

The representation is given by writing the anode on left hand side followed by its ion with its molar concentration. It is followed by a salt bridge. Then the cathodic ion with its molar concentration is written and then the cathode.

Oxidation reaction is defined as the reaction in which a substance looses its electrons. The oxidation state of the substance increases.

Anode : $Co(s)\rightarrow Co^{2+}(aq)+2e^{-}$

Reduction reaction is defined as the reaction in which a substance gains electrons. The oxidation state of the substance gets reduced.

Cathode : $Ag^+(aq)+e^{-1}\rightarrow Ag(s)$ $\times 2$

The number of electrons lost must be equal to the number of electrons gained , thus overall equation will be :

$Co(s)+2Ag^+(aq)\rightarrow Co^{2+}(aq)+Ag(s)$

5 0

2 years ago