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4 years ago

H. Identity of unknown metal

Chemistry

1 answer:

OLEGan [10]4 years ago

8 0

Answer:

you can identify an unknown substance by measuring its density and comparing your results to a list of known densities. Density=mass/volume. Assume that you have to identify an unknown metal. You can determine the mass of the metal on a scale.

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An unknown compound contains only carbon, hydrogen, and oxygen (). Combustion of 7.50 of this compound produced 11.0 of carbon d

Sergeu [11.5K]

<span>Firstly, we know that M= m/n, the main formula which shows the relationship between m, n, and M. The nknown compound contains only carbon, hydrogen, and oxygen, so we can get n(C)=m/M, from M(C)= m(C)/n (C), besides the stoechiometric equality, we have </span>

n( C)= m(C)/M(C ) = m(CO2)/ M(CO2)=11/44, because m(CO2)=11.0, M(CO2)=44.01

so n(C )= 0.24moles,

5 0

4 years ago

Name the following compound: CH₂-C- CH₂-CH₂-CH₂-C - H butanal O2-butanol O propyl hydrogen ether 1-propanone

bulgar [2K]

Answer:

butanal

Explanation:

It's not 1-propanone because otherwise you have OH bonded to C not O and H bonded to C separately. 2-butanol means that the O and H would be bonded to a different C atom, and it's not propyl hydrogen ether because the O would be in the chain of bonded Cs.
Hope this helps :)

6 0

3 years ago

It is expected that a chemical reaction will occur when copper metal is combined with aqueous zinc sulfate. Explain why there wi

REY [17]

Answer:

$E_{cell}$ = +ve, reaction is spontaneous

$E_{cell}$ = -ve, reaction is non spontaneous

$E_{cell}$ = 0, reaction is in equilibrium

Case 1: when copper metal is combined with aqueous zinc sulfate.

$Cu+ZnSO_4\rightarrow Zn+CuSO_4$

Here copper is undergoing oxidation ad thus acts as anode and zinc is undergoing reduction , thus acts as cathode.

$E^o_{cell}$ = standard electrode potential = $E^0_{cathode}- E^0_{anode}$

Where both $E^0$ are standard reduction potentials.

$E^0_{[Cu^{2+}/Cu]}= +0.34V$

$E^0_{[Zn^{2+}/Zn]}= -0.76V$

$E^0=E^0_{[Zn^{2+}/Zn]}- E^0_{[Cu^{2+}/Cu]}$

$E^0=-0.76-(+0.34)=-1.10V$

Thus as $E_{cell}$ is negative , the reaction is non spontaneous.

Case 2: when zinc metal and aqueous copper sulfate solution are combined.

$Zn+CuSO_4\rightarrow Cu+ZnSO_4$

Here zinc is undergoing oxidation ad thus acts as anode and copper is undergoing reduction , thus acts as cathode.

$E^o_{cell}$ = standard electrode potential = $E^0_{cathode}- E^0_{anode}$

Where both $E^0$ are standard reduction potentials.

$E^0_{[Cu^{2+}/Cu]}= +0.34V$

$E^0_{[Zn^{2+}/Zn]}= -0.76V$

$E^0=E^0_{[Cu^{2+}/Cu]}- E^0_{[Zn^{2+}/Zn]}$

$E^0=+0.34-(-0.76)=+1.10V$

Thus as $E_{cell}$ is positive , the reaction is spontaneous.

3 0

4 years ago

Ammonium nitrate, a common fertilizer, is used as an explosive in fireworks and by terrorists. It was the material used in the t

Tamiku [17]

Answer:

The explosive decomposition of 98.4 kg of ammonium nitrate produces 58498.8 L of nitrogen, 29249.4 L of oxygen and 116997.6 L of water vapor.

Explanation:

To find how many liters of gas are formed from the explosive decomposition of ammonium nitrate it is necessary to follow these steps (check the attachment for better understanding):

1st) Balance the equation:

Write the decomposition equation and then find the correct coefficients to make sure that it goes according to the "Law of conservation of mass" (the mass of the reactants side must be equal to the mass of the products side). So, 2 moles of ammonium nitrate produces 2 moles of nitrogen, 1 mole of oxygen and 4 moles of water vapor.

2nd) Find the Ammonium nitrate molar mass:

The ammonium nitrate mass it is calculated by adding de molar mass of each atom that forms the ammonium nitrate molecule. You can find the elements molar mass in the Periodic Table.

In this example I decided to round the number to simplify tha calculus, for example: the oxygen molar mass in the periodic table is 15.9994 but I use 16. You can use the complete number if you want.

By doing this, the ammonium nitrate molar mass is 80 g/mol.

The statement says that there is 98.4 kg of ammonium nitrate. In ordder to use the same units in all the calculus sometimes it is usefull to convert the kg to g, so it is the same as 98400g. You can do it the other way around if you prefer (g to kg).

3rd) Find the number of moles of each gases and aqua vapor formed:

It is important to know the amount of each compound formed by the decomposition reaction, that's why we need to pay attention to the coefficients of the balanced reaction.

The amount of each compound is easily found by using the "rule of three".

To use the rule of three we need to think using the balanced reaction so:

If 160g (2 moles) of ammonium nitrate produces 2 moles of nitrogen gas, the 98400g that we have of ammonium nitrate will produce an X amount of nitrogen gas. With this information we multiply 98400g by 2 moles and then we divide the result by 160g. The final result it is 1230 moles of nitrogen.

In the same way we use the rule of three to calculate the number of moles of oxygen and water.

4th) Find the liters (volume) of each gas and aqua vapor formed:

Finally, to find the liters from the number of moles, it is necessary to apply the "Ideal gases law", that relates the pressure (atm), volume (L), moles number and temperature (Kelvin) with the R gas constant in the formula:

PxV = nxRxT

It is important to use the correct units because the R gas constant is equal to 0.082 atm.L/mol.K.

As we need to calculate the liters (volume) we pass the pressure dividing to the other side and then we just have to replace the information:

V = (nxRxT)/P

As you can see in the attachment, doing this last step for each compound, we can find the liters produced of them.

8 0

4 years ago

Did I do everything right?

nikitadnepr [17]

WHAT DO YOU MEAN?????

5 0

3 years ago