# Which allotrope of carbon is most recently discovered, so the uses are fairly unknown? A. graphite B. diamond C. fullerene D. coal

2 months ago

## Solution 1

Guest #4838
2 months ago
C-fullerene
It can be used as a drug delivery method :)

## 📚 Related Questions

Question
What mass (g) of nan3 is required to provide 26.5 l of n2 at 22.0 °c and 1.10 atm?
Solution 1
NaN₃ (sodium azide) produces N₂ gas when it is heated. The balanced equation is,
2NaN₃ → 2Na + 3N₂

To find out mass of NaN₃, we have to find the moles of N₂ formed by using given data.

Let's assume that N₂ gas is an ideal gas. Then we can use ideal gas equation,
PV = nRT
Where,
P = Pressure of the gas (Pa)
V = volume of the gas (m³)
n = number of moles (mol)
R = Universal gas constant (8.314 J mol⁻¹ K⁻¹)
T = temperature in Kelvin (K)

The given data for N
₂ gas is,
P = 1.10 atm = 111457.5 Pa
V = 26.5 L = 26.5 x 10⁻³ m³
T = (273 + 22) K = 295 K
R = 8.314 J mol⁻¹ K⁻¹
n = ?

By applying the formula,

111457.5 Pa x  26.5 x 10⁻³ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 295 K
n   = 1.204 mol

Hence, produced N₂ gas = 1.204 mol

The stoichiometric ratio between NaN₃ and N₂ is 2 : 3

hence moles of NaN₃ = 1.204 mol x 2 / 3
= 0.803 mol

Molar mass of NaN₃ = 65 g/mol

Hence, required mass of NaN₃ = 0.803 mol x 65 g/mol
= 52.195 g
Question
What about uranium room temperature is it "SOLID LIQUID or GAS)?
Solution 1
It is a solid 100%............
Question
If the number of moles of a gas initially contained in a 2.10 l vessel is doubled, what is the final volume of the gas in liters? (assume the pressure and temperature remain constant.) none of the above 8.40 4.20 1.05 6.30
Solution 1
Avagadro's law gives the relationship between volume of gas and number of moles of the gas. When the temperature and pressure are constant, volume is directly proportional to number of moles of gas.

V -volume and n - number of moles of gas.
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
initial number of moles n1 - N
n2 - 2N as the number of moles are doubled
V1 - initial volume - 2.10 L
V2 - final volume - V
substituting the values in the equation

V = 4.20 L
final volume is 4.20 L
Question
Household sugar sucrose has the molecular formula c12h22o11. what is the carbon in sucrose by mass
Solution 1
%age of Carbon is 42.10 %.

Explanation:
In order to find the %age of elements present in a molecule we will use following formula,

%age of Element  =  (Mass of Element / Mass of Molecule) × 100

The molecular mass of Sucrose is.

=  (12)12  +  (1)22  +  (16)11
Where;
12  =  Atomic mass of Carbon

1  =  Atomic Mass of Hydrogen

16  =  Atomic Mass of Oxygen

=  144  +  22  +  176

=   342 g/mol
%age of C;
=  (144 / 342) ×100  =  42.10 %

%age of H;
=  (22 / 342) ×100    =  06.42 %

%age of O;
=  (176 / 342) ×100  =  51.46 %
Question
If 20.60 ml of 0.0100 m aqueous hcl is required to titrate 30.00 ml of an aqueous solution of naoh to the equivalence point, what is the molarity of the naoh solution?
Solution 1

Answer : The molarity of the NaOH solution is, 0.00687 M.

Explanation :

The balanced chemical reaction will be:

Using neutralization method :

where,

are the n-factor, molarity and volume of acid which is

are the n-factor, molarity and volume of base which is NaOH.

We are given:

Putting values in above equation, we get:

Hence, the molarity of the NaOH solution is, 0.00687 M.

Solution 2
The balanced equation for the above reaction is as follows;
NaOH + HCl -- > NaCl + H₂O
stoichiometry of NaOH to HCl is 1:1
number of HCl moles reacted - 0.0100 mol/L x 0.02060 L = 0.000206 mol
according to 1:1 molar ratio
number of NaOH moles reacted - 0.000206 mol
number of NaOH moles in 30.00 mL - 0.000206 mol
therefore number of NaOH moles in 1000 mL - 0.000206 mol / 0.03000 L = 0.00687 mol
molarity of NaOH is 0.00687 M
Question
Electrons in an excited state are more likely to enter into chemical reactions. a. true b. false
Solution 1

False

Explanation:

The electrons  are excited because:

1.- They posses more energy than the basal form, but this doesn´t mean that they can easily react, remember, the electrons are the outside part of an atom.

2.-  By the time, this electrons are loosing their energy, that mean that they return to the basal form.

Question
When you add so much solute that no more dissolves, you have a ______
Solution 1
When you add so much solute that no more dissolves, you have a saturated solution.
Question
Differences in which property allows the separation of a sample of sand and sea water by filteration
Solution 1

Filtration is a type of process in which components of a mixture are separated based in the size. The sea water and sand can be separated by filtration due to different particle size.

Seawater is a mixture of several substances, including salt, sand, and stones. The substances of seawater can be separated via filtration. The filter paper in this technique separates the substances based on the size of particles.

The filtration separates sand and seawater as:

• Seawater and sand are dissolved in each other. When the filtration technique is performed, it separates out the water from sand.
• The size of the sand particles are larger than the pore size of filter paper, thus, it easily separates the sand from seawater.

Therefore, difference in the particle size allows the separation of seawater and sand.

brainly.com/question/9468294?referrer=searchResults

Solution 2
Differences in Particle Size allows the separation of a sample of sand and sea water by filtration.

Explanation:
This is a very simple example which can be conducted at home. When Salt dissolved in water is filtered all the solution get passed through filter paper, while solution of sand and water on filtration separates the sand from water. This is because of the particle size. The size of Na⁺ and Cl⁻ ions are very small and can easily pass through the pores of filter paper, while the size of sand particles are very large and fails to pass through the pores of filter paper.
Question
Which formulas and their respective names match? I) Li3N lithium nitride II) H2S hydrogen sulfite III) SO2 sulfur dioxide IV) AlF3 aluminum trifluoride a.I and II b.I and III c.II and IV d.I, II, and III?
Solution 1

Explanation:

Ionic compound is defined as the compound which is formed by the complete transfer of electrons from one atom to another atom. These compounds are usually formed between metals and non-metals.

Covalent compounds are defined as the compounds which are formed by the sharing of electrons between two atoms. These compounds are formed between two non-metals.

The nomenclature of covalent compound is given by:

1. The less electronegative element is written first.
2. The more electronegative element is written then, and a suffix is added with it. The suffix added is '-ide'.
3. If atoms of an element is greater than 1, then prefixes are added which are 'mono' for 1 atom, 'di' for 2 atoms, 'tri' for 3 atoms and so on..

The nomenclature of ionic compounds is given by:

1. Positive ion is written first.
2. The negative ion is written next and a suffix is added at the end of the negative ion. The suffix added is '-ide'.

For the given compounds:

(I) : This is an ionic compound and hence, its name is lithium nitride.

(II) : This is a covalent compound and hence, its name is hydrogen sulfide.

(III) : This is a covalent compound and hence, its name is sulfur dioxide.

(IV) : This is an ionic compound and hence, its name is aluminium fluoride.

Hence, the correct answer is Option b.

Solution 2
b.I and III
hope this helps!!
Question
Tritium (31h) is an isotope of hydrogen that is sometimes used to make the hands of watches glow in the dark. the half-life of tritium is 12.3 years. after 49 years, approximately how much of the original tritium remains?
Solution 1
The answer IS 6.25%.  Tritium (31h) is an isotope of hydrogen that is sometimes used to make the hands of watches glow in the dark. the half-life of tritium is 12.3 years. after 49 years, approximately 6.25% of the original tritium remains?
Solution 2

6.31%

Explanation:

In order to solve this, you need to use the expression to calculate half life, which is the following:

C = Co e^-t λ  (1)

Where:

C: concentration after t has passed

Co: initial concentration

t: time that has passed

λ: lambda which relation half life time

λ this can be calculated with the following expression:

λ = ln2 / t(1/2)   (2)

So, let's calculate λ first and then, the concentration. In this case, we will assume that we begin with a concentration at 100%.

The value of lambda is:

λ = ln2 / 12.3

λ = 0.0564

Now, let's use (1) to calculate the concentration after 49 years:

C = 100 e^ (-49 * 0.0564)

C = 100 e^(-2.7636)

C = 6.31 %

And this will be the tritium remaining after 49 years

2647929
842281
748681
586256
406852
368373
348603
324927
199835
130075
112100
106146
77164
23213
22589
19607
17108
13966
10987
3389