How many molecules are in 2.6 grams H4C

Answers

Answer 1
Answer: Molar mass of H4C = 1 * 4 + 12 = 16 g/mol

Number of moles:

n = m / mm

n = 2.6 / 16

n = 0.1625 moles

Number of molecules:

number of moles * 6.02x10²³

0.1625 * 6.02x10²³

= 9.78x10²² molecules

hope this helps!




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What is the mass of 8.12 × 1023 molecules of CO2 gas? (Atomic mass of carbon = 12.011 u; oxygen = 15.999 u.)

Answers

Answer:

8306.76

Explanation:

you just calcuate 8.12 x 1023 and that will give you the answer

PLZ mark me as a BRAINLIEST

Which of the following sets represents a pair of isotopes? 14C and 14N 206Pb and 208Pb O2 and O3 32S and 32S2-

Answers

Isotopes are substances that have the same number of protons but differ in the number of neutrons. Hence, the pair of isotopes above should be of the same element. In the given choices, 14C is not an isotope of 14N. 206Pb is an isotope of 208 Pb. O2 and O3 differ in molecular formula but still made up of same kind of atom, hence they are allotropes, while 32S and 32S2- are not isotopes.  

Which type of reaction occurs when nonmetal atoms become negative nonmetal ions?A. oxidation
B. reduction
C. substitution
D. condensation

Answers

Answer: Option (B) is the correct answer.

Explanation:

When an atom gains electrons then it becomes rich in electrons and therefore it acquires a negative charge.

Also, it is known that a reduction reaction is a reaction in which an element gain electron(s).

For example, Br_(2) + 2e^(-) \rightarrow 2Br^(-) is a reduction reaction.

Thus, we can conclude that when nonmetal atoms become negative nonmetal ions then type of reaction which occurs is reduction reaction.

The answer is B. reduction. The atom becoming negative ions means it gains electrons. The reaction which gain electrons is reduction reaction. So the answer is B.

Write the atomic symbols for isotopes with the following characteristics? 1)an oxygen atom with 10 neutrons 2)a neon atom with twelve neutrons 3)25 electrons and 28 neutrons 4)a mass number of 24 and 13 neutrons 5)a titanium atom with 25 neutrons

Answers

Isotopes are elements that differ in the number in neutrons while the numner of protons are the same. The answers of the questions are: 1) 18/8 O 2) 22/10 Ne 3) 53/25 Mn 4) 37/24 Cr 5) 106/81 Ti.

What is the solute and the solvent in a carbonated drink?

Answers

The solvent is water. The solutes may include fruit juice, natural and artificial flavors, natural and artificial colorings, preservatives, carbon dioxide, and lots and lots of sugar.
The solvent is: Water 
The solute is : sugar any sweetener in the drink 
:P

What is the freezing point of an aqueous solution that boils at 105.0 ∘C? Express your answer using two significant figures.

Answers

The boiling point of standard water is 100 degree Celsius, with the addition of solute the boiling point is elevated. The freezing point of the solution will be -18.04 degree Celsius.

What is boiling point?

The boiling point is the temperature at which the liquid is converted to vapor. The change in boiling point of the aqueous solution gives the molality of the solution as:

\rm \Delta T=ebuliloscopic constant\;*\;molality\;*\;von't\;hoff\;factor\n105^\circ C-100^\circ C=0.512^\circ C.kg/mol\;*\;1\;*\;m\n9.7\;mol/kg=m

The depression in freezing point from molality is given as;

\rm \Delta T=K_f\;*\;molality\;*\;i\n\Delta\;T=1.86\;^\circ C/m\;*\;9.7\;*\;1\n\Delta T=18.04\;^\circ C\n

The freezing point of aqueous water is zero degree Celsius. The freezing point of the solution will be:

\rm \Delta T=0^\circ\;C-New\;freezing\;point\n18.04^\circ\;C=0-New\;freezing\;point\nNew\;freezing\;point=-18.04^\circ C

The freezing point of the solution is -18.04 degree Celsius.

Learn more about boiling point, here:

brainly.com/question/2153588

Answer:

T°fussion of solution is -18°C

Explanation:

We have to involve two colligative properties to solve this. Let's imagine that the solute is non electrolytic, so i = 1

First of all, we apply boiling point elevation

ΔT = Kb . m . i

ΔT = Boiling T° of solution - Boiling T° of pure solvent

Kb =  ebuliloscopic constant

105°C - 100° = 0.512 °C kg/mol  . m . 1

5°C / 0.512 °C mol/kg = m

9.7 mol/kg = m

Now that we have the molality we can apply, the Freezing point depression.

ΔT = Kf . m . i

Kf =  cryoscopic constant

0° - (T°fussion of solution) = 1.86 °C/m  . 9.76 m . 1

- (1.86°C /m . 9.7 m) = T°fussion of solution

- 18°C = T°fussion of solution