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3 years ago

Intermolecular forces allow which of the following substances to be the least fluid?

2 answers:

4 0

The correct option is calcite. It is a carbonate mineral of calcium carbonate. Its quite a hard mineral and it not fluid in nature at all. It exhibits both ionic and covalent chemical bonds.

slava [35]3 years ago

3 0

Out of the options, glass is the least fluid. The proof of this also lies in the fact that glass is the most difficult to melt out of all of the mentioned substances, and melting point gives us a rough estimate of the strength of intermolecular forces.

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34 atoms of carbon (C) react with 22 molecules of hydrogen gas (H2). How many molecules of methane (CH4) will be formed, and wha

kolbaska11 [484]

Answer:

11 molecules of CH4.

23 atoms of C is the leftover.

Explanation:

Hello!

In this case, for the formation of methane:

$C+2H_2\rightarrow CH_4$

We can see there is an excess of carbon based on their stoichiometry, because the needed amount of hydrogen gas molecules would be:

$molecules _{H_2}=34atomC*\frac{2molec\ H_2}{1atomC} =68molec\ H_2$

Thus, the formed molecules of methane are computed below:

$molec\ CH_4=22molec\ H_2 *\frac{1molec\ CH_4}{2molmolec\ H_2} \\\\molec\ CH_4=11molec\ CH_4$

In such a way, the leftover of carbon atoms are:

$atoms \ C^{left over}=34-22molec\ H_2*\frac{1atoms C}{2molec\ H_2} \\\\atoms \ C^{left over}=23 atoms C$

Best regards!

4 0

2 years ago

Which oh these statements describes a chemical property?

Gelneren [198K]

Iron rusts when exposed to air → chemical property

3 0

2 years ago

What is the percent by mass of carbon in acetone, c 3 h 6 o?

Tatiana [17]

In the problem, we are tasked to solved for the amount of carbon (C) in the acetone having a molecular formula of C 3 H 6 O. We need to find first the molecular weight if Carbon (C), Hydrogen (H), Oxygen (O).

Molecular Weight:
C=12 g/mol
H=1 g/mol
O=16 g/mol

To calculate for the percent by mass of acetone, we assume 1 mol of acetone.
%C= $( \frac{3(12 g/mol)}{3(12 g/mol)+6(1 g/mol)+16 g/mol} ) x 100%$
%C=62.07%
Therefore, the percent by mass of carbon in acetone is 62.07%

5 0

2 years ago

Read 2 more answers

Sodium hydrogen carbonate and citric acid Word equation

timofeeve [1]

Answer:

the right answer is:

Explanation:

How to Balance: NaHCO3 + HC2H3O2 = NaC2H3O2 + CO2 + H2O|

5 0

2 years ago

Read 2 more answers

A sample consisting of 1.00 mol of perfect gas molecules at 27 °C is expanded isothermally from an initial pressure of 3.00 atm

Evgesh-ka [11]

Answer:

a) reversibly

ΔU = 0

q = 2740.16 J

w = -2740.16 J

ΔH = 0

ΔS(total) = 0

ΔS(sys) =9.13 J/K

ΔS(surr) = -9.13 J/K

b) against a constant external pressure of 1.00 atm

ΔU = 0

w = -1.66 kJ

q = 1.66 kJ

ΔH = 0

ΔS(sys) = 9.13 J/K

ΔS(surr) = -5.543 J/K

ΔS(total) = 3.587 J/K

Explanation:

Step 1: Data given:

Number of moles = 1.00 mol

Temperature = 27.00 °C = 300 Kelvin

Initial pressure = 3.00 atm

Final pressure = 1.00 atm

The gas constant = 8.31 J/mol*K

(a) reversibly

Step 2: Calculate work done

For ideal gases ΔU depends only on temperature. So as it is an isothermal (T constant).

Since the temperature remains constant:

ΔU = 0

ΔU = q + w

q = -w

w = -nRT ln (Pi/Pf)

⇒ with n = the number of moles of perfect gas = 1.00 mol

⇒ with R = the gas constant = 8.314 J/mol*K

⇒ with T = the temperature = 300 Kelvin

⇒ with Pi = the initial pressure = 3.00 atm

⇒ with Pf = the final pressure = 1.00 atm

w =- 1*8.314 *300 * ln(3)

w = -2740.16 J

q = -w

q = 2740.16 J

Step 3: Calculate change in enthalpy

Since there is no change in energy, ΔH = 0

Step 4: Calculate ΔS

for an isothermal process

ΔS (total) = ΔS(sys) + ΔS(surr)

ΔS(sys) = -ΔS(surr)

ΔS(sys) = n*R*ln(pi/pf)

ΔS(sys) = 1.00 * 8.314 * ln(3)

ΔS(sys) = 9.13 J/K

ΔS(surr) = -9.13 J/K

ΔS (total) = ΔS(sys) + ΔS(surr) = 0

(b) against a constant external pressure of 1.00 atm

Step 1: Calculate the work done

w = -Pext*ΔV

w = -Pext*(Vf - Vi)

⇒ with Vf = the final volume

⇒ with Vi = the initial volume

We have to calculate the final and initial volume. We do this via the ideal gas law P*V=n*R*T

V = (n*R*T)/P

Initial volume = (n*R*T)/Pi

⇒ Vi = (1*0.08206 *300)/3

⇒ Vi = 8.206 L

Final volume = (n*R*T)/Pf

⇒ Vf = (1*0.08206 *300)/1

⇒ Vf = 24.618 L

The work done w = -Pext*(Vf - Vi)

w = -1.00* ( 24.618 - 8.206)

w = -16.412 atm*L

w = -16 .412 *(101325/1atm*L) *(1kJ/1000J)

w = -1662.9 J = -1.66 kJ

Step 2: Calculate the change in internal energy

ΔU = 0

q = -w

q = 1.66 kJ

ΔH = 0 because there is no change in energy

Step 3: Calculate ΔS

ΔS(sys) = n*R*ln(3)

ΔS(sys) = 1.00 * 8.314 * ln(3)

ΔS(sys) = 9.13 J/K

ΔS(surr) = -q/T

ΔS(surr) = -1662.9J/300K

ΔS(surr) = -5.543 J/K

ΔS(total) = ΔS(surr) +ΔS(sys) = -5.543 J/K + 9.13 J/K = 3.587 J/K

4 0

3 years ago