In which of these would a nonpolar covalent bond be present? HBr H2O HCI Br2

Which of these compounds would you expect to have the lowest melting point?

Leya [2.2K]

Answer:

Hmmm. #3

Explanation:

5 0

3 years ago

The Density of pure carbon in Diamond form is 3.52 g/cm^3. How many cubic inches would 23.7 moles of pure diamond occupy?

Vedmedyk [2.9K]

Answer : The volume of pure diamond is $0.493inch^3$

Explanation : Given,

Density of pure carbon in diamond = $3.52g/cm^3$

Moles of pure diamond = 23.7 moles

Molar mass of carbon = 12 g/mol

First we have to calculate the mass of carbon or pure diamond.

$\text{ Mass of carbon}=\text{ Moles of carbon}\times \text{ Molar mass of carbon}$

Molar mass of carbon = 12 g/mol

$\text{ Mass of carbon}=(23.7moles)\times (12g/mole)=284.4g$

Now we have to calculate the volume of carbon or pure diamond.

Formula used:

$Density=\frac{Mass}{Volume}$

Now putting all the given values in this formula, we get:

$3.52g/cm^3=\frac{284.4g}{Volume}$

Volume = $80.8cm^3$

As we know that:

$1cm^3=0.061inch^3$

So,

Volume = $0.061\times 80.8inch^3$

Volume = $0.493inch^3$

Therefore, the volume of pure diamond is $0.493inch^3$

5 0

2 years ago

X(g)+4Y(g)→2Z(g), ΔH∘=−75.0 kJ Before the reaction, the volume of the gaseous mixture was 5.00 L. After the reaction, the volume

vlabodo [156]

Isobaric transition, first law: H=ΔU+w for a gas expansion: w=Pext∗ΔV to convert to joules, you need the gas constants. R = 0.08206 L atm/mol*K, R=8.314 J/mol*K w=Pext∗ΔV∗8.314 J/mol∗K0.08206 L atm/mol∗K ΔU=ΔH−[Pext∗ΔV∗8.314 J/mol∗K0.08206 L atm/mol∗K] ΔU=−75000 J−[(43.0atm)∗(2−5)L∗8.314 J0.08206 L atm] Then you need to convert to kJ. by the way U=E, internal energy.

7 0

2 years ago

Calculate the poh of this solution. round to the nearest hundredth. ph = 1.90 poh =

Viktor [21]

Answer:

The answer is 5.10

Explanation:

<h3>Given;</h3>

pH = 1.9

pOH = ?

We know that

pH + pOH = 7

pOH = 7 – pH

pOH = 7 – 1.90

pOH = 5.10

Thus, The pOH of the solution is 5.10

6 0

2 years ago

Calculate the approximate volume of a 0.6000mol sample of gas at 288.15K and a pressure of 1.10atm.

11111nata11111 [884]

Answer:

The volume of the sample of the gas is found to be 12.90 L.

Explanation:

Given pressure of the gas = P = 1.10 atm

Number of moles of gas = n = 0.6000 mole

Temperature = T = 288.15 K

Assuming the volume of the gas to be V liters

The ideal gas equation is shown below

$\textrm{PV} =\textrm{nRT} \\1.10 \textrm{ atm}\times V \textrm{ L} = 0.6000 \textrm{ mole}\times 0.0821 \textrm{ L.atm.mol}^{-1}.K^{-1}\times 288.5\textrm{K} \\\textrm{V} = 12.90 \textrm{ L}$

Volume occupied by gas = 12.90 L

6 0

3 years ago