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3 years ago

Evaluate the carbon dioxide molecule. Explain how to determine if double or triple bonds exist in the molecule.

Chemistry

1 answer:

S_A_V [24]3 years ago

4 0

Answer:

It contain double Bond.

Explanation:

To determine weather the bond is double or triple simply check the electron involved in mutual sharing of an electron if 2 electron takes parts it said to be double or if 3 it said to be triple.

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B. If the initial pressure of gas inside the balloon was 1.65 atm, what is the new pressure?

Orlov [11]

Question 1 :

V1/T1 = V2/T2

3.0L/273K = V2/373K

To get the value of Z, cross multiply

3.0L x 373K = 273K x V2

1119 = 273V2

Divide both sides by 273

1119/273 = 273V2/273

4.10L = V2

The new volume is 4.10 liters

Question 2 :

P1/T1 = P2 /T2

P1 = 880 kPA= 880 *10^3 Pa

T1 = 250 K

T2 = 303 K

P2 =?

Substituting for P2

P2 = P1 T2/ T1

P2 = 880 kPa * 303 / 250

P2 = 266,640 kPa/ 250

P2 = 1066.56 kPa.

The new pressure of the gas is 1066.56 kPa

Question 3 :

Given that:

Volume of gas V = 4.80L

(since 1 liter = 1dm3

4.80L = 4.80dm3)

Temperature T = 62°C

Convert Celsius to Kelvin

(62°C + 273 = 335K)

Pressure P = 2.9 atm

Number of moles of gas N = ?

Apply ideal gas equation

pV = nRT

2.9atm x 4.8dm3 = n x (0.0082 atm dm3 K-1 mol-1 x 335K)

13.92 atm dm3 = nx 2.747 atm dm3 mol-1

n = 13.92/2.747

n = 5.08 moles

There are 5.08 moles of gas contained in the sample

Question 4 :

Volume of gas V = 3.47L

(since 1 liter = 1dm3

3.47L = 3.47dm3)

Temperature T = 85.0°C

Convert Celsius to Kelvin

(85.0°C + 273 = 358K)

Pressure P = ?

Number of moles of gas N = 0.100 mole

Apply ideal gas equation

pV = nRT

p x 3.47dm3 = 0.10 x (0.0082 atm dm3 K-1 mol-1 x 358K)

p x 3.47dm3 = 0.29 atm dm3

p = (0.29 atm dm3 / 3.47 dm3)

p = 0.085 atm

If 1 atm = 760 mm Hg

0.085atm = 0.085 x 760

= 64.6 mm Hg

The pressure of the gas is 64.6 mm hg

8 0

3 years ago

By pushing backwards a roller skater moves forward. This is an example of Newton's

joja [24]

Answer:

newtons law of roller skated duh

7 0

4 years ago

PLEASE HELP ME ASAP! CHEMISTRY TUTOR SEE ATTACHED

Masteriza [31]

Answer:

$\large \boxed{\text{-827.4 kJ}}$

Explanation:

We have three equations:

1. 2H₂S(g) + O₂(g) ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ

2. S(s, rhombic) + O₂(g) ⟶ SO₂(g); ∆H = -296.8 kJ

3. PbO(s) + H₂S(g) ⟶ PbS(s) + SO₂(g); ∆H = -104.3 kJ

From these, we must devise the target equation:

4. 2PbS(s) + 3O₂(g) ⟶2PbO(s) + 2SO₂(g); ΔH = ?

The target equation has PbS(s) on the left, so you reverse Equation 3 and double it.

When you reverse an equation, you reverse the sign of its ΔH.

When you double an equation, you double its ΔH.

5. 2PbS(s) + 2H₂O(g) ⟶ 2PbO(s) + 2H₂S(g); ∆H = 208.6 kJ

Equation 5 has 2H₂O on the left. That is not in the target equation.

You need an equation with 2H₂O on the right, so you copy Equation 1.

6. 2H₂S(g) + O₂(g) ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ

Equation 6 has 2S(s, rhombic) on the right. That is not in the target equation.

You need an equation with 2S(s, rhombic) on the left, so you double Equation 2.

7. 2S(s, rhombic) + 2O₂(g) ⟶ 2SO₂(g); ∆H = -593.6 kJ

Now, you add equations 5, 6, and 7, cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you add their ΔH values.

You get the target equation 4:

5. 2PbS(s) + 2H₂O(g) ⟶ 2PbO(s) + 2H₂S(g); ∆H = 208.6 kJ

6. 2H₂S(g) + O₂(g) ⟶ 2S(s) + 2H₂O(g) ; ∆H = -442.4 kJ

7. 2S(s) + 2O₂(g) ⟶ 2SO₂(g); ∆H = -593.6 kJ

4 . 2PbS(s) + 3O₂(g) ⟶ 2PbO(s) + 2SO₂(g); ΔH = -827.4 kJ

$\Delta H \text{ for the reaction is $ \large \boxed{\textbf{-827.4 kJ}}$}$

8 0

3 years ago

What is the oxidation number of tin (Sn) in the compound Na2SnO2? A. -2 B. 0 C. +2 D. +3

Allisa [31]

O.N. of Na = +1

O.N. of O = -2

Let, O.N. of Tin = x

1*2 + x + -2*2 = 0

2+x-4 = 0

x-2 = 0

x = 2

SO OPTION C IS YOUR ANSWER......

7 0

3 years ago

Read 2 more answers

Radio waves travel at the speed of light which is 3.00 x 10^8. How many minutes does it take for a radio message to reach saturn

ryzh [129]

Answer:

43.89 min

Explanation:

Given that:-

The speed of light = $3.00\times 10^8\ m/s$

The distance = $7.9\times 10^8\ km$

The conversion of distance in km to distance into m is shown below as:-

1 km = 1000 m

So,

Distance = $7.9\times 10^8\times 1000\ m=7.9\times 10^{11}\ m$

The relation between speed distance and time is shown below as:-

$Speed=\frac{Distance}{Time}$

Thus,

$3.00\times 10^8=\frac{7.9\times 10^{11}}{Time}$

$300000000\times time=10^{11}\times \:7.9\ s$

Time = 2633.33 seconds

Also, 1 s = 1/60 min

So,

Time= $\frac{2633.33}{60}\ min=43.89\ min$

3 0

3 years ago