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3 years ago

Calculate the relative density of carbon monoxide and carbon dioxide with respect to air.

1 answer:

3 0

Density is proportional to molar mass, assuming pressure and temperature remain constant. Therefore, since CO has a molar mass of 28 and CO2 has a molar mass of 44:
The relative density of CO vs air is 28/29 = 0.9655.
The relative density of CO2 vs air is 44/29 = 1.517.

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Help please I need help with these two questions answer truthfully get brainliest?!!

irina1246 [14]

Answer:

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Value of x=7

2. 4x-5=23

value of x=7

5 0

2 years ago

Read 2 more answers

I need help with 1,2,3, and 4

Schach [20]

Answer:

Problem 1: 1.85atm

Problem 2: 110mL

Problem 3: 290 mL

Problem 4: 1.14 atm

Explanation:

Problem 1

1. Data

a) P₁ = 3.25atm

b) V₁ = 755mL

c) P₂ = ?

d) V₂ = 1325 mL

r) T = 65ºC

2. Formula

Since the temeperature is constant you can use Boyle's law for idial gases:

$PV=constant\\\\P_1V_1=P_2V_2$

3. Solution

Solve, substitute and compute:

$P_1V_1=P_2V_2\\\\P_2=P_1V_1/V_2$

$P_2=3.25atm\times755mL/1325mL=1.85atm$

Problem 2

1. Data

a) V₁ = 125 mL

b) P₁ = 548mmHg

c) P₁ = 625mmHg

d) V₂ = ?

2. Formula

You assume that the temperature does not change, and then can use Boyl'es law again.

$P_1V_1=P_2V_2$

3. Solution

This time, solve for V₂:

$P_1V_1=P_2V_2\\\\V_2=P_1V_1/P_2$

Substitute and compute:

$V_2=548mmHg\times 125mL/625mmHg=109.6mL$

You must round to 3 significant figures:

$V_2=110mL$

Problem 3

1. Data

a) V₁ = 285mL

b) T₁ = 25ºC

c) V₂ = ?

d) T₂ = 35ºC

2. Formula

At constant pressure, Charle's law states that volume and temperature are inversely related:

$V/T=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

The temperatures must be in absolute scale.

3. Solution

a) Convert the temperatures to kelvins:

T₁ = 25 + 273.15K = 298.15K

T₂ = 35 + 273.15K = 308.15K

b) Substitute in the formula, solve for V₂, and compute:

$\dfrac{V_1}{T_1`}=\dfrac{V_2}{T_2}\\\\\\\\\dfrac{285mL}{298.15K}=\dfrac{V_2}{308.15K}\\\\\\V_2=308.15K\times285mL/298.15K=294.6ml$

You must round to two significant figures: 290 ml

Problem 4

1. Data

a) P = 865mmHg

b) Convert to atm

2. Formula

You must use a conversion factor.

1 atm = 760 mmHg

Divide both sides by 760 mmHg

$\dfrac{1atm}{760mmHg}=\dfrac{760mmHg}{760mmHg}\\\\\\1=\dfrac{1atm}{760mmHg}$

3. Solution

Multiply 865 mmHg by the conversion factor:

$865mmHg\times \dfrac{1atm}{760mmHg}=1.14atm\leftarrow answer$

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3 years ago

Please help me with this question please!!!

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Answer:

Explanation:

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2 years ago

Maltase is an enzyme that breaks down maltose. If a maltase enzyme has just completed catalyzing the decomposition of maltose, t

netineya [11]

Yes it is available. It will continue catalyzing the reactions until it becomes completely consumed. That's how enzymes work. They work and are eventually consumed in the process completely without altering the reaction in any way other than speeding it up.

7 0

3 years ago

Balance the equations. Zn+ HCl → ZnCl2 +

motikmotik

Answer:

Zn + 2HCl → ZnCl2 + H2

Explanation:

Zn + HCl → ZnCl2 +

The complete equation is given below:

Zn+ HCl → ZnCl2 + H2

Now we can balance the equation by doing the following:

There are 2 atoms of Cl and 2 atoms of H on the left. This can be balanced by putting 2 in front of HCl as shown below:

Zn + 2HCl → ZnCl2 + H2

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3 years ago