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ryzh [129]
3 years ago
11

A 200.0mL closed flask contains 2.000mol of carbon monoxide gas and 2.000mol of oxygen gas at the temperature of 300.0K. How man

y moles of oxygen have to react with carbon monoxide in order to decrease the overall pressure in the flask by 10%? Assume ideal gas behavior. The reaction of carbon monoxide and oxygen gas is described by the following equation.
2CO(g)+O2(g)⟶2CO2(g)
Chemistry
1 answer:
max2010maxim [7]3 years ago
6 0

Answer:

There will react 0.400 moles of oxygen.

Explanation:

<u>Step 1:</u> Data given

Volume of the closed flask = 200.00 mL = 0.2 L

Number of moles of CO = 2.000 mol

Number of moles of O2 = 2.000 mol

Temperature = 300.0 K

Pressure decreases with 10%

<u>Step 2</u>: The balanced equation

2CO(g)+O2(g)⟶2CO2(g)

<u>Step 3</u>: Calculate the initial pressure of the flask before the reaction

P = nRT/V

⇒ with n = the number of moles (2.000 moles CO + 2.000 moles O2 = 4.000 moles)

⇒ R is gas constant (0.08206 atm*L/mol*K)

⇒T = the  temperature = 300.0K

⇒ V = the volume = 200.0 mL = 0.2 L

P = (4 * 0.08206*300)/0.2

P = 492.36 atm

<u>Step 4:</u> When the pressure is 10 % decreased:

The final pressure = 492.36 - 49.236 = 443.124 atm

<u>Step 5:</u> Calculate the number of moles

n = PV/RT

⇒ with n = the number of moles

⇒ with P = the pressure = 443.124 atm

⇒ V = the volume = 200.0 mL = 0.2 L

⇒ R is gas constant (0.08206 atm*L/mol*K)

⇒T = the  temperature = 300.0K

n =(443.124*0.2)/(0.08206*300)

n = 3.6 moles = total number of moles

<u>Step 6:</u> Calculate number of moles

For the reaction :2CO(g) + O₂(g) ⟶ 2CO₂(g)

For each mole of O2 we have 2 moles of CO, to produce 2 moles of CO2

Moles CO = (2 -2X) moles

Moles O2 = (2-X) moles

Moles CO2 = 2X

The total number of moles (4 -X)= 3.6 moles

Where X are moles that react

X = 0.400 moles

There will react 0.400 moles of oxygen.

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laiz [17]

Answer:

0.1 g/dl

Explanation:

The standard curve is a graph that relates the absorbance at 400 nm with the concentration of haemoglobin in mg/dl. To obtain the concentration from the absorbance value, we enter in the x-axis (absorbance at 400 nm) with the value 0.40 (the line between 0.2 and 0.6), we extrapolate the line to the curve and read the correspondent value on y-axis (concentration in mg/dl): 100 mg/dl.

So, we convert the concentration from mg/dl to g/dl by dividing into 1000:

100 mg/dl x 1 g/1000 mg = 0.1 g/dl

Therefore, the concentration of haemoglobin of the patient is 0.1 g/dl.

6 0
3 years ago
1. When the temperature of a rigid hollow sphere containing 685 L of helium gas is
mote1985 [20]
<h3>Answer:</h3>

250.756 moles He

<h3>Explanation:</h3>

From the question we are given;

Volume, L = 685 L

Temperature, T = 621 K

Pressure, P = 189 × 10 kPa

We are required to calculate the number of moles of the gas,

Using the Ideal gas equation,

PV = nRT, where P is the pressure, V is the volume, T is the temperature, n is the number of moles, and R is the ideal gas constant.

We can replace the known variables and constant in the equation to get the unknown variable, n.

Using ideal gas constant as 8.3145 L.kPa/K/mol

n=\frac{PV}{RT}

n=\frac{(1890)(685)}{(8.3145)(621)}

n = 250.7557 moles

   n = 250.756 moles

The moles of helium contained in the sphere is 250.756 moles

6 0
3 years ago
Calculate the pH of each solution at 25∘C
JulijaS [17]

Answer: pH of HCl =5, HNO3 = 1,

NaOH = 9, KOH = 12

Explanation:

pH = -log [H+ ]

1. 1.0 x 10^-5 M HCl

pH = - log (1.0 x 10^-5)

= 5 - log 1 = 5

2. 0.1 M HNO3

pH = - log (1.0 x 10 ^ -1)

pH = 1 - log 1 = 1

3. 1.0 x 10^-5 NaOH

pOH = - log (1.0 x 10^-5)

pOH = 5 - log 1 = 5

pH + pOH = 14

Therefore , pH = 14 - 5 = 9

4. 0.01 M KOH

pOH = - log ( 1.0 x 10^ -2)

= 2 - log 1 = 2

pH + pOH = 14

Therefore, pH = 14 - 2 = 12

8 0
3 years ago
Rutherford’s experiment showed that some substances allowed alpha particles to pass through them Frequently.he theorized this ha
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Made of smaller subatomic particles.
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Water has a specific heat of 4.186 J/g*C. How much energy would be required to raise the temperature of 10 g of water by 10 C?
Sladkaya [172]

Energy(heat) required to raise the temperature of water : 418.6 J

<h3>Further explanation  </h3>

Heat can be calculated using the formula:  

Q = mc∆T  

Q = heat, J  

m = mass, g  

c = specific heat, joules / g ° C  

∆T = temperature difference, ° C / K  

Specific heat of water = 4.186 J/g*C.

∆T(raise the temperature) : 10° C  

mass = 10 g

Heat required :

\tt Q=m.c.\Delta T\\\\Q=10\times 4.186\times 10\\\\Q=418.6~J

8 0
3 years ago
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