What are follicles?

If the partial pressure of oxygen in a scuba air tank is 73.44 atm, the partial pressure of nitrogen is 128.52 atm, and the part

Ksenya-84 [330]

To answer this, we use Raoult's Law where the partial pressure is equal to the product of the fraction of the gas in the mixture and its total pressure. Also, we use Dalton's Law of Partial Pressure where the total pressure is equal to the sum of the partial pressure of the gases in the mixture.

Ptotal = 73.44 + 128.52 + 2.04 = 204 atm

201.96 = x (204)
x = 0.99

4 0

3 years ago

A 10.0 mL sample of 0.75 M CH3CH2COOH(aq) is titrated with 0.30 M NaOH(aq) (adding NaOH to CH3CH2COOH). Determine which region o

Llana [10]

Answer:

5.75

Explanation:

Weak acid and strong base,

Make it a two part problem

The first will be Partial neutralization of the weak acid, while the second is the equilibrium and final pH

1.Parameters

HC2H3O2 =10mL x 0.75M = 7.5 mmol

NaOH =22mL x 0.30M = 6.6 mmol

R HC2H3O2 + OH- --> C2H3O2- + H2O

I 7.5 mmol 6.6 mmol 0 mmol ignore

C -6.6 mmol -6.6 mmol +6.6 mmol

E 0.9 mmol 0.0 mmol 6.6 mmol

2.) HC2H3O2 --> H+ + C2H3O2-

Initial concentrations

[HC2H3O2]: 0.9 mmol / 32mL = 0.0281M

[H+]:

[C2H3O2-]: 6.6 mmol x 32mL = 0.206M

Concentrations at equilibrium

[HC2H3O2]: 0.0281M - x

[H+]: [C2H3O2-]: 0.206M + x

If x is small and can be ignored except [H+]

Substitute and solve

1.3 x 10-5 = (x)(0.206)

(0.0281)

X= 1.77 x 10-6M => pH = 5.75

3 0

3 years ago

The formula of nitrogen oxide is NO, of nitrogen dioxide is NO_2. Write a balanced equation for the reaction of nitrogen oxide w

denis23 [38]

Answer:

a) 0,5 mol O₂; 1 mol NO₂

b)

Liters of NO = 22,4 L

Liters of O₂ = 11,2 L

Liters of NO₂ = 22,4 L

c) NO = 30g

O₂ = 16g

NO₂ = 46g

d) 7,41g HCl

e) 107,7 g/mol

f) 0,0312 moles of O₂; 0,999g of O₂; 699 mL at STP or 803 mL

ii. 51%

g) 32,6L

Explanation:

a) For the reaction:

2 NO + O₂ → 2 NO₂

For 1 mole of NO there are consumed:

1 mol NO × $\frac{1molO_{2}}{2molNO}$ = 0,5 mol of O₂

And produced:

1 mol NO × $\frac{2molNO_{2}}{2molNO}$ = 1 mol of NO₂

b) By ideal gas law:

V = nRT/P

Where n is moles of each compound; R is gas constant (0,082atmL/molK); T is temperature (273,15 K at state conditions); P is pressure (1 atm at STP) and V is volume in liters. Replacing each moles for each compound:

Liters of NO = 22,4 L

Liters of O₂ = 11,2 L

Liters of NO₂ = 22,4 L

c) The mass of each compound are:

1 mol NO× $\frac{30 g}{1molNO}$ = 30g

0,5 mol O₂× $\frac{32 g}{1molO_{2}}$ = 16g

1 mol NO₂× $\frac{46 g}{1molNO_{2}}$ = 46g

d) Using:

n = PV / RT

Moles of 4,55 L of HCl (using the values of P = 1 atm; R = 0,082atmL/molK; T = 273,15K) are:

0,203 moles of HCl. In grams:

0,203 mol HCl× $\frac{36,46 g}{1molHCl}$ = 7,41 g of HCl

e) Using:

δRT/P = MW

Where δ is density in g/L (4,81 g/L); R is gas constant (0,082atmL/molK); T is temperature (273,15K); P is pressure (1 atm)

And MW is molecular mass: 107,7 g/mol

f) For the reaction:

2 KClO₃ → 2 KCl + 3 O₂

2,550 g of KClO₃ are:

2,550 g of KClO₃× $\frac{1mol}{122,55 gKClO_{3}}$ = 0,0208 moles of KClO₃

When these moles reacts completely produce:

0,0208 moles of KClO₃× $\frac{3 mol O_{2}}{2 molKClO_{3}}$ = 0,0312 moles of O₂

In grams:

0,0312 moles of O₂ × $\frac{32g}{1 molO_{2}}$ = 0,999g of O₂

V = nRT/P

At STP, n = 0,0312 mol; R = 0,082atmL/molK;T= 273,15K; P = 1atm; V = 0,699L ≡ 699mL

At 29 °C (302,15K) and 732 torr (0,963 atm)

V = 0,803L ≡ 803mL

ii. 182 mL ≡ 0,182L of O₂ are:

n = PV/RT

moles of O₂ are 7,07x10⁻³. Moles of KClO₃ are:

7,07x10⁻³ moles of O₂× $\frac{2 mol KClO_{3}}{3 molO_{2}}$ = 0,0106 mol KClO₃. In grams:

0,0106 moles of KClO₃× $\frac{122,55 g}{1molKClO_{3}}$ = 1,300 g of KClO₃.

Thus, percent by mass of KClO₃ in the mixture is:

1,300g/2,550g ×100 = 51%

g. Combined gas law says that:

$\frac{P_{1}V_{1}}{T_{1}} =\frac{P_{2}V_{2}}{T_{2}}$

Where:

P₁ = 755 torr; V₁ = 35,9L; T₁ = 26°C (299,15 K); P₂ = 760 torr (STP): T₂ = 273,15K (STP) V₂ = 32,6 L

I hope it helps!

4 0

3 years ago

How many moles of co2 are produced when 35.0 mol c2h2 react completely?

Setler [38]

The balanced chemical reaction will be:

C2H2 + 3/2O2 = 2CO2 +H2O

We are given the amount of C2H2 to be burned. This will be our starting point.

35.0 mol C2H2 ( 2 mol CO2/1 mol C2H2) = 70.0 mol O2

Therefore, the CO2 produced is 70.0 moles.

7 0

3 years ago

Read 2 more answers

chlorine boils at 239 k. what is the boiling point of chlorine expressed in degrees celsius? a. 93°c b. 34°c c. –61°c d. –34°c

tankabanditka [31]

Answer: Option (d) is the correct answer.

Explanation:

It is given that boiling point of chlorine is 239 K. To convert temperature in degree Celsius, use the formula as follows.

Temperature in Kelvin = Temperature in degree Celsius + 273

or, Temperature in degree Celsius = Temperature in Kelvin - 273

= 239 - 273

= $- 34^{0}C$

Thus, we can conclude that when chlorine boils at 239 K then boiling point of chlorine expressed in degrees Celsius is $- 34^{0}C$ .

7 0

4 years ago

Read 2 more answers