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igor_vitrenko [27]
3 years ago
8

Use the ideal gas law to calculate the pressure in atmospheres of 0.21 mol of helium (He) at 16°C & occupying 2.53 L. You mu

st show all of your work to earn credit. **Don't forget to convert your Celsius temperature to Kelvin**
*picture attached !!*

Chemistry
1 answer:
chubhunter [2.5K]3 years ago
4 0

Answer:

The answer to your question is 2.32 atm

Explanation:

Data

P = ?

n = 0.214

V = 2.53 L

T = 61°C

R = 0.082 atm L/mol°K

Formula

PV = nTR

solve for P

P = nRT/V

Process

1.- Calculate the temperature in K

°K = °C + 273

°K = 61 + 273

    = 334

2.- Substitution

P = (0.214 x 0.082 x 334) / 2.53

3.- Simplification

P = 5.86/2.53

4.- Result

P = 2.32 atm

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pentagon [3]

Answer:

1.  H₂SO₄  +  Na₂CO₃  ⇒  Na₂SO₄  +  H₂O  +  CO₂

sulfuric acid + sodium carbonate ⇒ sodium sulfate + water + carbon dioxide

2.  2 HCl  +  Mg(HCO₃)₂  ⇒  MgCl₂  +  2 H₂O  +  2 CO₂

hydrochloric acid + magnesium hydrogencarbonate ⇒ magnesium chloride + water + carbon dioxide

Explanation:

1.  Sulfuric acid is H₂SO₄.  Sodium carbonate is Na₂CO₃.

H₂SO₄  +  Na₂CO₃  ⇒  Na₂SO₄  +  H₂O  +  CO₂

sulfuric acid + sodium carbonate ⇒ sodium sulfate + water + carbon dioxide

2.  Hydrochloric acid is HCl.  Magnesium hydrogencarbonate is Mg(HCO₃)₂.

2 HCl  +  Mg(HCO₃)₂  ⇒  MgCl₂  +  2 H₂O  +  2 CO₂

hydrochloric acid + magnesium hydrogencarbonate ⇒ magnesium chloride + water + carbon dioxide

3 0
3 years ago
What is a nonpolar bond?
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8 0
3 years ago
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Why do thermistors increase in conductivity when heated? What happens in normal metals? Explain on the atomic level.
Marysya12 [62]

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7 0
2 years ago
The respiration rate of a goldfish is measured. The goldfish is then placed in cold water and the respiration rate is measured a
Ivenika [448]

Answer:

Temperature of the water

Explanation:

In every study, there must be independent and dependent variables. An independent variable is the variable that is changed in order to obtain a response. In this case, the temperature of the water is being changed, the response in this experiment is the respiration rate of the goldfish.

Thus the respiration rate of the goldfish is the dependent variable because it is controlled by the temperature of the water and changes accordingly.

Summarily, the independent variable is the temperature of the water while the dependent variable is the respiration rate of the goldfish.

4 0
3 years ago
A 20.0 mL 0.100 M solution of lactic acid is titrated with 0.100 M NaOH.
yan [13]

Answer:

(a) See explanation below

(b) 0.002 mol

(c) (i) pH = 2.4

(ii) pH = 3.4

(iii) pH = 3.9

(iv) pH = 8.3

(v) pH = 12.0

Explanation:

(a) A buffer solution exits after addition of 5 mL of NaOH  since after reaction we will have  both the conjugate base lactate anion and unreacted weak  lactic acid present in solution.

Lets call lactic acid HA, and A⁻ the lactate conjugate base. The reaction is:

HA + NaOH ⇒ A⁻ + H₂O

Some unreacted HA will remain in solution, and since HA is a weak acid , we will have the followin equilibrium:

HA  + H₂O ⇆ H₃O⁺ + A⁻

Since we are going to have unreacted acid, and some conjugate base, the buffer has the capacity of maintaining the pH in a narrow range if we add acid or base within certain limits.

An added acid will be consumed by the conjugate base A⁻ , thus keeping the pH more or less equal:

A⁻ + H⁺ ⇄ HA

On the contrary, if we add extra base it will be consumed by the unreacted lactic acid, again maintaining the pH more or less constant.

H₃O⁺ + B ⇆ BH⁺

b) Again letting HA stand for lactic acid:

mol HA =  (20.0 mL x  1 L/1000 mL) x 0.100 mol/L = 0.002 mol

c)

i) After 0.00 mL of NaOH have been added

In this case we just have to determine the pH of a weak acid, and we know for a monopric acid:

pH = - log [H₃O⁺] where  [H₃O⁺] = √( Ka [HA])

Ka for lactic acid = 1.4 x 10⁻⁴  ( from reference tables)

[H₃O⁺] = √( Ka [HA]) = √(1.4 x 10⁻⁴ x 0.100) = 3.7 x 10⁻³

pH = - log(3.7 x 10⁻³) = 2.4

ii) After 5.00 mL of NaOH have been added ( 5x 10⁻³ L x 0.1 = 0.005 mol NaOH)

Now we have a buffer solution and must use the Henderson-Hasselbach equation.

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.0005                0

after rxn    0.002-0.0005                  0                  0.0005

                        0.0015

Using Henderson-Hasselbach equation :

pH = pKa + log [A⁻]/[HA]

pKa HA = -log (1.4 x 10⁻⁴) = 3.85

pH = 3.85 + log(0.0005/0.0015)

pH = 3.4

iii) After 10.0 mL of NaOH have been ( 0.010 L x 0.1 mol/L = 0.001 mol)

                             HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.001               0

after rxn        0.002-0.001                  0                  0.001

                        0.001

pH = 3.85 + log(0.001/0.001)  = 3.85

iv) After 20.0 mL of NaOH have been added ( 0.002 mol )

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.002                 0

after rxn                 0                         0                   0.002

We are at the neutralization point and  we do not have a buffer anymore, instead we just have  a weak base A⁻ to which we can determine its pOH as follows:

pOH = √Kb x [A⁻]

We need to determine the concentration of the weak base which is the mol per volume in liters.

At this stage of the titration we added 20 mL of lactic acid and 20 mL of NaOH, hence the volume of solution is 40 mL (0.04 L).

The molarity of A⁻ is then

[A⁻] = 0.002 mol / 0.04 L = 0.05 M

Kb is equal to

Ka x Kb = Kw ⇒ Kb = 10⁻¹⁴/ 1.4 x 10⁻⁴ = 7.1 x 10⁻¹¹

pOH is then:

[OH⁻] = √Kb x [A⁻]  = √( 7.1 x 10⁻¹¹ x 0.05) = 1.88 x 10⁻⁶

pOH = - log (  1.88 x 10⁻⁶ ) = 5.7

pH = 14 - pOH = 14 - 5.7 = 8.3

v) After 25.0 mL of NaOH have been added (

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn           0.002                  0.0025              0

after rxn                0                         0.0005              0.0005

Now here what we have is  the strong base sodium hydroxide and A⁻ but the strong base NaOH will predominate and drive the pH over the weak base A⁻.

So we treat this part as the determination of the pH of a strong base.

V= (20 mL + 25 mL) x 1 L /1000 mL = 0.045 L

[OH⁻] = 0.0005 mol / 0.045 L = 0.011 M

pOH = - log (0.011) = 2

pH = 14 - 1.95 = 12

7 0
3 years ago
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