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Advocard [28]
1 year ago
8

The equation for the pH of a substance is pH = –log[H+], where H+ is the concentration of hydrogen ions. A basic solution has a

pH of 11.2. An acidic solution has a pH of 2.4. Based from the given, what is the approximate difference in the concentration of hydrogen ions between the two solutions? A. 1.6*10^-9 B. 4.0*10^-3 C. 6.7*10^-1 D. 1.6*10^-11
Chemistry
2 answers:
nordsb [41]1 year ago
5 0

Answer:

graph b

Explanation:

Rzqust [24]1 year ago
3 0
For the basic solution:
11.2 = -log[H+]
[H+] = 6.31 x 10⁻¹²
For the acidic solution:
2.4 = -log[H+]
[H+] = 3.98 x 10⁻³
The difference:
3.98 x 10⁻³ - 6.31 x 10⁻¹²
≈ 4.0 x 10⁻³
The answer is B
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A 100. mL sample of 0.200 M aqueous hydrochloric acid is added to 100. mL of 0.200 M aqueous ammonia in a calorimeter whose heat
Free_Kalibri [48]

Answer:

-154KJ/mol

Explanation:

mole of 100ml sample of 0.2M aqueous HCl = Molarity × volume in Liter

= 0.2 × 100 / 1000  ( 1L = 1000 ml) = 0.02 mol and 0.02 mole of HCl solution require 0.02 mole of ammonia according to the mole ratio in the balanced equation.

Heat loss by the reaction = heat gain by calorimeter = mcΔT + 480 J/K

where m is the mass of water = 100g + 100g = 200g since mass of 100ml of water = 100g and it is in both of them and specific heat capacity of water 4.184 J/gK

heat gain by calorimeter  = (4.184 × 200 + 480) × 2.34 = 3081.3 J

ΔH per mole = heat loss / number of mole = 3081.3 / 0.02 = 154065.6 = -154KJ/mol

8 0
1 year ago
Ammonia gas is compressed from 21°C and 200 kPa to 1000 kPa in an adiabatic compressor with an efficiency of 0.82. Estimate the
Evgen [1.6K]

Explanation:

It is known that efficiency is denoted by \eta.

The given data is as follows.

     \eta = 0.82,       T_{1} = (21 + 273) K = 294 K

     P_{1} = 200 kPa,     P_{2} = 1000 kPa

Therefore, calculate the final temperature as follows.

         \eta = \frac{T_{2} - T_{1}}{T_{2}}    

         0.82 = \frac{T_{2} - 294 K}{T_{2}}    

          T_{2} = 1633 K

Final temperature in degree celsius = (1633 - 273)^{o}C

                                                            = 1360^{o}C

Now, we will calculate the entropy as follows.

       \Delta S = nC_{v} ln \frac{T_{2}}{T_{1}} + nR ln \frac{P_{1}}{P_{2}}

For 1 mole,  \Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}

It is known that for NH_{3} the value of C_{v} = 0.028 kJ/mol.

Therefore, putting the given values into the above formula as follows.

     \Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}

                = 0.028 kJ/mol \times ln \frac{1633}{294} + 8.314 \times 10^{-3} kJ \times ln \frac{200}{1000}

                = 0.0346 kJ/mol

or,             = 34.6 J/mol             (as 1 kJ = 1000 J)

Therefore, entropy change of ammonia is 34.6 J/mol.

3 0
1 year ago
I NEED HELP ASAP, WILL MARK BRAINLEST!
Andre45 [30]

Answer:

1. 90%

2. 217.4 g O₂

3. 95.0%

4. Trial 2 ratios

Explanation:

Original: SiCl₄ + O₂ → SiO₂ + Cl₂

Balanced: SiCl₄ + O₂ → SiO₂ + 2Cl₂

Trial        SiCl₄                   O₂                    SiO₂

 1           120 g                  240 g              38.2 g

 2           75 g                   50 g                25.2 g

<u>Percentage yield for trial 1</u>

We need to get actual yield (38.2 g) and theoretical yield, in grams.

Mass to moles:

 molar mass SiCl₄: 28.09 + 4(35.45) = 169.9 g/mol

 120 g SiCl₄ x 1 mol/169.9 g = .706 mol SiCl₄

Moles to moles:

 For each mole SiCl₄, we have one mol SiO₂ based on the balanced rxn.

 .706 mol SiCl₄ = .706 mol SiO₂

Moles to mass:

 molar mass SiO₂: 28.09 + 2(16.00) = 60.09 g/mol

 .706 mol SiO₂ x 60.09g/mol = 42.44 g SiO₂

Theoretical yield:

 actual/theoretical x 100

 38.2 / 42.44 = .900 = <u>90.0% yield</u>

<u>Leftover reactant for trial 1</u>

We know oxygen is the excess reactant.

Mass to moles:

 molar mass O₂ = 32.00 g/mol

 240 g O₂ x 1 mol/32.00 g = 7.5 mol O₂

We used .706 mol SiO₂, so we also used .706 mol O₂.

 7.5 - .706 = 6.8 moles left over

Moles to mass:

 6.8 mol O₂ x 32.00g/mol =<u> 217.4 g O₂</u>

<u />

<u>Percentage yield for trial 2</u>

Mass to moles:

 molar mass SiCl₄: 169.9 g/mol

 75 g SiCl₄ x 1 mol/169.9 g = .441 mol SiCl₄

Moles to moles:

 For each mole SiCl₄, we have one mol SiO₂ based on the balanced rxn.

 .441 mol SiCl₄ = .441 mol SiO₂

Moles to mass:

 molar mass SiO₂: 60.09 g/mol

 .441 mol SiO₂ x 60.09g/mol = 26.5 g SiO₂

Theoretical yield:

 actual/theoretical x 100

 25.2 / 26.5 = .950 = <u>95.0% yield</u>

Because the percentage yield of trial 2 is higher than that of trial 1, we know that the ratio of reactants in trial 2 is more efficient! We got a result closer to our theoretical yield.

6 0
2 years ago
The pka of hf is 3.2 determine the pkb of hf?
Julli [10]

Well, first we must remember that

pK_{a}+pK_{b}=14

This is because

K_{a}*K_{b}=10^{-14}

-log(K_{a}*K_{b})=-log(10^{-14})\\-logK_{a}+-logK_{b}=-log(10^{-14})\\pK_{a}+pK_{b}=14

So then

pK_{b}=14-pK_{a}=14-3.2=1.8

7 0
2 years ago
Read 2 more answers
This same chemistry student has a weight of 155 lbs. What is the student’s weight in
Virty [35]

Answer:

the same chemistry student has a weight of 155lbs what is the student weight in grams? (16 oz= 1 lb, 1 oz= 28.34g)

Explanation:1 lb = 16oz, so multiply your pounds by 16 to get you ounces of the student, then multiply by 28.34 to get grams

155 X 16 X 28.34 = 70283.2

8 0
1 year ago
Read 2 more answers
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