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larisa [96]
4 years ago
15

onsider two aqueous solutions of nitrous acid (HNO2). Solution A has a concentration of [HNO2]= 0.55 M and solution B has a conc

entration of [HNO2]= 1.55 M . You may want to reference (Page 743) Section 16.6 while completing this problem. Part A Which statement about the two solutions is true? Which statement about the two solutions is true? Solution A has the higher percent ionization and solution B has the higher pH. Solution B has the higher percent ionization and the higher pH. Solution B has the higher percent ionization and solution A has the higher pH. Solution A has the higher percent ionization and the higher pH.
Chemistry
2 answers:
azamat4 years ago
8 0

Answer:

Solution A has the higher percent ionization and the higher pH.

Explanation:

Percent ionization depends on the concentration of acid in a solution. If the solution having more concentration of acid so the percent ionization will be lower while if the solution have low amount of acid i. e. dilute solution so the percent ionization will be higher. In solution A, the concentration of HNO2 is lower which is an acid so the percent ionization is higher and the pH of the solution is also higher as compared to solution B.

Aleonysh [2.5K]4 years ago
4 0

Answer:

Solution A has the higher percent ionization and the higher pH

Explanation:

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4 0
3 years ago
Please show work
sasho [114]

Answer:

The pressure is 16, 9 atm

Explanation:

We use the formula PV=nRT. The temperature in Kelvis is: 273 + 25 = 298K

PV=nRT  P =(nRT)/V

P= (0, 450 mol x 0,082 l atm/K mol x 298)/0,650 l= 16, 91723077atm

3 0
4 years ago
Please somebody give me the answers
RSB [31]

Answer:

1. 9.4 grams of methane produce<u> 25.85</u> grams of CO2

2.Grams of water produced = <u>11.81 grams</u>

3.Mass of Methane produced by 10.1 gram of O2 = <u>2.52 grams</u>

4.Amount of methane consumed = <u>46.9 grams</u>

5. Grams of Co2 produced =<u> 8.32 grams</u>

<u></u>

Explanation:

Molar masses :

Methane = CH4 = mass of C + 4x (mass of H)

CH4 = 12 +4(1) = 16 grams

<u>1 mole of CH4 = 16 gram</u>

Oxygen O2 = 2 x (mass of O) = 2x(16) = 32 gram (1 mole of O2 =32 gram)

Carbon Dioxide =CO2 = mass of C + 2(mass of O)

= 12 + 2(16)

= 44 grams <u>(1 mole of CO2 = 44 gram )</u>

Water = H2O = 18 grams ( 1 mole of H2O = 18 gram)

1 mole of each molecule is equal to their molar masses

The balanced equation is :

1CH_{4}(g)+2O_{2}\rightarrow 1CO_{2}+2H_{2}O(l)

According to Stoichiometry :

1 mole of CH4 = 2 Mole of O2 = 1 mole of CO2 = 2 mole of H2O

1. From the equation ,

1 mole of methane produce  =1 mole of CO2

16 gram of methane = 44 gram of CO2

1 gram of methane =

\frac{44}{16} gram of CO2

9.4 gram of CH4 =

\frac{44}{16}\times 9.4 gram of CO2

= 25 .85 gram of CO2

2.

2 mole of O2 produces = 2 mole of H2O(water)

1 mole of O2 produces = 1 mole of H2O

32 gram of O2 = 18 gram of water

1 gram of O2 =

\frac{18}{32}

21 gram of O2 =

\frac{18}{32}\times 21

11.81 gram of water

3. 1 mole of CH4 = 2 mole of O2

16 gram of CH4 = 2(32)  = 64 grams of O2

64 gram of O2 needs = 16 grams of CH4

1 gram of O2 needs =

\frac{16}{64}

10.1 gram of O =

\frac{16}{64}\times 10.1 of CH4

= 2.52 gram

4.

1 mole of CO2 is produced from = 1 mole of CH4

44 gram of CO2 is produced from 16 gram of CH4

1 gram CO2 =

\frac{16}{44} gram of CH4

129 gram of CO2 =

\frac{16}{44}\times 129 gram of CH4

= 46.90 grams

5.

2 mole of O2  produce = 1 mole of CO2

2x 32 gram of O2 = 44 gram of CO2

1 gram of O2 =

\frac{44}{64} of CO2

12.1 gram of O2 produce=

\frac{44}{64}\times 12.1 of CO2

= 8.318 gram

Note : Write the quantity give on left side of "="

write the substance asked on right side of "="

8 0
3 years ago
Calculate the mass of glucose metabolized by a 60.0 −kg person in climbing a mountain with an elevation gain of 1550 m . Assume
lbvjy [14]

Answer:

Mass of glucose = 515.34 g

Explanation:

We are given;

Mass; m = 60 kg

Elevation; h = 1550 m

Acceleration due to gravity; 9.8 m/s²

Now, work performed to lift 60kg by 1550m is given by the formula;

W = mgh

W = 60 × 9.8 × 1550

W = 911400 J

We are told the actual work is 4 times the one above.

Thus;

Actual work = 4W = 4 × 911400 = 3,645,600 J

Now,

Molar mass of Glucose(C6H12O6) = 180 g/mol

We are given standard enthalpy of combustion = -1273.3 KJ/mol = -1273300

Moles of glucose = 3645600/1273300 = 2.863mol

Mass of glucose = 2.863 mol × 180 g/mol

Mass of glucose = 515.34 g

4 0
4 years ago
I need help with that question please ASAP
Sonbull [250]
Load is the answer..
3 0
3 years ago
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