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scZoUnD [109]
3 years ago
15

A strong acid with a concentration of 0.01 Mol/L would have a pH of __________.

Chemistry
1 answer:
BARSIC [14]3 years ago
5 0

Answer:

2

Explanation:

the pH of the strong acid solution will be

pH = -log[H+] = -log[0.01] =  2

You might be interested in
Boyles Law P1V1 = P2V2
arsen [322]

Answer:

A. The balloons will increase to twice their original volume.

Explanation:

Boyle's law states that the pressure exerted on a gas is inversely proportional to the volume occupied by the gas at constant temperature. That is:

P ∝ 1/V

P = k/V

PV = k (constant)

P = pressure, V = volume.

P_1V_1=P_2V_2

Let the initial pressure of the balloon be P, i.e. P_1=P, initial volume be V, i.e. V_1=V. The pressure is then halved, i.e. P_2=\frac{P}{2}

P_1V_1=P_2V_2\\\\P*V=\frac{P}{2} *V_2\\\\V_2=\frac{2*P*V}{P}\\\\V_2=2V

Therefore the balloon volume will increase to twice their original volume.

3 0
3 years ago
Match the following aqueous solutions with the appropriate letter from the column on the right.1. 0.19 m AgNO3 2. 0.17 m CrSO4 3
vichka [17]

Answer:

0.13 m of Mn(NO_3)_2 → Highest boiling point

0.19 m of AgNO_3 → Second  Highest boiling point

0.17 m of CrSO_4 → Third highest boiling point

0.31 m Sucrose (nonelectrolyte)  → Lowest boiling point

Explanation:

Elevation in boiling is given by :

\Delta T_b=i\times k_b\times m

Where :

i = van't Hoff factor

k_b= Molal Elevation constant of solvent

m = molaity of the solution

1) 0.19 m of AgNO_3

AgNO_3\rightarrow Ag^++NO_3^{-}

i = 2 (electrolyte)

Molality of the solution = 0.19

Elevation is boiling point of solution:

\Delta T_b=2\times k_b\times 0.19 m

\Delta T_b=0.38 m\times k_b

2) 0.17 m of CrSO_4

CrSO_4\rightarrow Cr^{2+}+SO_4^{2-}

i = 2 (electrolyte)

Molality of the solution = 0.17

Elevation is boiling point solution :

\Delta T_b=2\times k_b\times 0.17 m

\Delta T_b=0.34 m\times k_b

3) 0.13 m of Mn(NO_3)_2

Mn(NO_3)_2\rightarrow Mn^{2+}+2NO_3^{-}

i = 3 (electrolyte)

Molality of the solution = 0.13

Elevation is boiling point solution :

\Delta T_b=3\times k_b\times 0.13 m

\Delta T_b=0.39 m\times k_b

4) 0.31 m Sucrose (nonelectrolyte)

i = 1 ( non electrolyte)

Molality of the solution = 0.31 m

Elevation is boiling point solution :

\Delta T_b=1\times k_b\times 0.31 m

\Delta T_b=0.31 m\times k_b

Higher the value of elevation in temperature higher will be the boiling point of the solution .

The decreasing order of solution from highest boiling point to lowest boiling point is :

0.39 m\times k_b>0.38 m\times k_b>0.34 m\times k_b>0.31 m\times k_b

0.13 m of Mn(NO_3)_2 → Highest boiling point

0.19 m of AgNO_3 → Second  Highest boiling point

0.17 m of CrSO_4 → Third highest boiling point

0.31 m Sucrose (nonelectrolyte)  → Lowest boiling point

6 0
3 years ago
Ionic compounds involve a (share, tranfer) of electrons and typically involve atoms that have (Large, similar, very different) e
Juliette [100K]

Answer:

2 Transfer; very different

Explanation:

Ionic compound are compound that are formed from transfer of electron. The cations donate the electron(s) while the anion receive the electron(s). The compound possess a positively charged end and a negatively charge end. Example of ionic compound is NaCl . The sodium donates one electron to the chlorine  to fulfill the octet rule. The sodium atom becomes positively charged  as it donate electron to the chlorine atom. The chlorine atom becomes negatively charged as it receive electron from Sodium atom.

The atoms that are involve in forming ionic compound have very different electronegativity . Electronegativity  is the measure of the ability of an atom to attract a bond pair of electron. For example the NaCl compound, the two atom has verse differences in the electronegativity. Cl atom is far more electronegative than Sodium atom.

6 0
3 years ago
2. Matt went to his friend’s party. He ate a big meal and drank a keg of beer. He felt heartburn after the meal and took Tums to
evablogger [386]

Answer:

390.85mL

Explanation:

Step 1:

Data obtained from the question.

Initial pressure (P1) = 780 torr

Initial volume (V1) = 400mL

Initial temperature (T1) = 40°C = 40°C + 273 = 313K

Final temperature (T2) = 25°C = 25°C + 273 = 298K

Final pressure (P2) = 1 atm = 760torr

Final volume (V2) =?

Step 2:

Determination of the final volume i.e the volume of the gas outside Matt's body.

The volume of the gas outside Matt's body can be obtained by using the general gas equation as shown below:

P1V1/T1 = P2V2/T2

780 x 400/313 = 760 x V2 /298

Cross multiply to express in linear form

313 x 760 x V2 = 780 x 400 x 298

Divide both side by 313 x 760

V2 = (780 x 400 x 298) /(313 x 760)

V2 = 390.85mL

Therefore, the volume of the gas outside Matt's body is 390.85mL

3 0
3 years ago
You suspect one of the products of a single replacement is hydrogen gas. How might you experimentally confirm your prediction? C
Nataly_w [17]
The best option i think is b would be my best answer
6 0
3 years ago
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