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3 years ago

How much did one drop of HNO3 cause the pH of water to change?

Chemistry

2 answers:

Scrat [10]3 years ago

7 0

Answer:

-4

Explanation:

It went from a 7 to a 2 and 7-2 is 4, so it changed by 4 and it went down by 4.

ankoles [38]3 years ago

6 0

Assuming that there is missing information in this question I’m just going to go ahead and say that it’s lower.

HNO3 (nitric acid) is a very strong acid which should be found on the lower end of the pH scale. While water, on the other hand, is neutral and found at a pH of 7.

So if you add one drop of a strong acid into a neutral solution of 100ml let’s say, one can assume that the pH of that solution is now very low and probably found between 1-3 on a pH scale.

Reminder:

-Acids are found between 1-6 on the pH scale. The lower the number the stronger the acid

-Bases are found between 8-14 on the pH scale. The higher the number the stronger the base

You might be interested in

What is the electron configuration for N (nitrogen)?

ollegr [7]

Answer:

$1s^22s^22p^3$

Explanation:

Nitrogen has the atomic number = 7

So, No. of electrons = 7

Electronic Configuration:

$1s^22s^22p^3$

Remember that:

s sub shell holds upto 2 electrons while p sub shell upto 6

7 0

3 years ago

PLEASE HELP ME ASAP! CHEMISTRY TUTOR SEE ATTACHED

Masteriza [31]

Answer:

$\large \boxed{\text{-827.4 kJ}}$

Explanation:

We have three equations:

1. 2H₂S(g) + O₂(g) ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ

2. S(s, rhombic) + O₂(g) ⟶ SO₂(g); ∆H = -296.8 kJ

3. PbO(s) + H₂S(g) ⟶ PbS(s) + SO₂(g); ∆H = -104.3 kJ

From these, we must devise the target equation:

4. 2PbS(s) + 3O₂(g) ⟶2PbO(s) + 2SO₂(g); ΔH = ?

The target equation has PbS(s) on the left, so you reverse Equation 3 and double it.

When you reverse an equation, you reverse the sign of its ΔH.

When you double an equation, you double its ΔH.

5. 2PbS(s) + 2H₂O(g) ⟶ 2PbO(s) + 2H₂S(g); ∆H = 208.6 kJ

Equation 5 has 2H₂O on the left. That is not in the target equation.

You need an equation with 2H₂O on the right, so you copy Equation 1.

6. 2H₂S(g) + O₂(g) ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ

Equation 6 has 2S(s, rhombic) on the right. That is not in the target equation.

You need an equation with 2S(s, rhombic) on the left, so you double Equation 2.

7. 2S(s, rhombic) + 2O₂(g) ⟶ 2SO₂(g); ∆H = -593.6 kJ

Now, you add equations 5, 6, and 7, cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you add their ΔH values.

You get the target equation 4:

5. 2PbS(s) + 2H₂O(g) ⟶ 2PbO(s) + 2H₂S(g); ∆H = 208.6 kJ

6. 2H₂S(g) + O₂(g) ⟶ 2S(s) + 2H₂O(g) ; ∆H = -442.4 kJ

7. 2S(s) + 2O₂(g) ⟶ 2SO₂(g); ∆H = -593.6 kJ

4 . 2PbS(s) + 3O₂(g) ⟶ 2PbO(s) + 2SO₂(g); ΔH = -827.4 kJ

$\Delta H \text{ for the reaction is $ \large \boxed{\textbf{-827.4 kJ}}$}$

8 0

3 years ago

What is the mass of 1.72 moles of sodium nitrate? Use the periodic table and the polyatomic ion resource. A. 85.0 g B. 91.2 g C.

Mrac [35]

The answer is C. 146g because you add all of the masses of the individual elements and then mulyiply by 1.72 to get your answer.

7 0

3 years ago

Read 2 more answers

What temperature would 3.54 moles of xenon gas need to reach to exert a pressure of 1.57 atm at a volume of 34.6 l

ira [324]

Answer:

186.9Kelvin

Explanation:

The ideal gas law equation is PV = n R T

where

P is the pressure of the gas

V is the volume it occupies

n is the number of moles of gas present in the sample

R is the universal gas constant, equal to 0.0821 atm L /mol K

T is the absolute temperature of the gas

Ensure units of the volume, pressure, and temperature of the gas correspond to R ( the universal gas constant, equal to 0.0821 atm L /mol K )

n = 3.54moles

P= 1.57

V= 34.6

T=?

PV = n R T

PV/nR = T

1.57 x 34.6/3.54 x 0.0821

54.322/0.290634= 186.908620464= T

186.9Kelvin ( approximately to 1 decimal place)

5 0

3 years ago

How many atoms would fit across a human hair? *

kow [346]

A typical human hair is about 1 million carbon-12 atoms JUST IN WIDTH.

4 0

3 years ago

Read 2 more answers