Classify each as acidic or basic - pH 9 - pH 2.3 - pH 11 - pH 5 - solution where the concentration of hydronium ions [H ] is 0.0056 M - solution where the concentration of hydronium ions [H ] is 3.45 E-9 M.

Answers

Answer 1

Answer:

pH = 9: Basic; pH = 2.3: acidic; pH = 11: basic; pH = 5: Acidic; [H⁺] = 0.0056M: Acidic; [H⁺] = 3.45E-9M: Basic

Explanation:

A solution is defined as acidic when pH < 7 and as basic when pH > 7.

Also, pH = -log[H⁺].

Thus:

pH = 9: >7 → Basic

pH = 2.3: <7 → Acidic

pH = 11: >7 → Basic

pH = 5: <7 → Acidic

[H⁺] = 0.0056M, pH = -log0.0056M = 2.25: <7: Acidic

[H⁺] = 3.45E-9M, pH = 8.46: > 7: Basic


Related Questions

A compound is made of 6.00 grams of oxygen, 7.00 grams of nitrogen, and 20.00grams of hydrogen. Find the percent composition of the compound.

A O-18.18%, N-21.21%, H-60.60%
B O-11.18%, N-22.21%, H-69.60%
C O-20%, N-30%, H-50%
D O-60.60%, N-21.21%, H-18.18%

Answers

The percent composition of the compound.

A O-18.18%, N-21.21%, H-60.60%

Further explanation

Given

6.00 grams of oxygen,

7.00 grams of nitrogen,

20.00 grams of hydrogen.

Required

The percent composition

Solution

Total mass :

= mass of O + mass of N + mass of H

= 6 + 7 + 20

= 33 g

% O = 6/33 x 100%= 18.18%

% N = 7/33 x 100%=21.21%

% H = 20/33 x 100% = 60.6 %

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Answers

Answer:

I don't get it is it even a question?

16. Using the average atomic masses given in the inside front cover of this book, calculate the indicated quantities.

d. the number of moles of cobalt represented by 5.99 x 1021 cobalt atoms e. the mass of 4.23 mol of cobalt

f. the number of cobalt atoms in 4.23 mol of cobalt

g. the number of cobalt atoms in 4.23 g of cobalt

Answers

Answer:

d. 9.95 × 10⁻³ mol

e. 249 g

f. 2.55 × 10²⁴ atoms

g. 4.32 × 10²² atoms

Explanation:

d. the number of moles of cobalt represented by 5.99 x 10²¹ cobalt atoms

We will use Avogadro's number: there are 6.02 × 10²³ atoms of cobalt in 1 mole of atoms of cobalt.

5.99 x 10²¹ atoms × 1 mol/6.02 × 10²³ atoms = 9.95 × 10⁻³ mol

e. the mass of 4.23 mol of cobalt

The molar mass of cobalt is 58.93 g/mol.

4.23 mol × 58.93 g/mol = 249 g

f. the number of cobalt atoms in 4.23 mol of cobalt

We will use Avogadro's number: there are 6.02 × 10²³ atoms of cobalt in 1 mole of atoms of cobalt.

4.23 mol × 6.02 × 10²³ atoms/1 mol = 2.55 × 10²⁴ atoms

g. the number of cobalt atoms in 4.23 g of cobalt

First, we will calculate the moles of cobalt using the molar mass of cobalt.

4.23 g × 1 mol/58.93 g = 0.0718 mol

Then, we will calculate the number of cobalt atoms using Avogadro's number.

0.0718 mol × 6.02 × 10²³ atoms/1 mol = 4.32 × 10²² atoms

Helium on the Moon was found to be 0.420% 2He, 2.75% 3He, and 96.83% 4He. What is the average atomic mass of helium on the Moon?

Answers

Answer:

Average atomic mass  = 3.9 amu

Explanation:

Given data:

Percent abundance of He-2 = 0.420%

Percent abundance of He-3 = 2.75%

Percent abundance of He-4 = 96.83%

Average atomic mass = ?

Solution:

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass)  / 100

Average atomic mass = (0.420×2)+(2.75×3) +(96.83×4)/100

Average atomic mass =  0.84 + 8.25 +387.32 / 100

Average atomic mass = 396.41 / 100

Average atomic mass  = 3.9 amu.

How can you model the cycling of matter in the Earth system?

Answers

Answer:

The cycling of matter is important to many Earth processes and to the survival of organisms the existing matter must cycle continuously for this planet to support life Water, carbon, nitrogen, phosphorus, and even rocks move through cycles If these materials did not cycle, Earth could not support life.

Explanation:

Earth activities depend on matter cycling, and for organisms to survive, this planet's surface must cycle with the flow of matter.

What is Earth system?

Rocks, as well as water, carbon, nitrogen, and phosphorus, go through cycles. The planet Earth could not support life if these materials did not cycle.

Subsystems exist within the Earth system. These subsystems include the exosphere, atmosphere, hydrosphere, lithosphere and geosphere, also referred to as the lithosphere, and the living environment (biosphere).

These systems are powered by energy that comes from both the Sun and the interior of the Earth. Through processes known as biogeochemical cycles, nutrients and elements also move through these systems along with energy.

Therefore,  Earth activities depend on matter cycling, and for organisms to survive, this planet's surface must cycle with the flow of matter.

To learn more about Earth, refer to the link:

https://brainly.com/question/1204146

#SPJ6

A species of desert plant produces flowers that only bloom at night. How does this enhance the survival of the species?

A. This allows the plants to conserve water and not bloom during the heat of the day.

B. This species relies on nocturnal animals like moths for pollination and reproduction

C. This species relies on moonlight for photosynthesis.

D.This allows flowers to stay closed durng the day when herbivores are more likely to eat them.

Od

Answers

Answer:

A. This allows the plants to conserve water and not bloom during the heat of the day

Explanation:

Most desert plants only bloom at night because they take advantage of animals like moths and insects that fly at night for pollination and reproduction.

Because these plants are in the desert and do not get enough water except from short occasional rainfalls, they conserve water and not bloom during the heat of the day. They bloom at night when the temperature is low and this enhances their water conservation and survival.

Why is observational evidence important in an experiment?

Answers

Answer:

Observational evidence is essential for investigating the way disease affects populations, the patterns and distribution of risk within them, and the emergence of trends in health and disease over time.

Answer:

It tests a prediction It supports the results. It asks a testable question It predicts what will happen

Explanation:

When preparing for work in the fume hood, be sure to gather all necessary tools, glassware, and chemicals _________ to minimize the number of times the hood sash is raised and lowered. Work as much as possible in the _________ of the work surface to keep the area tidy and promote air flow. If you need to step away from the experiment to obtain another item, _________ the sash during this time.

Answers

Answer:

In advance

middle

lower

Explanation:

These are the safety precautions needed when carrying out duties in the fume hood.

When planning and preparing to work in a fume hood (a locally designed area to reduce exposure to hazardous fumes). It is advisable to make all equipment readily available at your disposal in advance to reduce and minimize the raising and lowering of the hood sash at intervals.

It is also pertinent to understand that working in the middle of the work surface helps to promote the movement of air and keeps the area neat and tidy.

However, if any case where there is a need to get a new tool or equipment during the process of working in a fume hood, it is advisable to lower the sash at that point in time.

3.4 x 10-25 kg = ? microounces

Answers

Answer: 1.2 x 10^-17 microounces

Explanation:

Ounce = 28.5G microounce = 28.5*10^-6g

3.4*10^-25 kg = 3.4*10^-22 g = (3.4/2.85)*10^(-22+5) = 1.2*10-17

In biology class, we had to______ the parts of a plant cell under a microscope.

Answers

Answer:

In biology class we had to look at the parts of a plant in a microscope

water is a unique material in that the density of the solid is lower than the density of the liquid (which is why ice forms at the top of a pond and why ice floats in our drinks). if the density for ice at 0C is .917g/mL and the density for water at 0C is .999 g/mL, what is the calculated free space (as %) for each of these materials. you will need to estimate the volume of water as the sum of 2 H atoms and 1 O atom with radii 37 and 66 pm respectively. note that you will also have to assume a quantity of water to perform this exercise

Answers

Answer:

% Free space in water = [tex]\frac{9.945* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.45%

% Free space in ice  = [tex]\frac{9.98* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.8%

Explanation:

As given ,

Density for ice at 0⁰C = 0.917 g/ml

Density for water at 0⁰C = 0.999 g/ml

Radii of H atoms = 37 pm

Radii of O atoms = 66 pm

Now,

Consider 1 ml of water = 1 cm²

As , we know that mass of water in 1 cm² = 0.999 g

Moles of water = [tex]\frac{0.999}{18} = 0.056[/tex]

Volume of H₂O = 1.624×[tex]10^{-31}[/tex] m²

Now,

Volume occupied by water = 0.056×6.022×[tex]10^{23}[/tex]× 1.624×[tex]10^{-31}[/tex] m²

                                              = 5.48×[tex]10^{-9}[/tex] m²

⇒Volume occupied by water = 5.48×[tex]10^{-9}[/tex] m²

Now,

Free space = 1×[tex]10^{-6}[/tex]  - 5.48×[tex]10^{-9}[/tex] = 9.95×[tex]10^{-7}[/tex] m²

% Free space = [tex]\frac{9.945* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.45%

Now,

Consider 1 ml of ice  = 1 cm²

S.I unit of ice = 1×[tex]10^{-6}[/tex] m²

As , we know that mass of water in 1×[tex]10^{-6}[/tex] m² = 0.917 g

Moles of ice = [tex]\frac{0.917}{18} = 0.012[/tex]

Volume of H₂O = 6.022×[tex]10^{23}[/tex] ×0.012

Volume of ice unit = [tex]\frac{4}{3} \pi (37*10^{-12})^{3} *2 + \frac{4}{3} \pi (66*10^{-12})^{3} = 1.624*10^{-31}m^{3}[/tex]

Now,

Volume occupied by water = 0.012×6.022×[tex]10^{23}[/tex]× 1.624×[tex]10^{-31}[/tex] m²

                                              = 1.17×[tex]10^{-9}[/tex] m²

⇒Volume occupied by water = 1.17×[tex]10^{-9}[/tex] m²

Now,

Free space = 1×[tex]10^{-6}[/tex]  - 1.17×[tex]10^{-9}[/tex] = 9.98×[tex]10^{-7}[/tex] m²

% Free space = [tex]\frac{9.98* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.8%

How many cm 3 are in 0.014 in 3? (1 in = 2.54 cm)

Answers

Answer:

0.229 cm³.

Explanation:

The following data were obtained from the question:

Volume (in in³) = 0.014 in³

Volume (in cm³) =?

1 in = 2.54 cm

Next, we shall determine a conversion scale to convert from in³ to cm³. This can be obtained as follow:

1 in = 2.54 cm

Therefore,

1 in³ = 2.54³ cm³

1 in³ = 16.387 cm³

Finally, we shall convert 0.014 in³ to cm³. This can be obtained as follow:

1 in³ = 16.387 cm³

Therefore,

0.014 in³ = 0.014 in³ × 16.387 cm³ / 1 in³

0.014 in³ = 0.229 cm³

Thus, 0.014 in³ is equivalent to 0.229 cm³.

Two volumes of nitric oxide react with one volume of oxygen gas to form two volumes of a reddish-brown gas. Deduce the formula of this gas and sketch particle representations of its molecules.

Answers

Answer:

Explanation:

Nitric oxide is the gas NO, it reacts with oxygen as shown below;

2NO(g) + O2(g) -----> 2NO2(g)

Now the gas formed is the gas NO2 which is known to be reddish brown in colour.

A diagrammatic representation of this reaction is shown in the image attached to this answer.

Image credit: Chemlibretext

an unknown substance has a mass of 57.4 g and occupies a volume of 34.3 ml. what is the density in g/ml?

Answers

Answer:

1.6734 g\ml..hope it helps

What produces the magnetic force of an electromagnet?

O magnetic fields passing through the device

O static charged particles on the wire

O movement of charged particles through the wire

O positive and negative charges repelling each other

Answers

Answer:

movement of charged particles through the wire .

Explanation:

When electricity is passed through the wire of electromagnet , moving electrons of the wire produces magnetic field . This magnetic field in increased due to high permeability of soft iron of the electromagnet . It is this magnetic field which creates magnetic force .

identify which element is oxidized and which element is reduced.
PLZ HELPP...​

Answers

In the first three equations, Magnesium, sodium and aluminum are oxidized because they loose electrons. Sulfur, oxygen and chlorine are reduced because they gain electrons. In the last equation, magnesium is reduced because it gains hydrogen, and hydrogen is oxidized because it loses electrons.

If snails and crayfish die, what happens to the other species?

Answers

Crayfish are very intolerant of pollution and other human-generated fouling of their environment. ... This abundance may be due more to the acid-generated absence of fish which prey on crayfish than to a direct, positive influence of the acid on the crayfish itself.

Answer:

Other species will slowly die, because the species that eat snails and crayfish will not have anything to eat, then the thing that eats them will not have anything. This process will go on and on. So long story short populations will decrease, then off.

Suppose a 500.mL flask is filled with 0.40mol of N2 and 1.0mol of NO. The following reaction becomes possible:
N2g+O2g ->2NOg
The equilibrium constant K for this reaction is 5.93 at the temperature of the flask. Calculate the equilibrium molarity of N2. Round your answer to two decimal places.

Answers

Answer:

[N₂] = 1.1M

Explanation:

Based on the chemical reaction:

N₂(g) + O₂(g) ⇄ 2 NO(g)

Equilibrium constant, K, is defined as:

K = 5.93 = [NO]² / [N₂] [O₂]

Where [] are equilibrium concentrations of each specie

As initial concentrations are:

N₂ = 0.40mol / 0.500L = 0.8M

NO = 1mol / 0.500L = 2M

The equilbrium concentrations are:

[NO] = 2M - 2X

[N₂] = 0.8M +X

[O₂] = X

Replacing:

5.93 = [2 - 2X]² / [0.8+X] [X]

5.93 = 4 - 8X + 4X² / 0.8X + X²

4.744X + 5.93X² = 4 - 8X + 4X²

1.93X² + 12.744X - 4 = 0

Solving for X:

X = -6.9M → False solution. There are no negative concentrations

X = 0.3M. Real solution.

[N₂] in equilibrium is:

[N₂] = 0.8M +0.3M

[N₂] = 1.1M

How many moles of water can be formed from 0.57 moles of hydrogen gas?

Answers

Answer:

0.57 water

Explanation:

To solve this problem, we need to write the reaction expression first.

The reactants are oxygen gas and hydrogen gas.

They react to give a product of water

       2H₂    +    O₂   →   2 H₂O  

Given that;

Number of moles of hydrogen gas = 0.57moles

From the balanced reaction expression;

       2 moles of hydrogen gas produces 2 moles of water

   So;

    0.57mole of hydrogen gas will also produce 0.57 water

chemistry
Definition in your own words. I will check if you got it from online.

Word:
Malleable
(malleability)

Answers

mallebable- a material that is able to be hammered or pressed permanently without breaking .

A container holds 100.0 mL of nitrogen at 21° C and a pressure of 736 mm Hg. What will be its volume if the temperature increases by 35° C?

Answers

Answer:

V₂ = 104.76 mL

Explanation:

Given data:

Initial volume = 100.0 mL

Initial temperature = 21°C (21 + 273.15 K = 294.15 K)

Final temperature = 35°C (35 + 273.15 K = 308.15 k)

Final volume = ?

Solution:

Charles Law:

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ =100.0 mL × 308.15 K / 294.15 K

V₂ = 30815 mL.K /294.15 K

V₂ = 104.76 mL

Molecule A undergoes isomerization to molecule B in acetone. Using curved arrows, showing key intermediates and any formal charges, propose a detailed mechanism for this isomerization. Provide a brief explanation why this isomerization occurs.

Answers

Answer:

hello your question is incomplete attached below is the complete question

answer :

we use Isomerization because conjugated allylic carbocation  is more stable when compared to a Non-conjugated Allylic carbocation

Explanation:

Reason for the mechanism

we use Isomerization because conjugated allylic carbocation  is more stable when compared to a Non-conjugated Allylic carbocation

attached below is the detailed mechanism

1. Each substance written to the right of the arrow in a chemical equation is a

(1 point)

O catalyst

O reactant

O precipitate

O product

Answers

Answer: product

Explanation:

Each substance written to the right of the arrow in a chemical equation is referred to as a product.

When writing a chemical equation, the substance that's written to the left of arrow in the equation is the reactants.

On the other hand which is the right side is the product.

How many moles of hydrogen gas are present in 65.0 liters at STP?
1456 moles
1.45 moles
3.00 moles
2.90 moles

Answers

Answer:

2.9moles of hydrogen gas

Explanation:

convert liters to dm³

since 1liter= 1dm³

thus, 65.0liters = 65.0dm³

number of moles = volume given/22.4dm³

= 65.0/22.4

=2.9moles

Magnesium reacts with a silver nitrate solution
The reaction is represented by the ionic equation given
State why this reaction is considered a redox reaction

Answers

Answer: Mg oxidised to Mg++ and Ag+ reduced to Ag

Explanation:No ionic equation given!!

assume equation is Mg + 2Ag+ + 2NO3- —> Mg++ + 2Ag + 2NO3-

redox reaction: Ag gains an electron = reduced

Mg loseselectrons = oxidised

is C5H10 ionic or covalent?

Answers

Covalent because it is 5 and 10 so there even numbers I think
covalent. there is 5 c-c bonds 2 hydrogen atoms attach to each. total # of bonds is 15

definition of solubility
(science)

Answers

Answer:

th relative ability of a solute to devolve into a solvent

According to the Michaelis-Menten equation, when an enzyme is combined with a substrate of concentration s (in millimolars), the reaction rate (in micromolars/min) is

Answers

Answer:

The answer is "A"

Explanation:

Please find the complete question in the attachment file.

[tex]\to R(s)= \frac{As}{K+s}[/tex]

when the s in the approach, that is infinity R(s) tends

[tex]\to \frac{A}{\frac{K}{s}+1} \\\\ \to\frac{A}{0+1} \\\\ \to\frac{A}{1} \\\\ \to A[/tex]

A series of dilute NaCl solutions are prepared starting with an initial stock solution of 0.100 M NaCl. Solution A is prepared by pipeting 10 mL of the stock solution into a 250-mL volumetric flask and diluting to volume. Solution B is prepared by pipeting 25 mL of solution A into a 100-mL volumetric flask and diluting to volume. Solution C is prepared by pipeting 20 mL of solution B into a 500-mL volumetric flask and diluting to volume. What is the molar concentration of NaCl in solutions A, B and C

Answers

Answer:

Solution A: 0.00400M

Solution B: 0.00400M

Solution C: 4.00x10⁻⁵M

Explanation:

Solution A is diluting the 0.100M NaCl from 10mL to 250mL. That is:

250mL / 10mL = 25 times.

That means molar concentration of sln A is:

0.100M / 25 = 0.00400M

Solution B is obtained diluting 25mL to 100mL:

100mL / 25mL = 4 times

0.00400M / 4 times = 0.00100M

And solution C is obtained diluting the solution C from 20mL to 500mL:

500mL / 20mL = 25 times

Solution C:

0.00100M / 25 times = 4.00x10⁻⁵M

The formula for serial dilution can be used to obtain the molarity of solution A, B , C.

For solution A

M1V1 = M2V2

M2 = 0.100 M ×  10 mL/250-mL

M2 = 0.004 M

For solution B

M1V1 = M2V2

M2 = 0.004 M × 25 mL/100-mL

M2 = 0.001 M

For solution C

M1V1 = M2V2

M2 = 0.001 M × 20 mL/500-mL

M2 = 0.00004 M

Learn more about serial dilution: https://brainly.com/question/2167827

what is the formula for H-H

Answers

Answer:

H-H equation is written as follows:

pH=pK + log

{HCO3-}(base)

{H2CO3}(acid)

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