Which of the following is an example of a scientific theory?
Matter can be neither created nor destroyed.
Force equals mass times acceleration.
Matter behaves the way it does because it is made of atoms.
There is probably life on Mars.


C. Matter behaves the way it does because it is made of atoms.

Answers

Answer 1

Answer:

Matter can neither be created nor destroyed

Explanation:

Mark me. as brainliest


Related Questions

How can a Bose-Einstein condensate be formed? A. B super-heating a gas. B. By super-cooling certain types of solid. C. By super-cooling certain types of plasma. D. By super-heating a plasma

Answers

Answer:

C. By super-cooling certain types of plasma.

Explanation:

Bose-Einstein condensate is a state of matter whereby atoms or particles become cooled to a very low energy state leading to their condensation to give a single quantum state.

Note that plasma refers to atoms that have had some or even all of its electrons stripped away leaving only positively charged ions. Simply put, plasma is ionized matter.

When certain types of plasma are super cooled, Bose-Einstein condensate are formed.

A student prepares a aqueous solution of acetic acid . Calculate the fraction of acetic acid that is in the dissociated form in his solution. Express your answer as a percentage. You will probably find some useful data in the ALEKS Data resource.

Answers

Answer:

10.71%

Explanation:

The dissociation of acetic acid can be well expressed as follow:

CH₃COOH ⇄   CH₃COO⁻  + H⁺

Let assume that the prepared amount of the aqueous solution is 14mM since it is not given:

Then:

The I.C.E Table is expressed as follows:

                     CH₃COOH       ⇄   CH₃COO⁻        +           H⁺  

Initial              0.0014                       0                                0

Change            - x                           +x                               +x

Equilibrium   (0.0014 - x)                 x                                 x

Recall that:

Ka for acetic acid CH₃COOH  = 1.8×10⁻⁵

[tex]K_a = \dfrac{[x][x]]}{[0.0014-x]}[/tex]

[tex]1.8*10^{-5} = \dfrac{[x][x]]}{[0.0014-x]}[/tex]

[tex]1.8*10^{-5} = \dfrac{[x]^2}{[0.0014-x]}[/tex]

[tex]1.8*10^{-5}(0.0014-x) = x^2[/tex]

[tex]2.52*10^{-8} -1.8*10^{-5}x = x^2[/tex]

[tex]2.52*10^{-8} -1.8*10^{-5}x - x^2 =0[/tex]

By rearrangement:

[tex]- x^2 -1.8*10^{-5}x +2.52*10^{-8}= 0[/tex]

Multiplying through  by (-) and solving the quadratic equation:

[tex]x^2 +1.8*10^{-5}x-2.52*10^{-8}= 0[/tex]

[tex](-0.00015 + x) (0.000168 + x) =0[/tex]

x = 0.00015 or x = -0.000168

We will only consider the positive value;

so x=[CH₃COO⁻] = [H⁺] = 0.00015

CH₃COOH = (0.0014 - 0.00015) = 0.00125

However, the percentage fraction of the dissociated acetic acid is:

[tex]= \dfrac{ 0.00015}{0.0014}\times 100[/tex]

= 10.71%

Use the following key to classify each of the elements below in its elemental form:
A. Discrete atoms .. C. Metallic lattice
B. Molecules ... D. Extended, three-dimensional network
1. Magnesium
2. Nitrogen ...
3. Lithium
4. Potassium ...

Answers

Answer:

Magnesium - Metallic lattice

Nitrogen - Molecules

Lithium - Metallic lattice

Potassium - Metallic lattice

Explanation:

Metals exist in metallic lattices. In this lattice, metal ions are held together with a sea of electrons by strong electrostatic forces.

All metals possess this metallic lattice, hence; potassium, lithium and magnesium all consist of metal lattices.

Nitrogen is a nonmetal and consists of molecules of N2.

How many electrons are shown in the following electron
configuration: 1s22s22p63s 23p64s23d104p65s24d105p66s2 ?
Express your answer numerically as an integer.

Answers

Answer:

1s22s22p6

Explanation:

Neon is an element in the periodic table and has an atomic number of 10, which means it has 10 protons in its nucleus and thus since the number of protons and electrons is the same then it has 10 electrons.

Therefore, it has 2 electrons in the first energy shell and 8 electrons in the second energy shell. To elaborate further, the first shell has a single s-sub shell that contains a single s-orbital that can hold two electrons. The second energy shell has a single s-sub-shell whose s-orbital will occupy 2 electrons, and also has a p-orbital which can hold 6 electrons, making the second shell to have 8 electrons.

The reversible reaction: 2SO2(g) O2(g) darrow-tn.gif 2SO3(g) has come to equilibrium in a vessel of specific volume at a given temperature. Before the reaction began, the concentrations of the reactants were 0.060 mol/L of SO2 and 0.050 mol/L of O2. After equilibrium is reached, the concentration of SO3 is 0.040 mol/L. What is the equilibrium concentration of O2

Answers

Answer:

[tex][O_2]_{eq}=0.030M[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to solve this problem by firstly writing out the mathematical expression for the concentration of oxygen at equilibrium, given the initial one and the change due to the reaction extent:

[tex][O_2]_{eq}=0.050M-x[/tex]

Whereas [tex]x[/tex] can be found considering the equilibrium of SO3:

[tex][SO_3]_{eq}=2x=0.040M[/tex]

Which means:

[tex]x=\frac{0.040M}{2} =0.020M[/tex]

Thus, the equilibrium concentration of oxygen gas turns out:

[tex][O_2]_{eq}=0.050M-0.020M=0.030M[/tex]

Regards!

A sample of oxygen occupies 1.00 L. If the temperature remains constant, and the pressure on the oxygen is decreased to one third the original pressure, what is the new volume

Answers

Answer:

3.00 L

Explanation:

P₁V₁ = P₂V₂

V₁ = 1.00 L

P₁ = (x) atm

P₂ = [tex]\frac{1}{3}[/tex] · (P₁) = [tex]\frac{x}{3}[/tex]

V₂ = unknown

(x atm)(1.00 L) = ( [tex]\frac{x}{3}[/tex] atm)(V₂)

divide both sides by ( [tex]\frac{x}{3}[/tex] atm)

( 1.00x )( [tex]\frac{3}{x}[/tex] ) = V₂

x cancels out

(1.00)(3) = V₂

V₂ = 3.00 L

For a gas sample containing equimolar amounts of carbon monoxideand heliumat 300 K, heliumhas _____________average speed and _____________ average kinetic energy compared tocarbon monoxidegas.a.a lower; the same b. the same; the same c. a higher; the same d. a higher; higher

Answers

Answer:

Option C (a higher; the same) is the appropriate response.

Explanation:

Given:

Temperature,

T = 300 K (both [tex]N_2[/tex] and [tex]H_2[/tex])

As we know,

Average speed of a molecule,

⇒ [tex]\bar v=\sqrt{\frac{8RT}{\pi M} }[/tex]

Thus, the average speed of [tex]N_2[/tex] will be lower as its molar mass is greater than [tex]H_2[/tex].

Now,

⇒ [tex]Average \ kinetic \ energy = \frac{3}{2} \ KT[/tex] (not depend on molar mass)

Hence, it will be the same.

The other three alternatives aren't connected to the scenario given. So the above is the correct answer.

dentify the correct formula for the following ionic compounds. - sodium chloride - magnesium chloride - calcium oxide - lithium phosphide - aluminum sulfide - calcium nitride A. SCl B. LiP 3 C. AlS D. Li 3P E. CaN F. CaO G. Ca 3N 2 H. MgCl 2 I. NaCl J. CaO 2 K. CaN 2 L. LiP M. MnCl 2 N. Al 2S 3 O. AlS 3

Answers

Explanation:

The chemical formula of an ionic compound can be written by using the symbols of the respective cations and anions.

The overall charge on the molecule should be zero.

Hence, the total charge of cations=total charge of anions.

The symbols of the given molecules are shown below:

sodium chloride  ---- NaCl

magnesium chloride ---[tex]MgCl_2[/tex]

calcium oxide ---- CaO

lithium phosphide----[tex]Li_3P[/tex]

aluminum sulfide ----- [tex]Al_2S_3[/tex]

calcium nitride---- [tex]Ca_3N_2[/tex]

Calculate the mass of sodium phosphate in aqueous solution to fully react with 37 g of chromium nitrate(III) an aqueous solution?(report answer in grams and only three Sigg figs do not put the unit)

Answers

Answer:

41 g

Explanation:

The equation of the reaction is;

Cr(NO3)3(aq)+Na3PO4(aq)=3NaNO3(s)+CrPO4(aq)

Number of moles of chromium nitrate = 37g/ 146.97 g/mol = 0.25 moles

1 mole of sodium phosphate reacts with 1 mole of chromium nitrate

x moles of sodium phosphate react as with 0.25 moles of chromium nitrate

x= 1 × 0.25/1

x= 0.25 moles

Mass of sodium phosphate = 0.25 moles × 163.94 g/mol

Mass of sodium phosphate = 41 g

Help!!!!!!!!!
I'm using plato

Answers

Answer:

- Two black balls: they represent a diatomic molecule composed by two atoms of the same element.

- One black ball and two black balls: they represent a compound formed by two different elements.

- One gray ball and two black balls: they represent a compound formed by two different elements.

- Two black-dotted balls: they represent a diatomic molecule composed by two atoms of the same element.

Explanation:

Hey there!

In this case, according to the given information, we can firstly bear to mind the fact that each ball color represents a different element, for that reason we can tell the following:

- Two black balls: they represent a diatomic molecule composed by two atoms of the same element.

- One black ball and two black balls: they represent a compound formed by two different elements.

- One gray ball and two black balls: they represent a compound formed by two different elements.

- Two black-dotted balls: they represent a diatomic molecule composed by two atoms of the same element.

Regards!

Identify the highest energy molecular process that occurs when a molecule absorbs a microwave photon.

Answers

The question is incomplete, the complete question is;

Identify the highest energy molecular process that occurs when a molecule absorbs a microwave photon

a) electronic excitation

b) bond breakage

c) molecular vibration

d) molecular rotation

Answer:

molecular rotation

Explanation:

Microwaves are part of the electromagnetic spectrum. They are lower energy, lower frequency radiation.

When molecules absorb infrared radiation, they transition between the rotational states of the molecule.

Hence, the highest energy molecular process that occurs when a molecule absorbs a microwave photon is molecular rotation.

In the given range,at what temperature does oxy gen have the highest solubility?​

Answers

Water solubility of oxygen at 25oC and pressure = 1 bar is at 40 mg/L water. In air with a normal composition the oxygen partial pressure is 0.2 atm. This results in dissolution of 40 . 0.2 = 8 mg O2/L in water that comes in contact with air.
25oC
Solubility of oxygen and oxygen compounds

Water solubility of oxygen at 25oC and pressure = 1 bar is at 40 mg/L water. In air with a normal composition the oxygen partial pressure is 0.2 atm. This results in dissolution of 40 . 0.2 = 8 mg O2/L in water that comes in contact with air.

Calculate the molarity of a 17.5% (by mass) aqueous solution of nitric acid. Select one: a. 2.74 m b. 4.33 m c. 0.274 m d. 3.04 m e. The density of the solution is needed to solve the problem.

Answers

Answer:

Option e.

Explanation:

Molarity is the concentration that indicates moles of solute in 1 L of solution.

We have another concentration, percent by mass.

Percent by mass indicates mass of solute in 100 g of solution.

Our solute is HNO₃, our solvent is water.

17.5 g of nitric acid is the mass of solute. We can convert them to moles:

17.5 g . 1mol / 63g = 0.278 moles

We do not have volume of solution. We assume the mass is 100 g because the percent by mass but we need density to state the volume.

Density = Mass / Volume

Mass / Density = Volume

Once we have the volume, we need to be sure the units is in L, to determine molarity

M = mol /L

A hypothetical A-B alloy of composition 53 wt% B-47 wt% A at some temperature is found to consist of mass fractions of 0.5 for both and phases. If the composition of the phase is 92 wt% B-8 wt% A, what is the composition of the phase

Answers

Answer:

the composition of the ∝ phase C∝ = 14  or [ 14 wt% B-86 wt% A ]

Explanation:

Given the data in the question;

Co = 53 or [ 53 wt% B-47 wt% A ]

W∝ = 0.5 = Wβ

Cβ = 92 or [ 92 wt% B-8 wt% A ]

Now, lets set up the Lever rule for W∝ as follows;

W∝ = [ Cβ - Co ] / [ Cβ - C∝ ]

so we substitute our given values into the expression;

0.5 = [ 92 - 53 ] / [ 92 - C∝ ]

0.5 = 39 /  [ 92 - C∝ ]

0.5[ 92 - C∝ ] = 39

46 - 0.5C∝  = 39

0.5C∝ = 46 - 39

0.5C∝ = 7

C∝ = 7 / 0.5

C∝ = 14  or [ 14 wt% B-86 wt% A ]

Therefore, the composition of the ∝ phase C∝ = 14  or [ 14 wt% B-86 wt% A ]

Classify each molecule as an alcohol, ketone, or aldehyde based on its name. Propanone (acetone) Choose... Ethanal Choose... 3-phenyl-2-propenal Choose... Butanone Choose... Ethanol Choose... 2-propanol Choose...

Answers

Answer:

1.) Propanone (ketone)

2.) Ethanal( aldehyde)

3.) 3-phenyl-2-propenal (aldehyde)

4.) Butanone (ketone)

5.) Ethanol ( alcohol)

6.) 2-propanol (alcohol)

Explanation:

In organic chemistry, ALCOHOL ( also known as alkanol) are compounds in which hydroxyl groups are linked to alkyl groups. They can be considered as being derived from the corresponding alkanes by replacing the hydrogen atoms with hydroxyl groups. The hydroxyl group is the functional group of the alcohol as it is responsible for their characteristic chemical properties. A typical example of alcohol is ethanol and 2-propanol.

Alkanals or ALDEHYDES have the general formula RCHO while alkanones or KETONES have the general formula RR'CO where R and R' may be alkyl or aryl groups. The main similarity between these two classes of compounds is the presence of the carbonyl group. In aldehydes, there is a hydrogen atom attached to the carbon In the carbonyl group while there is none on the ketones.

Some common examples of ketones are Propanone, Butanone while examples of aldehydes are Ethanal and 3-phenyl-2-propenal

A sample of oxygen gas has a volume of 89.6 L at STP. How many moles of oxygen gas are present ?

Answers

Answer:

89,6/22,4 =4(mol)

Explanation:

There are approximately 1.089 moles of oxygen gas present in the sample at STP.

At STP (Standard Temperature and Pressure), the conditions are defined as follows:

Temperature (T) = 0 degrees Celsius = 273.15 Kelvin

Pressure (P) = 1 atmosphere (atm) = 101.325 kPa = 1013.25 hPa

Now, to find the number of moles of oxygen gas (O2) present in the sample, we can use the ideal gas law:

PV = nRT

Where:

P = pressure (in atm)

V = volume (in liters)

n = number of moles

R = ideal gas constant = 0.0821 L.atm/(mol.K)

T = temperature (in Kelvin)

Given:

V = 89.6 L (volume at STP)

T = 273.15 K (STP temperature)

Let's plug in the values and solve for n (number of moles):

n = PV / RT

n = (1 atm) × (89.6 L) / (0.0821 L.atm/(mol.K) × 273.15 K)

n = 1.089 moles

So, there are approximately 1.089 moles of oxygen gas present in the sample at STP.

To know more about oxygen gas:

https://brainly.com/question/31842859

#SPJ2

A rectangular piece of plastic has a width of 4.2 cm, a height of 1.9 cm and a length of 8.8 cm. If the mass of the plastic rectangle is 64.6 g, what is its density in g/mL?

Answers

Answer:

0.92g/mL

Explanation:

Density of a substance is calculated as follows:

Density = mass (m) ÷ volume (V)

According to this question, a rectangular piece of plastic has a width of 4.2 cm, a height of 1.9 cm and a length of 8.8 cm. Using the formula; L × W × H, the volume of the plastic can be calculated

V = L × W × H

V = 8.8 × 4.2 × 1.9

V = 70.2cm³

The mass of the plastic is 64.6g, hence, its density is:

Density = 64.6g ÷ 70.2cm³

Density of the rectangular plastic = 0.92g/cm³ or 0.92g/mL

What is true about the properties of liquids and gases?

Gas particles are much more densely packed than liquid particles.
The crystal lattice structure of liquids is more defined than in gases.
Liquids form amorphous crystals while gases do not.
There are strong intermolecular forces between particles that make up liquids, but not gases.

Answers

Answer:

There are strong intermolecular forces between particles that make up liquids, but not gases.

Explanation:

Solids, liquids and gases are the three states of matter that exists. However, they possess varying properties that distinguishes them from one another. One of these properties is the strength of the intermolecular forces that hold their molecules together.

The intermolecular forces of each state of matter becomes weak in this order: solid>liquid>gas.

- Intermolecular forces in solid molecules are very strong, hence making them compact and well attached to each other.

- Intermolecular forces in liquid molecules are not too strong, hence, cannot exist in a fixed position but tend to flow.

- Intermolecular forces in gaseous molecules are very weak, hence, gases can move easily and rapidly in any given space.

Write a balanced half-reaction for the oxidation of liquid water to aqueous hydrogen peroxide in basic aqueous solution. Be sure to add physical state symbols where appropriate.

Answers

Answer:

O2(g) + 2H2O(l) --------> 2H2O2(aq) + 2e

Explanation:

An oxidation reaction reaction refers to a reaction in which electrons are lost. In this case, we are about to see the full balanced half-reaction for the oxidation of liquid water to aqueous hydrogen peroxide in basic aqueous solution.

The full equation is;

O2(g) + 2H2O(l) --------> 2H2O2(aq) + 2e

So, two electrons were lost in the process.

1. Most of the chemicals included in your General Chemistry Lab kit can be discarded down a drain. Describe a situation in which you would need to neutralize a chemical before discarding down a drain.

Answers

Answer: Chemicals like acids and bases are harmful and must be neutralized before draining.

Explanation:

A strong acid or strong base is required to be diluted or neutralized before it is discarded in the drain as if is discarded without diluting and neutralization it can spill and splash from sink or drain and can harm people in chemistry lab, moreover the fumes of the discarded chemical on spilling can cause respiratory tract burning and can even cause fire hazard so it must be converted into less harmful form and then must be drained.  

Poly(ethylene terephthalate) (PET), which has glass transition (Tg) and crystalline melting (Tm) temperature of 69 and 267 °C, respectively, can exist in a number of different states depending upon temperature and thermal history. Thus, it is possible to prepare materials that are semicrystalline with amorphous regions that are either glassy or rubbery and amorphous materials that are glassy, rubbery or melts. Consider a sample of PET cooled rapidly from 300 °C (state A) to room temperature. The resulting material is rigid and perfectly transparent (state B). The sample is then heated to 100 °C and maintained at this temperature, during which time is gradually becomes translucent (state C). It is then cooled to room temperature, where it is again observed to be translucent (state D).

Answers

Answer:

Following are the solution to the given points:

Explanation:

For point A:

The sample cooking (PET) is between 300°C and room temperature.Now in nature, the substance is exceedingly stiff.Samples of PET up to 100°C were heated and stayed on equal footing.Now it has cooled off the same sample below 100° C and we may see how it is again TRASNEPARENT in nature.

For point B:

In point 3, the mixture was added to 100°C, which implies that the granular material flows and deforms, enabling it to become elongated. This is termed solid-state crystalline such that grains are flexible, but this material contaminates numerous little crystalline that has spheres when we cool down in point  4 polymers. It forms therefore an unstructured solid, which then in point 4 is higher in particles and less pliable in orderly atoms.

For point C:

In point 2, the specimen gets forced at room temperature to organize a huge molecule in an ordinary and crystal fashion and therefore is transparent due to highly crystalline atoms in point 2 of the PET sample.

In point 4, however, we notice how amorphous, firm but not crystalline develops. It's why light tends to disperse over many cereal limits, since many microscopic crystallines, therefore dispersion, PET in point 4 is translucent.

A uniform plastic block floats in water with 50.0 % of its volume above the surface of the water. The block is placed in a second liquid and floats with 23.0 % of its volume above the surface of the liquid.
What is the density of the second liquid?
Express your answer with the appropriate units.

Answers

Answer:

density of second liquid = 650 kg/m³

Explanation:

Given that:

The volume of the plastic block submerged inside the water  = 0.5 V

The force on the plastic block  = [tex]\rho_1V_1g[/tex]

[tex]= 0.5p_1 V_g[/tex]

when the block is floating, the weight supporting the force (buoyancy force) is:

W [tex]= 0.5p_1 V_g[/tex]

[tex]\rho Vg = 0.5p_1 V_g[/tex]

[tex]\rho = 0.5 \rho _1[/tex]

where;

water density [tex]\rho _1[/tex] = 1000

[tex]\rho = 0.5 (1000)[/tex]

[tex]\rho = 500 kg/m^3[/tex]

In the second liquid, the volume of plastic block in the water = (100-23)%

= 77% = 0.7 V

The force on the plastic block is:

[tex]= 0.77p_2 V_g[/tex]

when the block is floating, the weight supporting the force (buoyancy force) is:

[tex]W = 0.77p_2 V_g[/tex]

[tex]\rho Vg = 0.77 \rho_2 V_g \\ \\ \rho = 0.77 \rho_2 \\ \\ 500 = 0.77 \rho_2 \\ \\ \rho_2 = 500/0.77[/tex]

[tex]\mathbf{ \rho_2 \simeq 650 \ kg/m^3}[/tex]

2. How many joules of heat are released when 32g of water cools down from 71%
specific heat of water is 4.184 J/gºC)
How many kilojoules is this?

Answers

he says he doesnt know sorry

For each of the following compounds, indicate the pH at which 50% of the compound will be in a form that possesses a charge and at which pH more than 99% of the compound will be in a form that possesses a charge.

ClCH2COOH (pKa = 2.86)
CH3CH2NH+3 (pKa = 10.7)

Express your answer using two decimal places

a. Determine a pH at which 50% of ClCH2COOH will be in a form that possesses a charge.
b. Determine a pH at which pH more than 99% of ClCH2COOH will be in a form that possesses a charge.
c. Determine a pH at which 50% of CH3CH2NH+3 will be in a form that possesses a charge.
d. Determine a pH at which pH more than 99% of CH3CH2NH+3 will be in a form that possesses a charge.

Answers

Answer:

a. 2..86 b. 4.86 c. 10.7 d. 8.7

Explanation:

a. Determine a pH at which 50% of ClCH2COOH will be in a form that possesses a charge.

Using the Henderson-Hasselbalch equation,

pH = pKa + log[A⁻]/[HA]

where [A⁻] = concentration of conjugate base (or charged form) and [HA] = concentration of acid.

At 50% concentration, [A⁻] = [HA] ⇒ [A⁻]/[HA] = 1

So, pH = pKa + log[A⁻]/[HA]

pH = pKa + log1

pH = pKa = 2.86

b. Determine a pH at which pH more than 99% of ClCH2COOH will be in a form that possesses a charge.

Let x be the concentration of the acid. Since 99% of it should possess a charge, the basic concentration is 0.99x while the acidic concentration is remaining 1 % (1 - 0.99)x = 0.01x

Using the Henderson-Hasselbalch equation,

pH = pKa + log[A⁻]/[HA] where [A⁻] = concentration of conjugate base (or charged form) = 0.99x and [HA] = concentration of acid = 0.01x.

pH = pKa + log0.99x/0.01x

pH = pKa + log0.99/0.01

pH = 2.86 + log99

pH = 2.86 + 1.996

pH = 4.856

pH ≅ 4.86

c. Determine a pH at which 50% of CH3CH2NH+3 will be in a form that possesses a charge.

Using the Henderson-Hasselbalch equation,

pH = pKa + log[A⁻]/[HA]

where [A⁻] = concentration of conjugate base and [HA] = concentration of acid.

At 50% concentration, [A⁻] = [HA] ⇒ [A⁻]/[HA] = 1

So, pH = pKa + log[A⁻]/[HA]

pH = pKa + log1

pH = pKa = 10.7

d. Determine a pH at which pH more than 99% of CH3CH2NH+3 will be in a form that possesses a charge.

Let x be the concentration of the acid. Since 99% of it should possess a charge, the basic concentration is 0.01x while the acidic concentration is remaining 99 % (1 - 0.01)x = 0.99x (which possesses the charge).

Using the Henderson-Hasselbalch equation,

pH = pKa + log[A⁻]/[HA] where [A⁻] = concentration of conjugate base = 0.01x and [HA] = concentration of acid = 0.99x.

pH = pKa + log0.01x/0.99x

pH = pKa + log1/99

pH = 10.7 - log99

pH = 10.7 - 1.996

pH = 8.704

pH ≅ 8.7

What enzyme below is an exoenzyme?
A. Casease
B. Citrase
C. Catalase
D. Oxidase

Answers

Casease! Good luck!

Use the Ka values for weak acids to identify the best components for preparing buffer solutions with the given pH values.

Name Formula Ka
Phosphoric acid H3PO4 7.5 x 10^-3
Acetic acid CH3COOH 1.8 x 10^-5
Formic acid HCOOH 1.8 x 10^-4

pH 1.9 =_________
pH 5.0 = ________
pH 3.9= ________

Answers

Answer:

pH= 1.9 then [tex]H_{3} PO_{4}[/tex]

pH = 5.0 , [tex]CH_{3} COOH[/tex]

pH = 3.9 , HCOOH

As we know range left [tex]pH= pKa+/- 1[/tex]

A system receives 425 J of heat from and delivers 425 J of work to its surroundings. What is the change in internal energy of the system (in J)?

Answers

Answer:

0 J

Explanation:

Applying,

ΔE = q+w................ Equation 1

Where ΔE = change in internal energy of the system, q = Heat of the system, w = work of the system.

Note: q is positive, while w is negative

From the question,

Given: q = 425 J, w = -425 J

Substitute these values into equation 1

ΔE = 425-425

ΔE = 0 J

Hence the change in internal energy of the system is 0 J

What is the major product in this reaction

Answers

Answer:

I think option A is right answer

What is the energy change when 78.0 g of Hg melt at −38.8°C

Answers

Answer:

The correct answer is - 2.557 KJ

Explanation:

In this case, Hg is melting, the process is endothermic, so the energy change will have a positive sign.

we can calculate this energy by the following formula:

Q = met

where, m = mass,

e = specific heat

t = temperature

then,

Q = 78*0.14* (273-38.8)

here 0.14 = C(Hg)

= 2.557 Kj

complete the following steps.
Remember to follow lower numbered rules first.
Na2CO3(aq) + Pb(OH)2(aq) → NaOH (?) + PbCO3(?)
a. Write a balanced chemical equation. (1 pt)
b. If a reaction occurs, write the balanced
chemical equation with the proper states of matter
(i.e. solid, liquid, aqueous) filled in. If no reaction
occurs, write “No reaction.” (1 pt)
c. If a reaction occurs, write the net ionic equation
for the reaction. If no reaction occurs, write "no
reaction.” (1 pt)

Answers

Answer:

See explanation

Explanation:

a) The balanced reaction equation is;

Na2CO3(aq) + Pb(OH)2(aq) -----> 2 NaOH + PbCO3

b) When we include states of matter;

Na2CO3(aq) + Pb(OH)2(aq) -----> 2 NaOH(aq) + PbCO3 (s)

c) Complete ionic equation;

2Na^+(aq) + CO3^2-(aq) + Pb^2+(aq) + 2OH^-(aq) ----> 2Na^+(aq) + 2OH^-(aq) + PbCO3(s)

Net Ionic equation;

Pb^2+(aq) + CO3^2-(aq) ----> PbCO3(s)

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