Anyone know this
Please help me

Answer:

C

Explanation:

A physical change is one in which no new substance is formed. The moistening of food by saliva is purely a physical change because no new substance is formed in the process.

A chemical change leads to the formation of a new substance. The action of the enzyme converts starch into sugar which means that a new substance is formed. This is a chemical change.

Answer:

(A) Gamma

Explanation:

They are emitted during radioactive decay and have no mass and no electrical charge meaning no change in atomic number or mass number.

Hope it helps :)

Answer:

A. Gamma

Explanation:

Answers Via Educere/Founders Education.

Answer:

382.2g of H3PO4 are needed

Explanation:

Based on the reaction, 3 moles of Mg(OH)2 react with 2 moles of H3PO4.

To solve this question we must find the moles of Mg(OH)2 using Avogadro's number. Then, using the reaction we can find the moles of H3PO4 and its mass as follows:

Moles Mg(OH)2:

3.523x10²⁴ molecules * (1mol / 6.022x10²³ molecules) = 5.850 moles Mg(OH)2

Moles H3PO4:

5.850 moles Mg(OH)2 * (2 mol H3PO4 / 3 mol Mg(OH)2) = 3.90 moles H3PO4

Mass H3PO4:

Molar mass:

3H = 1.01g/mol*3 = 3.03

P = 30.97g/mol*1 = 30.97

4O = 16.00g/mol*4 = 64.00g/mol

3.03 + 30.97 + 64.00 = 98.00g/mol

3.90 moles H3PO4 * (98.00g / mol) =

Answer:

12.5 L

Explanation:

Part 1 - Finding Number of moles

First thing we need to find out is number of moles of oxygen gas. We can achieve this using the formula: PV=nRT

Pressure (Pa) x Volume (cubic meters) = no. moles x gas constant (8.31 J/K*mol) x Temperature (K)

We can rearrange this formula to solve for no. moles:[tex]n=\frac{PV}{RT}[/tex] and plug in the values to solve.

1 atm = 1.01325 x 10^5 Pa, therefore 1.76atm = [tex]1.76*1.01325*10^{5} =178332Pa[/tex]

1 L = 1 x 10^-3 m^3, therefore 5L = [tex]5*1*10^{-3}=5*10^{-3}m^{3}[/tex]

Finally Kelvin = Celsius + 273.15, therefore 37.0C = [tex]37.0+273.15=310.15K[/tex]

Substituting this all back into the formula we get:

[tex]n=\frac{(178332)(5*10^{-3})}{(8.31)(310.15)}[/tex]

[tex]n=\frac{891.66}{2577.3465}[/tex]

[tex]n=0.346[/tex] moles (3.s.f)

Part 2 - Balancing Equation

The equation given is not balanced so we need to balance it to find the correct molar ratio. We can start by balancing the oxygens as there are less of them. We can see that there are 2 oxygen atoms on the left as denoted by 'O2' but only one on the right in 'H2O', therefore we can add a two in front of the H2O on the right side to balance the oxygen atoms:

[tex]H_{2(g)} +O_{2(g)} - > 2H_{2} O_{(g)}[/tex]

Now we have four hydrogens on the right side so we have a 2 in front of H2 on the left hand side to balance the hydrogens:

[tex]2H_{2(g)} +O_{2(g)} - > 2H_{2} O_{(g)}[/tex]

And voila! The equation is balanced and shows the correct molar ratio.

Part 3 - Calculating moles of water

As we can see from the molar ratio in the balanced equation, every mole of O2 will produce two moles of H2O- the ratio is 1:2. As we have an excess of hydrogen, we only need to worry about the amount of oxygen we have. Therefore to calculate the number of moles of water that can be made we simply need to multiple by the ratio:

[tex]0.346*2=0.692[/tex] moles of water

Part 4 - Converting to Liters

Finally, to convert our amount of moles into volume, we can use water's molecular mass to find the mass and then its density to find its volume. Water's molecular mass is 18.02gmol^-1 (2x1.01+16.00) and its density is 997kg/m^3.

m=nM (mass = no. moles x molecular mass): m=[tex]0.692*18.02=12.46984g[/tex]

rho = m/v (density = mass/volume): [tex]997=12.46984/v[/tex], therefore [tex]v=0.0125m^{3}[/tex]

Finally as mentioned above, 1 L = 1 x 10^-3 m^3, therefore the volume of water that can be made is 12.5 Liters.

Answer:

The number of ATP molecules made per NADH and FADH2 oxidation differ by 1.

The amount of ATP generated per mole of glucose when the glycerol-3-phosphate shuttle would be 30 instead of 32.

Explanation:

The glycerol-3-phosphate shuttle can transport cytosolic NADH equivalents into the mitochondrial matrix. In this shuttle, the protons and electrons are donated to FAD, which is reduced to FADH2. These protons and electrons are subsequently donated to coenzyme Q in the electron transport chain. Given that the number of ATP molecules made per NADH and FADH2 oxidation differ by 1, the amount of ATP generated per mole of glucose when the glycerol-3-phosphate shuttle would be 30 instead of 32.

FADH2 generates 1.5 ATP per molecule unlike NADH which generates 2.5 ATP per molecule. This is because electron transfer via FADH2 is not coupled to proton pumping unlike electron transfer reactions involving NADH. Thus, two moles of NADH from the oxidation of 2 moles of glyceraldehyde-2-phosophate to two moles of 1,3-bisphosphoglycerate will yield 3 moles of ATP rather than 5 moles when shuttled through the glycerol-3-phosphate shuttle. The glycerol-3-phosphate shuttle of cytosolic NADH shuttling occurs mainly in the brain and skeletal muscles and does not involve membrane transporters.

The mass of copper nitrate produced in the reaction 3.65g of copper and 5.16g of nitric acid​ is 3.84g.

Nitric acid reacts with copper

4 HNO3(l) + Cu(s) ==> Cu(NO3)2(s and aq) + 2 NO2(g) + 2 H2O(l)

Deep blue colour characterises the copper nitrate salt that develops. Nitrogen dioxide is the gas that is created when strong nitric acid and copper interact. This maroon mist is nitrogen dioxide.

Copper :

=> mass copper nitrate

=> 3.65g Cu x 1mol of Cu/63.546 g Cu x 1 mol copper nitrate / 1 mol cu x 187.56 g/mol copper nitrate

=> 10.77g

HNO3 (Nitric acid):

=> Mass copper nitrate =3.65 g HNO3 X 1 mol HNO3/63.546 g HNO3 x 1mol of copper nitrate /4 mol HNO3 x 187.56 g/mol copper nitrate.

=> 3.84g.

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Explanation:

[tex]KOH→K {}^{ + } + OH {}^{ - } \\ [OH {}^{ - } ] = 3.4 \times {10}^{ - 3} \\ k _{w} = [OH {}^{ - } ] [H {}^{ + } ] \\ 1 \times {10}^{ - 14} = 3.4 \times {10}^{ - 3} \times [H {}^{ + } ] \\ [H {}^{ + } ] = 2.94 \times {10}^{ - 12} \: M[/tex]

Answer:

2.9 X 10^-12 M

Explanation:

Answer:

D. Quaternary

Explanation:

I'm going to attach a picture that can help you understand how amines are classified!

I'll explain how amines are classified below. You can skip it if you feel that you already understand that!

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So we know that amines are composed of nitrogen, at least one R group, and hydrogens. The way that we categorize amines as primary, secondary, tertiary, or quarternary is by counting the number of R groups.

Now, how do we identify the R groups?

Well, basically anything that isn't a hydrogen. You will see that the nitrogen in an amine standardly only makes 3 bonds. If all those three bonds are with a hydrogen, then it will be ammonia. If it is a primary amine, then one of those hydrogens will be replaced with another compound (like [tex]CH_{3}[/tex] in this case). This other compound is called an R group. R groups will change based on the amine. A secondary amine will have two hydrogens replaced with groups. A tertiary amine will have all three hydrogens replaced with R groups.

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Now, by this logic, we can just count the number of R groups in [tex](CH_{3} )_{4} N^{+}[/tex] and we should get our answer! We don't see any hydrogens. That means we can immediately rule out both A. and B.

Well, let's see. It looks like [tex]CH_{3}[/tex] is our R group. The subscript for it is also 4.

But we've used up our 3 bond spots, how is a quaternary amine possible? And that is why you see that your nitrogen has a positive charge in [tex](CH_{3} )_{4} N^{+}[/tex]. Since the nitrogen is bonded with more groups than it usually does, the electrons of nitrogen and therefore the electrical charge will be shared between more compounds and elements, leading to less charge for the nitrogen. This will cause it to have a positive charge. Making your amine an ion will allow for 4 bonds.

That is why even though nitrogen usually only has 3 bonds, we can still classify [tex](CH_{3} )_{4} N^{+}[/tex] as a quaternary amine.

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The last part got a little lengthy, but I hope I was able to explain this question in-depth! If you had any trouble understanding what I wrote, feel free to leave a comment. I will reply as soon as possible!

Answer:

SiO2

Explanation:

Si + O2 → SiO2

Answer:

It's D the number of organisms an ecosystem can support.

Check if it's right first then if it is, happy to help

Answer:

D: The number of organisms an ecosystem can support

Explanation:

D is the correct answer.

1.4 x 1023 molecules are present in 2.3 moles of oxygen.

We know that if we multiply the 2.3 moles of oxygen with Avogadro's number which is 6.023 x 1023 then this will give us the number of molecules which is present in the 2.3 moles of oxygen and which is 13.85 x 1023.

Mole is a standard scientific unit for measuring large quantities of very small entities like atoms, molecules, or other specific particles. The mole entitles an extremely large number of units about 6.02214076 × 1023.

So we can conclude that by multiplying Avogadro's number with the 2.3 moles of oxygen then it will give us 1.4 x 1023 molecules.

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Explanation:

The other half of the equation involves the hydrogen ions (initially bonded to the chloride ion in the hydrochloric acid). The hydrogen ions gain the electrons lost by the zinc atom, and bond together to form hydrogen gas. Since the hydrogen ions gain electrons, it is a reduction reaction.

Answer:

Answer:

The answer is D.

Explanation:

I know because I am currently doing this topic in Chemistry as well. I hope it was enough.

Answer:

A

smooth muscle is responsible

[tex]\huge{\textbf{\textsf{{\color{pink}{An}}{\red{sw}}{\orange{er}} {\color{yellow}{:}}}}}[/tex]

D. Two or more of the same type or different types of atoms.

Answer:

Dithiosulfatosilver

Explanation:

Answer:

Paraffin Wax

Explanation:

Paraffin wax is one of the most widely used waxes for candle making. A cheaper alternative, it is a derivable from petroleum, coal or oil. This wax is white and odorless and will keep a consistent appearance throughout your candle's life.

Answer:

.. ..

:O = C - Cl:

.. | ..

:Cl:

..

Answer:

70 180 000 kg

Explanation:

We can solve this problem by multiplying the amount of garbage each Canadian throws out each day (2.2 kg) by the number of Canadians:

Thus according to this problem, all Canadians throw out 70 180 000 kg of garbage each day.

Answer:

Eventually, these individual laws were combined into a single equation—the ideal gas ... We find that temperature and pressure are linearly related, and if the ... then P and T are directly proportional (again, when volume and moles of gas are ... of the variables, and they are more difficult to use in fitting theoretical equations ...

Explanation:

Answer:

reactivity, flammability, and if they rust

Answer: The statement, gases have no defined volume but a defined mass accurately depicts the mass and volume of a gas according to the kinetic molecular theory.

Explanation:

According to the kinetic molecular theory there exists constant motion between the particles of gas and perfectly elastic collisions also exist there between these particles.

Some postulates of Kinetic molecular theory are as follows.

1). The exists constant random motion in the particles of a gas.

2). There is negligible combined volume but a definite mass.

3). There exists no force of attraction between the particles of a gas.

4). Collisions which take place are completely elastic in nature.

Thus, we can conclude that the statement, gases have no defined volume but a defined mass accurately depicts the mass and volume of a gas according to the kinetic molecular theory.

The statement that accurately depicts the mass and volume of a gas according to the kinetic molecular theory is that gases have no defined volume but a defined mass.

The kinetic molecular theory of gases states constant motion exists between the particles of gas and perfectly elastic collisions also exist there between these particles.

The postulates of the kinetic molecular theory are as follows:

Based on these postulates, it can be said that gases have no defined volume but a defined mass.

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Answer:

Number of mole in HCL = 2.56 x 10⁻³ mol HCL

Explanation:

Given:

Volume of HCL = 25.62 ml

Molarity of HCL = 0.1 M

Find;

Number of mole in HCL

Computation:

Volume of HCL = 25.62 ml

Volume of HCL (in liter) = 25.62 / 1000

Volume of HCL (in liter) = 0.02562 L

Number of mole = Molarity x volume in liter

Number of mole in HCL = Volume of HCL (in liter) x Molarity of HCL

Number of mole in HCL = 0.02562 x 0.1

Number of mole in HCL = 0.002562

Number of mole in HCL = 2.56 x 10⁻³ mol HCL

The molecular formula of the unknown compound, given that it contains carbon, hydrogen, and oxygen is C₆H₁₂O₆

First, we'll begin by obtaining the mass of carbon, hydrogen and oxygen. Details below:

For Carbon

Mass of C = (12 / 44) × 1.649

Mass of C = 0.45 g

For Hydrogen

Mass of H = (2 / 18) × 0.675

Mass of H = 0.075 g

For Oxygen

Mass of O = (mass of compound) - (mass of C + mass of H)

Mass of O = 1.125 - (0.45 + 0.075)

Mass of O = 0.6 g

Next, we shall determine the empirical formula. details below:

Divide by their molar mass

C = 0.45 / 12 = 0.0375

H = 0.075 / 1 = 0.075

O = 0.6 / 16 = 0.0375

Divide by the smallest

C = 0.0375 / 0.0375 = 1

H = 0.075 / 0.0375 = 2

O = 0.0375 / 0.0375 = 1

Thus, the empirical formula of the compound is CH₂O

Finally, we shall determine the molecular formula. Details below:

Molecular formula = empirical × n = mass number

[CH₂O]n = 180

[12 + (2×1) + 16]n = 180

30n = 180

Divide both sides by 30

n = 180 / 30

n = 6

Molecular formula = [CH₂O]n

Molecular formula = [CH₂O]₆

Molecular formula = C₆H₁₂O₆

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Answer:

Option A. It has stayed the same.

Explanation:

To answer the question given above, we assumed:

Initial volume (V₁) = V

Initial temperature (T₁) = T

Initial pressure (P₁) = P

From the question given above, the following data were:

Final volume (V₂) = 2V

Final temperature (T₂) = 2T

Final pressure (P₂) =?

The final pressure of the gas can be obtained as follow:

P₁V₁/T₁ = P₂V₂/T₂

PV/T = P₂ × 2V / 2T

Cross multiply

P₂ × 2V × T = PV × 2T

Divide both side by 2V × T

P₂ = PV × 2T / 2V × T

P₂ = P

Thus, the final pressure is the same as the initial pressure.

Option A gives the correct answer to the question.

Natural vegetative propagation uses bulbils, aerial shoots, roots, underground stems, and subaerial stems. Utilizing unique vegetative components, artificial vegetative propagation is accomplished.

The production and growth of a new plant through asexual means or from a plant fragment is known as vegetative propagation. There is no gamete production or fertilization throughout the process of propagation.

In this procedure, tubers—modified roots—grow into new plants. In some plant species, adventitious buds even grow on the roots. If the correct circumstances are present, these buds develop into new plants or sprouts. Sweet potato and dahlia are two examples.

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Answer:  I believe the answer is A - the total growth of the plants.

Explanation:  The total growth of the plants depends on the hours of sunlight.    So, this would be your dependent variable, while the hours (amount) of sunlight would be your independent variable.  

Answer:

A

Explanation:

since that's what's being measured

According to the claim, Balloon A, which has more blue and less orange, exhibits accurate effusion at a relative effusion rate of 1.303. (orange is 1.303 times more effused than blue).Relative rate of effusion for the orange to blue spheres = 1.531.

The process of effusion is when a gas escapes from a container through a hole that is significantly smaller in diameter than the molecules' mean free path.

Gas molecules flow through a small opening in one container and into another by effusion. Graham's law allows rates of effusion to be compared at the same temperature. Diffusion is the random molecular motion-based movement of gas molecules through one or more other types of gas.

Rate of effusion of Orange / Rate of effusion of blue

= [Mblue / Morange]^1/2

Vrms = sqrt [(3RT/M)]

Vorange / Vblue = [ Mblue / Morange]^1/2

Rate of effusion of Orange / Rate of effusion of blue

= 565/ 369

= 1.531

Thus,  Balloon A, which has more blue and less orange, exhibits accurate effusion at a relative effusion rate of 1.303.

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