Which probing question lies within the scope of physics?
O A.
Are fish in the open ocean
attracted by underwater sounds?
O B.
Does increasing the saltiness of
ocean water affect the speed
of sound in the water?
O C.
What effect does the release of industrial wastewater have on
the acidity of oceans?
D. What is the effect of rising sea temperatures on
ocean currents?

Answers

Answer 1

Answer:

answer number C is the correct answer for this


Related Questions

Me please answer as follows

Answers

Answer:

no reaction occurs .that is no product

A chemist determines by measurements that 0.0800 moles of bromine liquid participate in a chemical reaction. Calculate the mass of bromine liquid that participates. Round your answer to 3 significant digits.

Answers

Answer:

The mass of bromine liquid that participates in  a chemical reaction=12.8 g

Explanation:

We are given that

Total number of moles of bromine liquid participate in chemical reaction=0.0800 moles

We have to find the mass of bromine liquid that participates.

Atomic mass of Br=79.9 g

1  mole of bromine liquid=2 atomic mass of bromine (Br)

1 mole of bromine liquid ([tex]Br_2[/tex]) =[tex]2\times 79.9=159.8 g[/tex]

0.0800 moles of  bromine liquid=[tex]159.8\times 0.0800[/tex] g

0.0800 moles of  bromine liquid=12.784 g

0.0800 moles of  bromine liquid[tex]\approx 12.8[/tex] g

Hence, the mass of bromine liquid that participates in  a chemical reaction=12.8 g

Draw the Lewis structure for the polyatomic formate anion. Be sure to include all resonance structures that satisfy the octet rule.

Answers

Answer:

Lewis structure of polyatomic formate anion.

Explanation:

To draw Lewis structure for any chemical species,

1)Count the total number of valence electrons present in it.

This can be obtained by adding valence electrons of each constituent atom.

2)Arrange those valence electrons in such a way that each atom should attain eight electrons around it to satisfy octet theory.

The structure of formate ion and its Lewis structure are shown below:

HCOO- is the formate ion.

It has total:

1+4+6+6+1 = 18 valence electrons.

Since, hydrogen has one, carbon has four and oxygen has six valence electrons and the charge of the anion is one.

Arrange this 18 electrons in such a way that each atom should get 8 electrons around it.

Resonance structures of formate ion:

what is the machine used to check melting point called?​

Answers

Answer:

Melting-point apparatus

Which is used to measure conc. HCl for preparation of 0.1M HCl solution?

a. Volumetric flask c. Measuring cylinder
b. Pipette d. Wash bottle​

Answers

Answer:

option a

hope helps you

have a great day

What molecule is this

Answers

Answer:

That is a " ball and stick " model which represents carbon compounds.

Explanation:

This is the answer. Hope it helps you find what you're looking for.

Identify a process that is NOT reversible. Identify a process that is NOT reversible. melting of snow baking of bread deposition of carbon dioxide freezing water melting of aluminum

Answers

Answer:

Identify a process that is NOT reversible.

Melting of snow

baking of bread

deposition of carbon dioxide

freezing water

melting of aluminum

Explanation:

A physical change is the one in which there is a change only in its physical state, color, the appearance of the substance. But the chemical composition of the substance remains unchanged.

It is a temporary change and can be reversed easily.

For example:

melting, freezing, deposition etc.

Baking is a permanent change and the chemical composition of the substance changes.

Hence, among the given options, baking of bread is not a reversible change.

Out of the following all are physical changes except baking of bread and physical changes are reversible so the process  which is not reversible is baking of bread.

What are physical changes?

Physical changes are defined as changes which affect only the form of a substance but not it's chemical composition. They are used to separate mixtures in to chemical components but cannot be used to separate compounds to simpler compounds.

Physical changes are always reversible using physical means and involve a change in the physical properties.Examples of physical changes include melting,boiling , change in texture, size,color,volume and density.Magnetism, crystallization, formation of alloys are all reversible and hence physical changes.

They involve only rearrangement of atoms and are often characterized to be changes which are reversible.

Learn more about physical changes,here:

https://brainly.com/question/17931044

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Consider the synthesis of water as shown in Model 3. A container is filled with 10,0 g of H, and
5.0 g of Oz
Which reactant (hydrogen or oxygen) is the limiting reactant in this case?

Answers

Answer:

Oxygen, O₂ is the limiting reactant

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2H₂ + O₂ —> 2H₂O

Next, we shall determine the masses of H₂ and O₂ that reacted from the balanced equation. This can be obtained as follow:

Molar mass of H₂ = 2 × 1 = 2 g/mol

Mass of H₂ from the balanced equation = 2 × 2 = 4 g

Molar mass of O₂ = 16 × 2 = 32 g/mol

Mass of O₂O from the balanced equation = 1 × 32 = 32 g

SUMMARY:

From the balanced equation above,

4 g of H₂ reacted with 32 g of O₂.

Finally, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

4 g of H₂ reacted with 32 g of O₂.

Therefore, 10 g of H₂ will react with

= (10 × 32)/4 = 80 g of O₂.

From the calculations made above, we can see that a higher mass (i.e 80 g) of O₂ than what was given (i.e 5 g) is required to react completely with 10 g of H₂. Therefore, O₂ is the limiting reactant.

Oxygen has been the limiting reactant in the reaction.

A limiting reactant can be defined as the reactant in the reaction in which the product concentration has been dependent.

The balanced equation for the formation of water has been:

[tex]\rm 2\;H_2\;+\;O_2\;\rightarrow\;2\;H_2O[/tex]

For the formation of reaction to form 2 moles of water, 2 moles of hydrogen reacts with 1 mole of oxygen.

The moles can be calculated as:

Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]

The moles of Hydrogen in 10 g [tex]\rm H_2[/tex]:

Moles = [tex]\rm \dfrac{10}{2}[/tex]

Moles of hydrogen = 5 mol.

Moles of Oxygen in 5 grams Oxygen:

Moles = [tex]\rm \dfrac{5}{32}[/tex]

Moles of oxygen = 0.156 mol.

For the reaction with 2 moles of Hydrogen 1 mole of Oxygen has been required.

For reacting with 5 mol of Hydrogen, moles of oxygen required are:

Moles of oxygen = [tex]\rm \dfrac{1}{2}\;\times\;5[/tex]

Moles of oxygen required = 2.5 moles.

The available oxygen = 0.156 moles.

Since the moles of oxygen available is lesser than required, the formation of the product has been dependent on the concentration of the oxygen.

Thus, oxygen has been the limiting reactant in the reaction.

For more information about the limiting reactant, refer to the link:

https://brainly.com/question/14225536

The graph below shows how the temperature and volume of a gas vary when
the number of moles and the pressure of the gas are held constant. How can
the volume of the gas be increased if the pressure is constant?
v
т
A. By increasing the temperature
B. By letting the gas expand over time
C. By letting the gas contract over time
D. By decreasing the temperature

Answers

Answer:

D

Explanation:

PV =nRT

So V and T are inversely proportional

In a pure metal, the electrons can be thought of as [ Select ] throughout the metal. Using molecular orbital theory, there [ Select ] an energy gap between the filled molecular orbitals and empty molecular orbitals. The [ Select ] orbitals are typically higher in energy and are mostly [ Select ] .

Answers

Answer:

Explanation:

In a pure metal, the electrons can be thought of as [concentrated] around atoms throughout the metal. Using molecular orbital theory, there [is ] an energy gap between the filled molecular orbitals and empty molecular orbitals. The [antibonding] orbitals are typically higher in energy and are mostly (filled]

A quantity of 1.435 g of naphthalene , was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the water rose from 20.28oC to 25.95oC If the heat capacity of the bomb plus water was , calculate the heat of combustion of naphthalene on a molar basis; that is, find the molar heat of combustion.

Answers

Answer:

molar heat of combustion = -5156 *10³ kJ/mol

Explanation:

A quantity of 1.435 g of naphthalene , was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the water rose from 20.28oC to 25.95oC If the heat capacity of the bomb plus water was 10.17 kJ/°C, calculate the heat of combustion of naphthalene on a molar basis; that is, find the molar heat of combustion.

Step 1: Data given

Mass of naphthalene = 1.435 grams

Initial temperature of water = 20.28 °C

Final temperature of water = 25.95 °C

heat capacity of the bomb plus water was 10.17 kJ/°C

Molar mass naphtalene = 128.2 g/mol

Step 2:

Qcal = Ccal * ΔT

⇒with Qcal =the heat of combustion

⇒with Ccal = heat capacity of the bomb plus water = 10.17 kJ/°C

⇒with ΔT = the difference in temperature = T2 - T1 = 25.95 - 20.28 = 5.67°C

Qcal = 10.17 kJ/°C * 5.67 °C

Qcal = 57.7 kJ

Step 3: Calculate moles

Moles naphthalene = 1.435 grams / 128.2 g/mol

Moles naphthalene = 0.01119 moles

Step 4: Calculate the molar heat of combustion

molar heat of combustion = Qcal/ moles

molar heat of combustion = -57.7 kJ/ 0.01119 moles

molar heat of combustion = -5156 *10³ kJ/mol

One student measured a spectrum and observed double yellow lines. He claimed that it must be Sodium. Please justify if he is correct. Why

Answers

Answer:

We know that the student was measuring a spectrum, and observed double yellow lines, he claimed that it was Sodium.

There are multiple elements with double yellow lines, like Mercury or Sodium, but Sodium has two bright yellow lines, so it is usually identified by them.

So when we look at a spectrum and we see a strong doublet in the yellow range, we can easily assume that it is Sodium.

Here we assume that the student only saw the yellow doublet, this would imply that the yellow doublet is way more intense than the other lines, that can't be seen (while for other elements with double yellow lines, we should see other lines with similar intensity) then we can conclude that it is Sodium.

The student is correct.

how to separate and purify the Flufenamic acid from the aqueous layer​

Answers

Answer:

Explanation:

H

The products obtained from hydroboration-oxidation of cis-2-butene are identical to the products obtained from hydroboration-oxidation of trans-2-butene. Draw the products and explain why the configuration of the starting alkene is not relevant in this case.

Answers

Answer:

a) Attached below

b)  The presence of racemic mixture found as product in both  cases shows that   products are identical ( i.e. they have same configuration

Explanation:

Diagrams of the products obtained from hydroboration-oxidation of cis-2-butene , hydroboration-oxidation of trans-2-butene.

attached below

The presence of racemic mixture found as product in both  cases shows that   products are identical ( i.e. they have same configuration )

Is it better to use graphite or carbon electrodes during electrolysis if I am trying to investigate volume of gas produced ?​

Answers

Answer:

yes

Explanation:

it is important for electrolysis

The element least likely to obey the octet rule in forming chemical bonds is Group of answer choices oxygen. carbon. nitrogen. neon. fluorine.

Answers

Answer:

neon

Explanation:

The octet rule states that atoms of elements are stable when they possess 8 electrons on their outermost shell.

Hence, atoms of elements participate in chemical reactions in order to attain this octet structure (eight electrons in the outermost shell).

Neon is a noble gas and already has eight electrons in its outermost shell. Hence, neon is least likely to obey the octet rule.

what is sterilization​

Answers

Answer:

Sterilization refers to any process that removes, kills, or deactivates all forms of life and other biological agents like prions present in a specific surface, object or fluid, for example food or biological culture media.

Plz help me ASAP in my final project I am ready to pay 20$

Answers

I’ll help you for free no $20

Ammonia is produced by the reaction of nitrogen and hydrogen: N2(g) + H2(g)  NH3(g)
(a) Balance the chemical equation.
(b) Calculate the mass of ammonia produced when 35.0g of nitrogen reacts with hydrogen.

Answers

Answer:

a) N2 (g) + H2 = 2 NH3

b) You have to state the mass of hydrogen

Which of the five type of equilibrium problems best applies to this question: Consider the following reaction at equilibrium. What effect will reducing the volume of the reaction mixture have on the system

Answers

Answer:

d

Explanation:

What is the mass number of an ion with 106 electrons, 157 neutrons, and a +1 charge?

Answers

Answer:

264 g/mol

Explanation:

#electrons equal #protons = 106

Plus 1 charge => m protons = 106 + 1 = 107

Mass number: 107 + 157 = 264 g/mol

g Calculate the theoretical yield (in grams) of your product if you start with 0.50 grams of E-stilbene. The molecular weight of E-stilbene is 180.25 g/mol, and the molecular weight of the product is 340.058 g/mol. 0.5109 grams 0.9433 g 0.2342 g 0.6312 g

Answers

Answer:

0.9433g

Explanation:

Theoretical yield is defined as the mass produced assuming all reactant reacts producing the product.

Assuming the reaction is 1:1, we need to find the moles of E-stilbene (Reactant). If all reactant reacts, the moles of E-stilbene = Moles of product.

Using the molar mass of the product we can find the theoretical yield as follows:

Moles E-stilbene:

0.50g * (1mol/180.25g) = 0.00277 moles = Moles Product

Mass Product = Theoretical yield:

0.00277 moles * (340.058g/mol) = 0.9433g

Will give brainliest answer please give explanation


33.5 cs=_________s

Answers

Answer: 3.4 × 10 -1 equal 33.5 cs. May I have brainiest? pls. (the -1 is to the power of the ten.

How many molecules make up 12.8g of N2O4​

Answers

Answer:

8.35 × 10 ²² molecules

Explanation:

gram molecular weight of N2O4

= 2 × 14 + 4 × 16

= 28 + 64

= 92g

1 mole N2O4 has 6 × 10²³ molecules of N2O4 and weighs 92g

therfore,

92 g has = 6 × 10²³ molecules

12.8 g has =

[tex] \frac{6 \times 10 {}^{23}}{92} \times 12.8[/tex]

= 0.835 × 10²³

= 8.35 × 10 ²² molecules

what the movement of the earth around the sun​

Answers

Answer:

Yan po Ang sagot NASA pic

Explanation:

pa heart po plss

at pa vote

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Answer:

The movement of the earth around the sun in a fixed path or orbit is called Revolution. The axis of the earth which is an imaginary line, makes an angle of 66½° with its orbital plane. ... The earth takes about 24 hours to complete one rotation around its axis.

Which compound contains both sigma and pi bonds... HCCl3, H2CO, H2S, or HBr?

Answers

Answer:

H2CO

Explanation:

Becuase it has 2 sigma bonds plus one pi bond and one sigma bond that consitute the double bond between C and O.

Answer:

B. H2CO

Explanation:

Please help help please

Answers

Answer:  The correct answer is B.

Explanation:  Segregate most organic acids from oxidizing mineral acids. Keep oxidizers away from other chemicals, especially flammables.

Answer:

Segregate most organic acids from oxidizing mineral acids. Keep oxidizers away from other chemicals, especially flammables, combustibles, and toxic materials. Keep corrosives away from substances that they may react with and release corrosive, toxic, or flammable vapors.

Please help meee! Chemistry!

Answers

P=18000000/6 zeros. not sure how to do rest

Explanation:

a) [tex]n = \dfrac{PV}{RT} = \dfrac{(1.8×10^7\:\text{Pa})(3\:\text{L})}{(8310\:\text{L•Pa/mol•K})(300\:\text{K})}[/tex]

[tex]\:\:\:\:\:\:\:= 21.7\:\text{mol}[/tex]

b) [tex]P = \dfrac{nRT}{V}[/tex]

[tex]\:\:\:\:\:\:\:\:\:= \dfrac{(50\:\text{mol})(8310\:\text{L•Pa/mol•K})(300\:K)}{(3\:L)}[/tex]

[tex]\:\:\:\:\:\:\:\:\:=4.2×10^7\:\text{Pa}[/tex]

what are the methods used to determine the boiling point of a compound?​

Answers

Answer:

There are a variety of methods by which a sample's boiling point can be determined, including distillation, reflux, and by using a Thiele tube. The most straightforward method uses a Thiele tube, and has the advantage of using less than 0.5mL of material.

Distillation Method

There are simpler methods than a distillation to measure a compound's boiling point, and it is recommended to explore other options (e.g. Thiele tube) if this is the only goal. However, if materials are limited, or if a purification is planned anyhow, a distillation can be used to determine a compound's boiling point. The distillation technique is discussed in great detail in Chapter 5.

A simple distillation should suffice for most situations and at least 5mL of sample should be used in the distilling flask along with a few boiling stones or stir bar. As the bulk of the material distills, the highest temperature noted on the thermometer corresponds to the boiling point. A major source of error with this method is recording too low a temperature, before hot vapors fully immerse the thermometer bulb. 5 Be sure to monitor the thermometer periodically, especially when the distillation is active. Record the barometric pressure along with the boiling point.

Reflux Method

A reflux setup can also be used to determine a compound's boiling point. Reflux is when a liquid is actively boiling and condensing, with the condensed liquid returning to the original flask. It is analogous to a distillation setup, with the main difference being the vertical placement of the condenser.

Thiele Tube Method

There are a variety of methods by which a sample's boiling point can be determined, including distillation, reflux and by using a Thiele tube. The most straightforward method uses a Thilele tube, and has the advantage of using less than 0.5ml of material.

If 11g of a gas occupies 5.6dm'3at s.t.p., calculate it's vapour density (1.0mol of a gas occupies 22.4dm'3at s.t.p.).​

Answers

Answer:

[tex]{ \boxed{ \bf{vapour \: density = 2 \times molecular \: mass}}} \\{ \tt{ PV= (\frac{m}{ m_{r}}) RT}} \\ { \tt{3 \times 5.6 = \frac{11}{m _{r}} \times 0.0831 \times 273}} \\ { \tt{m _{r} = 14.85 \: g}} \\ \\ { \bf{vapour \: density = 2 \times m _{r}}} \\ = 2 \times 14.85 \\ = 29.7 \: { \tt{g {dm}^{ - 3} }}[/tex]

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