which of the following describes all of the types of water found in the hydrosphere

Answer:

Q =  57934.8 J

Explanation:

Given data:

Mass of water = 180.0 g

Initial temperature = 19°C

Fina temperature = 96°C

Specific heat capacity of water = 4.18 J/g.°C

Heat absorbed = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 96°C - 19°C

ΔT = 77°C

Q = 180.0 g×4.18 J/g.°C×77°C

Q =  57934.8 J

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

Answer:

27.4°C

Explanation:

Using the equation:

Q = m*C*T

Where Q is heat added,

m the mass of water

C specific heat of water (4.18J/g°C)

And T the increase in temperature

We can solve for the increase in temperature and thus, the final temperature of water:

Q = 88200J; m = 6500g:

88200J = 6500g*4.18J/g°C*T

3.2°C = T = increase in temperature

Final temperature is:

24.2°C + 3.2°C =

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.

Answer:

Molecules/State

Explanation:

We know from the kinetic molecular theory that matter is made up of molecules. These molecules are in constant motion at various velocities and energies.

The total thermal energy refers to the total kinetic energies of these particles. It depends on the temperature of the system, the energies of the molecules present in the material and the arrangement of the object's molecules (states of matter).

answer:

In biology, the energy that is stored in the bonds between compounds is termed as the chemical energy. This chemical energy is released and converted into different forms according to the functions of the body.

Potential energy can be described as the energy that is stored in an object. In physics, the energy that an object possesses while at rest is termed as the potential energy.

The energy that can be used during work is termed as the kinetic energy. In physics, kinetic energy can be described as the energy that an object possesses while in motion.

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Answer: The stored energy is termed as potential energy while the working energy is termed as kinetic energy. The electricity used in our homes is also a form of energy because it is a form of usable power. The places from which the different energies are obtained are known as energy sources.

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.

Answer: Sol:-

Data provided in the question is :-

Atomic mass of isotope -1 = 64 amu

Atomic mass of isotope -2 = 66 amu

Atomic mass of isotope -3 = 67 amu

Atomic mass of isotope -4 = 68 amu

Atomic mass of isotope - 5 = 70 amu

Percentage abundace of isotope - 1 = 48.89 %

Percentage abundance of isotope -2 = 27.81 %

Percentage abundance of isotope - 3 = 4.11%

Percentage abundance of isotope-4 = 18.57%

Percentage abundance of isotope - 5 = 0.62 %

Formula used :-

Average atomic mass of an element =[ {(atomic mass of isotope-1 * percentage abundance of isotope-1) + ( atomic mass of isotope-2 * percentage abundance of isotope -2) + ( atomic mass of isotope -3 * percantege abundance of isotope-3 ) + ( atomic mass of isotope-4 * percentage abundance of isotope-4) + (atomic mass of isotope-5 * percentage abundance of isotope-5)} / 100]

Calculation :-

Put all the value in the formula :-

Average atomic mass of an element = [{(64 * 48.89) + (66 * 27.81) + (67 * 4.11) + (68 * 18.57) + (70 * 0.62)} / 100] amu

= [{(3128.96) + (1835.46) +(257.37) + (1262.76) + (43.4)} / 100] amu

= {(6528.04) / 100} amu

= 65.2804 amu

Average atomic mass of an element is = 65.2804 amu

Then this mass is approximatly equal to atomic mass of zinc so this element would be zinc

atomic mass of zinc = 65.38 \approx 65.2804 amu

Answer:

Electron (e-)

Explanation:

An electron is just one of the fundamental particles

Answer:heyyyy

Explanation:

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:

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

Answer:

An ionic bond is the electrostatic force that holds ions together when they form bonds

Explanation:

An ionic bond is formed when a metal looses electron(s) to a non metal leading to the formation of a positive ion and a negative ion.

An ionic compound is actually an ion pair, the ions are held together by strong electrostatic forces.

This strong electrostatic force that holds the ion pair together in ionic compounds is what we commonly call the IONIC BOND.  

Answer:

An ionic bond is the electrostatic force that holds ions together when they form bonds

Explanation:

I took this test and it was the correct answer :)

A gray element that borders on the zigzag line of the periodic table, is ductile and malleable, but is not a very good conductor of heat or electricity is known as metalloid. The correct option is option C.

In chemistry, a periodic table is an orderly arrangement of each of the chemical elements along order of atomic number is, the total amount of protons inside the atomic nucleus.

Whenever the chemical elements stand grouped in this manner, there is a recurrent pattern in their characteristics known as the "periodic law," in which elements within a single column (group) exhibit comparable qualities. A gray element that borders on the zigzag line of the periodic table, is ductile and malleable, but is not a very good conductor of heat or electricity is known as metalloid.

Therefore, the correct option is option C.

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Answer:

H-H equation is written as follows:

pH=pK + log

{HCO3-}(base)

{H2CO3}(acid)

Answer:

5.78 × 10¹⁴ Hz

Explanation:

Step 1: Given and required data

Step 2: Convert "λ" to meters

We will use the conversion factor 1 m = 10⁹ nm.

519. nm × 1 m/10⁹ nm = 5.19 × 10⁻⁷ m

Step 3: Calculate the frecquency of this light

We will use the following expression.

c = λ × ν

ν = c/λ

ν = (3.00 × 10⁸ m/s)/5.19 × 10⁻⁷ m

ν = 5.78 × 10¹⁴ s⁻¹ = 5.78 × 10¹⁴ Hz

Answer:

B. The microscope paved the way

Explanation:

This section from the article, "History of the Cell: Discovering the Cell", that emphasizes the role of technological development in the discovery of cells, is the subheading, "The microscope paved the way". In this subheading, the author narrated how Robert Hooke improved the existing microscope, and was able to discover the cell through it.

Further improvement of the simple microscope by Antonie van Leeuwenhoek led to the discovery of bacteria and protozoa.

Answer:

8.1 × 10² g

Explanation:

Step 1: Write the balanced equation

2 C₅₇H₁₁₀O₆ + 163 O₂ ⇒ 114 CO₂ + 110 H₂O

Step 2: Convert 1.6 lb of C₅₇H₁₁₀O₆ to g

We will use the conversion factor 1 lb = 453.592 g.

1.6 lb × 453.592 g/1 lb = 7.3 × 10² g

Step 3: Calculate the moles corresponding to 7.3 × 10² g of C₅₇H₁₁₀O₆

The molar mass of C₅₇H₁₁₀O₆ is 890.83 g/mol.

7.3 × 10² g × 1 mol/890.83 g = 0.82 mol

Step 4: Calculate the moles of water produced from 0.82 moles of C₅₇H₁₁₀O₆

The molar ratio of C₅₇H₁₁₀O₆ to H₂O is 2:110. The moles of H₂O produced are 110/2 × 0.82 mol = 45 mol

Step 5: Calculate the mass corresponding to 45 moles of H₂O

The molar mass of H₂O is 18.02 g/mol.

45 mol × 18.02 g/mol = 8.1 × 10² g

Answer:

1.2 cm

Explanation:

The reason how I got my answer is by dividing 150 by 125 and that gave me 1.2 cm.

Answer:

Acidic solution used to clean a crucible

Explanation:

This liquid dissolves alcoholic solvents such as crucible, that is why it was selected as the ideal for cleaning.

Although it would be ideal to know in detail which chemical compound is the one you want to clean so that the cleaning technique has better effectiveness.

WASTE CONTAINER refers to the solvent used to rinse chemicals out of a beaker, and it also refers to the acid solution used to clean a crucible. Water used to rinse the detergent out of a flask refer to the SINK.

In conclusion, WASTE CONTAINER refers to the solvent used to rinse chemicals out of a beaker, and it also refers to the acid solution used to clean a crucible. Water used to rinse the detergent out of a flask refer to the SINK.

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Answer:

look it updjk jkj vjx cv xckv

Explanation:

look it upvjcvjkcxjbvjkcvbjbcxkjvjdbfcvbdw uvd vjbvjfjv jdbv

Answer:

element :

a substance that is made up of only one type of atom

I hope it helps you ✌

Answer:

1.17 M

Explanation:

Step 1: Write the balanced equation

N₂ + 3 H₂ ⇒ 2 NH₃

Step 2: Calculate the rate of disappearance of H₂

We will use the following expression.

rH₂ = - Δ[H₂]/t = - (0.25 M - 0.50 M)/44 s = 0.0057 M/s

Step 3: Calculate the rate of disappearance of N₂

The molar ratio of N₂ to H₂ is 1:3.

0.0057 mol H₂/L.s × 1 mol N₂/3 mol H₂ = 0.0019 mol N₂/L.s

Step 4: Calculate the final concentration of N₂

We will use the following expression.

[N₂] = [N₂]₀ - rN₂ × t

[N₂] = 1.25 mol/L - 0.0019 mol/L.s × 44 s

[N₂] = 1.17 M

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]

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.

Answer:

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

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.

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.

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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.

Answer:

c = 0.528 J/g.°C

Explanation:

Given data:

Mass of titanium = 43.56 g

Heat absorbed = 0.476 KJ  = 476 j

Initial temperature = 20.5°C

Final temperature = 41.2°C

Specific heat capacity = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 41.2°C - 20.5°C

ΔT = 20.7 °C

476 J = 43.56 g × c × 20.7 °C

476 J = 901.692 g.°C × c

c = 476 J / 901.692 g.°C

c = 0.528 J/g.°C  

The percent composition of the compound.

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

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 %

The  uncertainties  of  the  delta  measurements  and  the  uncertainty  of  the  atomic  weight  derivedfrom  the  best  measurement  of  isotopic  abundances  constrain  the  number  of  significant  figures  in  theatomic-weight values of the upper and lower bounds. For carbon, the fifth digit after the decimal pointis uncertain because of the uncertainty value of 0.000 027. Therefore, the number of significant digitsin the atomic-weight value is reduced to four figures after the decimal point. The Commission may rec-ommend  additional  conservatism  and  reduce  the  number  of  significant  figures  further.  For  the  lowerbound of carbon, 12.009 635 is truncated to 12.0096. For an upper bound, the trailing digit is increasedto ensure the atomic-weight interval encompasses the atomic-weight values of all normal materials. Inthe case of carbon, the upper bound is adjusted from 12.011 532 to 12.0116 to express four digits afterthe decimal point. The lower and upper bounds are evaluated so that the number of significant digits ineach  is  identical.  If  a  value  ends  with  a  zero,  it  may  need  to  be  included  in  the  value  to  express  therequired number of digits. The following are examples of lower and upper atomic-weight bounds foroxygen that could be published by the Commission in its various tables.

---------------------------

Extracted from" Atomic weights of the elements 2009 (IUPAC Technical Report)"

Answer:

Liquid

Explanation: