Can anybody help it 8th grade science about the sun

Answer:

Earth's climate has changed throughout history. Just in the last 650,000 years there have been seven cycles of glacial advance and retreat, with the abrupt end of the last ice age about 11,700 years ago marking the beginning of the modern climate era — and of human civilization. Most of these climate changes are attributed to very small variations in Earth’s orbit that change the amount of solar energy our planet receives.

Explanation:

i might be right

When atoms of an element are energized, we see color because the electrons absorb the energy and jump to a higher energy level. When the electrons go back to their original energy level, they release the energy in the form of light.

The frequency and wavelength of the light depend on the amount of energy that was absorbed and released. Different amounts of energy result in different colors. The absorption of energy results in the promotion of an electron to a higher energy level. This electron is not stable at the higher level and quickly falls back to the original energy level.

As the electron drops, it releases energy in the form of a photon, which is a tiny packet of light. The energy of the photon is equal to the difference in energy between the two energy levels. Different energy levels produce different colors of light, which is why different elements emit different colors.

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Answer:

The answer is A. Decreased Fluidity

Explanation:

Taking the test on preformance matters

In evaporation, particles of liquids take energy in the form of sunlight from the surrounding and then convert into vapor phase. Increased kinetic energy causes certain liquid molecules to become vapor molecules.

Evaporation is a physical process in which the water in liquid state converts to gaseous state at the temperature below its boiling point. Evaporation happens before the boiling process.

Evaporation causes cooling effect in the surrounding. During summer when water is kept in earthen pot water cools, this is because of the evaporation process.

Evaporation is affected by temperature. As the temperature increases, kinetic energy of liquid molecules also increases and hence the rate of evaporation increases.

Therefore, increased kinetic energy causes certain liquid molecules to become vapor molecules.

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Answer:

Xe that is xenon

Explanation:

Xe that is xenon

In the above intermediate chemical equation, oxygen will appear as follows: O₂(g) as a reactant (option B).

A chemical equation in chemistry is a symbolic representation of a chemical reaction where reactants are represented on the left, and products on the right.

According to this question, an intermediate chemical equation is presented as follows:

As observed in the above chemical equation, oxygen will react in its gaseous form i.e. as a reactant.

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The concentration of CO (g) in the equilibrium mixture is 0.020 M. In other words, only a small amount of CO (g) is produced in this reaction at 1000°C. T 0.020 M.

The concentration of CO in an equilibrium mixture (in mol/L) is 5.8 M.

Given that the concentration of H2O (g) is [H2O] = 0.500 M at 1000°C, and  the reaction is:

C(s) + H2O(g) ⇄ CO(g) + H2(g) KC = 0.2 at 1000°C

We need to determine the concentration of CO in an equilibrium mixture (in mol/L)

if a reaction mixture initially contains only reactants.

We can solve this problem using the ICE table method as follows:

Let x be the change in concentration of H2O (g) and CO (g) when they reach equilibrium.

Then the equilibrium concentrations of CO (g) and H2 (g) are equal to x. Hence, the equilibrium concentration of H2O (g) is (0.500 - x) M. Substitute these values in the expression for Kc and solve for x.

Kc = [CO (g)] [H2 (g)] / [H2O (g)] [C (s)]

= 0.2[CO (g)] = Kc [H2O (g)] [C (s)] / [H2 (g)]

= 0.2 × (0.500 - x) / x

We can simplify this expression by cross-multiplication to get:

5x = 0.1 - 0.2xx = 0.02 M

Substituting x = 0.02 M in the expression for [CO (g)], we get:

[CO (g)] = 0.2 × (0.500 - 0.02) / 0.02 = 5.8 M (approx.)

Therefore, the concentration of CO (g) in an equilibrium mixture (in mol/L) is 5.8 M.                                                                                               The problem requires us to find the equilibrium concentration of CO (g) in a mixture that initially contains only reactants.

To solve this problem, we need to use the expression for the equilibrium constant Kc, which is given by:

Kc = [CO (g)] [H2 (g)] / [H2O (g)] [C (s)]

We can also use the ICE table method to solve this problem. In this method, we start with the initial concentration of the reactants and calculate the change in concentration of each species as they reach equilibrium.

We then use the equilibrium concentrations to calculate the value of Kc and solve for the unknowns. Here is how we can set up the ICE table for this problem: Reaction:

C(s) + H2O(g) ⇄ CO(g) + H2(g)

Initial: [C] = [H2]

= 0 M,

[H2O] = 0.500 M

Equilibrium: [C] = [H2] = x,

[H2O] = 0.500 - x,

[CO] = [H2] = x

Change: +x +x -x -x

Substituting the equilibrium concentrations into the expression for Kc, we get:

Kc = [CO] [H2] / [H2O] [C]

= x² / (0.500 - x)

= 0.2

Solving for x, we get: x = 0.020 M

Substituting this value of x into the expression for [CO], we get:

[CO] = x = 0.020 M

Therefore, the concentration of CO (g) in the equilibrium mixture is 0.020 M.

In other words, only a small amount of CO (g) is produced in this reaction at 1000°C. T 0.020 M.

The concentration of CO in an equilibrium mixture (in mol/L) is 5.8 M.

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To tell a Cu(NO3)2 solution from a Ni(NO3)2 solution, you can add KI to the solution. If the solution contains Cu(NO3)2, the brown precipitate will be formed, but if the solution contains Ni(NO3)2, no precipitate will be formed.

This is the best way to distinguish between these two solutions, since adding KOH or Na2C2O4 will result in blueish precipitates, being harder to identify if it is Cu(NO3)2 or Ni(NO3)2 and with Na2SO4 no reaction will happen.

It means that the answer is: By adding KI and seeing if a brown precipitate is formed.

1. LDFs

2. Intermolecular Forces

3. Intramolecular Forces

4. Linear

5. Tetrahedral

Edit: I'm new to this site and idk how to use it properly. I'm not sure about 2 and 3 currently because these forces are between molecules as well so INTERmolecular would be used twice (?)

Answer:

atomic weight or number of neutrons.

Explanation:

An isotope: is one of two or more forms of the same chemical element.

Different isotopes of an element have the same number of protons in the nucleus, giving them the same atomic number.

A different number of neutrons giving each elemental isotope a different atomic weight.

Answer:

• Reacting aqueous Sodium hydroxide solution with Magnesium chloride.

Explanation:

\({ \rm{2NaOH _{(aq)} + MgCl _{2(s)} \: \dashrightarrow \: 2NaCl _{(s)} + Mg(OH)_{2(aq)} }}\)

Answer:

Sodium hydroxide solution + Magnesium chloride = sodium chloride crystals

Explanation:

Answer:

The study of the human body as a machine for the performance of work has its foundations in three major areas of study—namely, mechanics, anatomy, and physiology; more specifically, biomechanics, musculoskeletal anatomy, and neuromuscular physiology. Explanation:

The total volumes of solution for the 1:5, 1:10, 1:20, and 1:40 dilutions are 10.4 mL, 11 mL, 14 mL, and 19 mL, respectively.

In the case study, four dilutions of an albumin stock solution were prepared: 1:5, 1:10, 1:20, and 1:40. The total volume of solution for each dilution can be calculated using the formula:

\( \text{Total volume} = \frac{\text{Volume of stock solution}}{\text{Dilution factor}} \)

For example, if 2 mL of the stock solution was diluted to 10 mL to create a 1:5 dilution, the calculation would be:

\( \text{Total volume} = \frac{2 \, \text{mL}}{5} = 0.4 \, \text{mL} \)

Thus, the total volume of solution for the 1:5 dilution is 10 mL + 0.4 mL = 10.4 mL.

Similarly, the total volumes for the other dilutions can be calculated:

1:10 dilution: total volume = 10 mL + 1 mL = 11 mL

1:20 dilution: total volume = 10 mL + 4 mL = 14 mL

1:40 dilution: total volume = 10 mL + 9 mL = 19 mL

The total volumes of solution for the 1:5, 1:10, 1:20, and 1:40 dilutions are 10.4 mL, 11 mL, 14 mL, and 19 mL, respectively.

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Answer:

light grey

Explanation:

cause it just got mixed

When mixed polycarboxylate cement is mixed correctly, it should have a syrup-like in consistency, which thickens as it sets.

One for the few dental fillings that exhibits real adherence to tooth structure is polycarboxylate cement. The liquid consists polyacrylic acid and a copolymer made from that acid, while the powder is predominantly zinc oxide. As filling substances, protective cavity lines, luting materials as crowns, bridges, inlays, and orthodontic appliances, as well as for root canal fillings and pulp capping, they are also employed in these procedures. When mixed polycarboxylate cement is mixed correctly, it should have a syrup-like in consistency, which thickens as it sets.

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Answer:

Leucippus and Democritus

Explanation:

Hope this helps~ :D

Thank you so much! :D

Answer:

(CH3COO)2 + redP + Cl2 → ClCH2COOH + HCl

Explanation:

This is an example of halogenation of carboxylic acids at alpha carbon atom. In this reaction, red phosphorus and chlorine are treated with carboxylic acids having alpha hydrogen atom followed by hydrolysis to form alpha chloro carboxylic acid.

It’s A) (0.1978)(10.01) + (0.8022)(11.01)

Relative and average atomic mass both describe properties of an element related to its different isotopes. Out of these two Relative atomic mas is more accurate. Therefore option A is correct option.

Mass defines the quantity of a substance. It is measured in gram or kilogram. Average mass is the mass of atoms of an element that are isotopes. It can be calculated by multiplying mass of a isotope to natural abundance of that isotope.

Mathematically,

average atomic mass=  ∑(relative abundance x mass of isotope)

average atomic mass boron= (relative abundance boron-10  x mass of isotope boron-10)+(relative abundance boron-11  x mass of isotope boron-11 )

average atomic mass boron=(0.1978)(10.01) + (0.8022)(11.01)

Therefore option A is correct option.

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The two types of hard water is temporary hard water and permanent hard water. Temporary hard water is softened by boiling.

What is hard water?

Hard water is known to be highly concentrated in calcium and magnesium largely due to to the ions of these elements dissolved in ground water that have the presence of these elements which could be sourced from rocks with the additional help with water from rain. A physical evidence that a water is hard water would be that it doesn't lather easily with soap.

In summary, temporary hard water responds to heating which makes the water much more easier to be used for laundry, and drinking.

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Answer:

1. The pressure in outer space is so low that many consider it as non-existant. It has a pressure of 1.322 × 10−11 Pa. Pressure may be detected from the molecule of air or water hitting you. Since there is very little air and hardly ever water hitting you in space, pressure is almost zero or negligible.

2. Standard sea-level pressure, by definition, equals 760 mm (29.92 inches) of mercury, 14.70 pounds per square inch, 1,013.25 × 103 dynes per square centimetre, 1,013.25 millibars, one standard atmosphere, or 101.325 kilopascals.

3. 0.46J/gC

explanation.The specific heat capacity of a material is given by:

C_s = \frac{Q}{m \Delta T}C s = mΔTQ

where

Q is the amount of heat absorbedm is the mass

\Delta TΔT is the variation of temperature

For the piece of iron in the problem:

m = 15.75 gm=15.75g

Q=1086.75 JQ=1086.75J

\Delta T=175 C-25 C=150^{\circ}ΔT=175C−25C=150∘

Substituting into the equation,

C_s = \frac{1086.75 J}{(15.75 g)(150^{\circ}C)}=0.46 J/gCC s = (15.75g)(150 ∘ C)1086.75J = 0.46J/gC

4. 207 J

Explanation:

The key to this problem lies with aluminium's specific heat, which as you know tells you how much heat is needed in order to increase the temperature of 1 g of a given substance by 1∘C.

#I hope it's help

Answer:

3.82 g.

Explanation:

To calculate the mass of sulfuric acid produced from 1.25 g of sulfur, we'll have to balance the chemical equations and use stoichiometry.

Starting with the first reaction:

2S(s) + 3O2(g) → 2SO3(g)

Since 1.25 g of sulfur is reacted, the number of moles of sulfur can be calculated as:

moles = mass / molar mass = 1.25 g / 32 g/mol = 0.03906 mol

Next, using the mole ratio from the balanced chemical equation, we can find the number of moles of sulfur trioxide produced:

2 moles of sulfur produce 2 moles of sulfur trioxide, so:

moles of SO3 = moles of S x (moles of SO3 / moles of S) = 0.03906 mol x (2 mol / 2 mol) = 0.03906 mol

Finally, we move on to the second reaction, the dissolution of sulfur trioxide in water:

SO3(g) + H2O(l) → H2SO4(l)

Using the mole ratio from this balanced equation, we can find the number of moles of sulfuric acid produced:

1 mole of sulfur trioxide reacts with 1 mole of water to produce 1 mole of sulfuric acid, so:

moles of H2SO4 = moles of SO3 x (moles of H2SO4 / moles of SO3) = 0.03906 mol x (1 mol / 1 mol) = 0.03906 mol

The mass of sulfuric acid produced can be calculated using the moles and the molar mass of sulfuric acid:

mass = moles x molar mass = 0.03906 mol x 98 g/mol = 3.82 g

Therefore, if 1.25 g of sulfur is reacted, the mass of sulfuric acid produced is approximately 3.82 g.

Quarts of pure antifreeze must be added to 5 quarts of a 20 antifreeze solution to obtain a 50 antifreeze solution is 3.

Antifreeze solution is a solute that lowers the freezing point of the liquid in a solution.

For calculating the amount of quarts of pure antifreeze solution added can be given by:

5×20/100 + x × 100% = (x+5) × 50/100

Where, x is the amount of antifreeze required for addition and pure antifreeze is 100% of the antifreeze taken.

1 + x = (x+5) 0.5

1 + x = 0.5x + 2.5

0.5x = 1.5

x = 3

So, 3 quarts of pure antifreeze must be added to 5 quarts of a 20 antifreeze solution to obtain a 50 antifreeze solution.

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If you react 1.000 g of lead ions with 1.250 g of the chemical chosen in part a, the theoretical amount of precipitate produced is 26.3 g.

To calculate the mass of precipitate that would be theoretically produced, we need to use the mole ratio between the lead ions and the chemical chosen in Part A. We will call the chemical chosen in Part A "X" for this example.

1.000 g of lead ions is equivalent to 1.000 g/207.2 g/mol = 0.0048 mol of lead ions.

1.250 g of X is equivalent to 1.250 g/molar mass of X = 0.0375 mol of X.

Now, we can calculate the mole ratio of lead ions to X, which is 0.0048 mol/0.0375 mol = 0.128.

We can use the mole ratio to calculate the mass of the precipitate, which is 0.128 x (207.2 g/mol of lead) = 26.3 g of precipitate.

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Using the ideal gas law PV =nRTPV=nRT , we find that the pressure will be P =\frac{nRT}{V}P=

V

nRT

​

 . Then, we'll substitute and find the pressure, using T = -25 °C = 248.15 K and R = 0.0821 \frac{atm\cdot L}{mol \cdot K}

mol⋅K

atm⋅L

​

 :

P =\frac{nRT}{V} = \frac{(0.33\,\cancel{mol})(0.0821\frac{atm\cdot \cancel{L}}{\cancel{mol \cdot K}})(248.15\,\cancel{K})}{15.0\,\cancel{L}} = 0.4482\,atmP=

V

nRT

​

=

15.0

L

​

(0.33

mol

)(0.0821

mol⋅K

atm⋅

L

​

​

)(248.15

K

​

)

​

=0.4482atm

In conclusion, the pressure of this gas is P=0.4482 atm.

Reference:

Chang, R. (2010). Chemistry. McGraw-Hill, New York.

Answer:

A covalent bond is a chemical bond that involves the sharing of electron pairs between atoms. These electron pairs are known as shared pairs or bonding pairs, and the stable balance of attractive and repulsive forces between atoms, when they share electrons, is known as covalent bonding.

Explanation:

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Answer:

the hydrogen atom of one water molecule and the lone pair of electrons on an oxygen atom of a neighboring water molecule.

Answer:

If it reflects all light, it looks white. Color can help identify a material and, while it is a physical property, it can be used together with chemical experiments when the experiments produce a known material with a specific color.

Explanation:

The ability to reflect light is the inherent and fundamental property of that substance.

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