What happens to particles when in a substance with a a low temperature

where??? coz there is nothing below

The mass of the asteroid is C. 1.2 x \(10^{12}\) Kg

To find the mass of the other asteroid, we can rearrange the equation for the gravitational force between two objects:

F = (G * m1 * m2) / \(r^{2}\)

where F is the force of gravity, G is the gravitational constant, m1 and m2 are the masses of the two asteroids, and r is the distance between them.

Given that the distance between the asteroids is 75000 m, the force of gravity between them is 1.14 N, and one asteroid has a mass of 8 x \(10^{7}\) kg, we can substitute these values into the equation and solve for the mass of the other asteroid (m2):

1.14 N = (6.67430 × \(10^{-11}\) N \(m^{2}\)/\(Kg^{2}\) * 8 x \(10^{7}\) kg * \(m2\)) / \((75000 m)^{2}\)

Simplifying and solving the equation, we find that the mass of the other asteroid (m2) is approximately 1.2 x \(10^{12}\) kg. Therefore, Option C is correct.

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Two asteroids are 75000 m apart one has a mass of 8 x \(10^{7}\) kg if the force of gravity between them is 1.14 what is the mass of the asteroid

A. 3.4 x \(10^{11}\) kg

B. 8.3 x \(10^{12}\) kg

C. 1.2 x \(10^{12}\) kg

D. 1.2 x \(10^{10}\) kg

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Answer:

Tetrahedral

Explanation:

The molecular geometry of SiCl₄ is a tetrahedral shape. According to the VESPR molecular theory, the compound has a shape of a tetrahedron.

Answer:

Answer:

the answer should be B

Explanation:

contour lines

Answer:

B- contour lines

Explanation:

I just took the test

YW

Answer:

6.76 mol/L

Explanation:

A chemist prepares a solution 0.607 kg of calcium bromide by measuring out of calcium bromide into a 450. mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the chemist's calcium bromide solution. Be sure your answer has the correct number of significant digits.

Step 1: Given data

Step 2: Calculate the moles of solute

The molar mass of calcium bromide is 199.89 g/mol.

607 g × 1 mol/199.89 g = 3.04 mol

Step 3: Convert "V" to liters

We will use the conversion factor 1 L = 1000 mL.

450. mL × 1 L/1000 mL = 0.450 L

Step 4: Calculate the concentration of calcium bromide in mol/L

[CaBr₂] = 3.04 mol/0.450 L = 6.76 mol/L

Answer:

089

Explanation:

I have no idea

Based on the data from the calorimetry experiment, the enthalpy change, ΔH of the reaction is -286 KJ .

The enthalpy change, ΔH of a reaction is the amount of heat evolved or absorbed when reactant molecules react to form products.

From the calorimetry experiment, the determined heat of combustion of 1.01 g of H2(g) reacting with excess O2(g), qsoln = 143 kJ.

Heat is evolved in the reaction, therefore the qsoln = -143 kJ

Equation of reaction is given as:

H2(g) + ½ O2(g) → H2O (l)

Molar mass of H = 1.01 g

Mass of H2 = 1.01 g × 2 = 2.02 g

ΔH of the reaction = -143 kJ × 2.02/1.01

ΔH of the reaction = -286 KJ

Therefore, the enthalpy change, ΔH of the reaction is -286 KJ .

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Answer:

DNA Repair Gene

Explanation:

If a person has an error in a DNA repair gene, mistakes remain uncorrected. Then, the mistakes become mutations. These mutations may eventually lead to cancer, particularly mutations in tumor suppressor genes or oncogenes. Mutations in DNA repair genes may be inherited or acquired.

The ideal gas law can be used to solve this issue:

PV = nRT

where R is the universal gas constant, n is the number of moles, P is the gas's pressure, V is its volume, and T is its temperature.

The following expression can be used to link the starting and final volumes and moles under the assumption that the gas's pressure and temperature don't change:

V1 / n1 = V2 / n2

where V1 and n1 represent the volume and number of moles at the beginning, and V2 and n2 represent the volume and number of moles at the end.

Inputting the specified values results in:

15 L/n2 = 3 L / 3 mol

When we simplify and account for n2, we obtain:

n2 = 3 mol (15 L / 3 L)

n2 = 15

As a result, 15 mol of the gas are present when the volume is changed to 15 L.

Using the equation V = nV m, where V is the volume in liters, n is the amount of gas in moles, and V m is the molar gas volume in liters per mole, is another technique to determine the solution. We may substitute the volume and molar gas volume from the question into this equation.

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Answer: D

Because when you balance the combustion equation

C4H8O2 + 5O2--> 4CO2 + 4H2O

this is now obeying the law of combustion meaning

both sides have 4 carbons

8 hydrogens

12 oxygens

Explanation:

The median reaction at the second end point in the HCl and NaOH titration is: HCl + NaOH → NaCl + H2O

In a titration between hydrochloric acid (HCl) and sodium hydroxide (NaOH), the reaction involved is the neutralization reaction between an acid and a base. The balanced equation for this reaction is:

HCl + NaOH → NaCl + H2O

In this reaction, one mole of HCl reacts with one mole of NaOH to form one mole of NaCl (sodium chloride) and one mole of water.

During the titration process, the reaction occurs gradually as the base is added to the acid solution.

The first end point of the titration is reached when the moles of HCl and NaOH are stoichiometrically equivalent, meaning they react in a 1:1 ratio. At this point, all the HCl has been neutralized by the NaOH, and no excess of either reagent remains.

However, if the titration is continued beyond the first end point, the reaction between HCl and NaOH can still occur, albeit in a different ratio.

The second end point refers to the point where the moles of NaOH added exceed the stoichiometrically required amount to neutralize the HCl completely. As a result, any excess NaOH added after the second end point reacts with the excess HCl in a 1:1 ratio.

Therefore, the median reaction at the second end point in the HCl and NaOH titration is:

HCl + NaOH → NaCl + H2O

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Rounding to three significant figures, potassium chlorate makes up 55.0% of the mixture by mass.

2 KClO3(s) → 2 KCl(s) + 3 O2(g)

2.04 g O2 × (1 mol O2 / 32.00 g O2) = 0.06375 mol O2

We can determine the quantity of KClO3 in the sample by calculating the amount of O2 produced by 2 moles of KClO3:

0.04250 mol KClO3 = 0.06375 mol O2 (2 mol KClO3 / 3 mol O2)

Percent by mass of KClO3 = (0.04250 mol KClO3 × 122.55 g/mol KClO3) / 9.51 g × 100% = 55.0%

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Answer:

A. Ca, Ba, Be, Sr

A. The synthesis of polymer A from monomers can be represented by the following equation:

Monomer + Monomer --> Polymer A

Where the monomers are represented by the structure –O(CH2)5CO– and the polymer is represented by –[–O(CH2)5CO–]n–.

B. The synthesis of polymer B from monomers can be represented by the following equation:

Monomer + Monomer --> Polymer B

Where the monomers are represented by the structure –OCH2CH2CH2– and the polymer is represented by –(–OCH2CH2CH2–)n–.

It is important to note that these equations are simplified representations of the overall chemical reactions involved in the synthesis of these polymers from monomers, and they do not take into account the specific chemical reactions and conditions that may be required in order to synthesize the polymers.

Answer:

CH2

Explanation:

The empirical formula is the lowest ratio form of a compound. You can divide both C and H by 3 to get to CH2.

Step 1 - Remembering what are acids and bases

Acids and bases are inorganic functions, i.e., classes of substances which share some common properties.

Bases usually liberate OH- when dissolved in water, while acids liberate H+ in the same conditions.

Both OH- and H+ ions have effects in the human body, specially because they can act as catalysts, i.e., they may accelerate or propitiate several chemical reactions.

Step 2 - Hazards of acids and bases

Being such strong catalysts, acids and bases may accelerate the decomposition of a lot of substances, resulting in corrosion or dismantlement of some tissues, including human tissues, such as skin, muscular tissues etc.

Of course, not all acids or bases are that agressive. It depends on their strenght, which can be measure by the pH scale. Extremes pHs, both high and low, are always dangerous, for they indicate a high amount of H+ or OH-.

Some acids, such as HF, do not harm the skin right away. We won't feel anything, in fact, if it drops on our hand. But it can very easily penetrate the skin, affecting muscular tissues and specially our calcium rich bones, "dissolving" them.

Some acids and bases can't be handled daily. Their sale are either controlled by the government of by the army. Some other substances demand training so you are aware of how to manipulate and deal with them.

At equilibrium, the number of moles of \(Cl_2\) (g) will be 0.2025 mol.

1: Write the balanced chemical equation:

\(C_O\)(g) + \(Cl_2\)(g) ⟶ \(C_OCl_2\)(g)

2: Set up an ICE table to track the changes in moles of the substances involved in the reaction.

Initial:

\(C_O\)(g) = 0.3500 mol

\(Cl_2\)(g) = 0.05500 mol

\(C_OCl_2\)(g) = 0 mol

Change:

\(C_O\)(g) = -x

\(Cl_2\)(g) = -x

\(C_OCl_2\)(g) = +x

Equilibrium:

\(C_O\)(g) = 0.3500 - x mol

\(Cl_2\)(g) = 0.05500 - x mol

\(C_OCl_2\)(g) = x mol

3: Write the expression for the equilibrium constant (Kc) using the concentrations of the species involved:

Kc = [\(C_OCl_2\)(g)] / [\(C_O\)(g)] * [\(Cl_2\)(g)]

4: Substitute the given equilibrium constant (Kc) value into the expression:

1.2 x \(10^3\) = x / (0.3500 - x) * (0.05500 - x)

5: Solve the equation for x. Rearrange the equation to obtain a quadratic equation:

1.2 x \(10^3\) * (0.3500 - x) * (0.05500 - x) = x

6: Simplify and solve the quadratic equation. This can be done by multiplying out the terms, rearranging the equation to standard quadratic form, and then using the quadratic formula.

7: After solving the quadratic equation, you will find two possible values for x. However, since the number of moles cannot be negative, we discard the negative solution.

8: The positive value of x represents the number of moles of \(Cl_2\)(g) at equilibrium. Substitute the value of x into the expression for \(Cl_2\)(g):

\(Cl_2\)(g) = 0.05500 - x

9: Calculate the value of \(Cl_2\)(g) at equilibrium:

\(Cl_2\)(g) = 0.05500 - x

\(Cl_2\)(g) = 0.05500 - (positive value of x)

10: Calculate the final value of \(Cl_2\) (g) at equilibrium to get the answer.

Therefore, at equilibrium, the number of moles of \(Cl_2\) (g) will be 0.2025 mol.

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Conversion of 7.6cm^3 to m^3 is 0.0000076m^3

Divide the volume by the conversion ratio to convert a cubic centimeter measurement to a cubic meter measurement. The cubic centimeters divided by 1,000,000 gives the volume in cubic meters.

The same attribute is expressed using a unit conversion, but in a different unit of measurement. For instance, time can be stated in minutes rather than hours, and distance can be expressed in kilometres, feet, or any other measurement unit instead of miles.

We need to convert between units in order to ensure accuracy and prevent measurement misinterpretation. For example, we do not measure a pencil's length in kilometers. In this situation, one must convert from kilometers (km) to centimeters (cm).

Conversion:

1m = 100cm

1m = 10^2cm

1cm = 10^-2m

1cm^3 = (10^-2)^3m^3

1cm^3 = 10^-6m^3

7.6cm^3 = 7.6 x 10^-6m^3

7.6cm^3 = 0.0000076m^3

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A student enters the canteen and buys iced tea. He then goes on to add more sugar to his iced tea. In this solution, iced tea is the solvent and sugar is the solute.

The ionization of the following acids can be represented as:

Tetraoxosulphate (VI) Acid (\(H_{2}SO_{4}\)) ionizes as H+ and SO4^2- ions.

Trioxonitrate (V) Acid (\(HNO_{3}\)) ionizes as H+ and \(NO_{3-}\) ions.

Ethanoic Acid (\(CH_{3}COOH\)) ionizes as H+ and \(CH_{3}COO^{-}\) ions.

Tetraoxosulphate (VI) Acid, also known as sulfuric acid (\(H_{2}SO_{4}\)), ionizes as follows:

\(H_{2}SO_{4}\) → \(H+\) + \(SO_{4}^{2-}\)

In this reaction, sulfuric acid donates two hydrogen ions (H+) to the solution, forming sulfate ions (\(SO_{4}^{2-}\)).

Trioxonitrate (V) Acid, commonly known as nitric acid (\(HNO_{3}\)), ionizes as follows:

\(HNO_{3}\) → \(H+_{}\) + \(NO_{3-}\)

Nitric acid dissociates to release one hydrogen ion (\(H+\)) and a nitrate ion  (\(NO_{3-}\)).

Ethanoic Acid, also known as acetic acid (\(CH_{3}COOH\)), ionizes as follows:

\(CH_{3}COOH\) → H+ + \(CH_{3}COO^{-}\)

Acetic acid donates a hydrogen ion (H+) to the solution, forming an acetate ion (\(CH_{3}COO^{-}\)).

In all cases, the acids dissociate in water, producing hydrogen ions (H+) as positively charged ions and their corresponding anions. The hydrogen ions are responsible for the acidic properties of these substances, while the anions contribute to the overall charge balance in the solution. The ionization of acids allows them to conduct electricity in aqueous solutions and react with other substances.

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Indicate the ionization of the following acids,

Tetraoxosulphate (VI) Acid

Trioxonitrate (V) Acid

Ethanoic Acid.​

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3 sigma and 3 pieee

okok kkk

The molar mass of MgCrO4 is approximately 140.30 g/mol.

To determine the molar mass of MgCrO4 (magnesium chromate), we need to calculate the sum of the atomic masses of each individual element in the compound.

The chemical formula MgCrO4 indicates that the compound consists of one magnesium atom (Mg), one chromium atom (Cr), and four oxygen atoms (O).

The atomic masses of the elements can be found on the periodic table:

Magnesium (Mg) has an atomic mass of approximately 24.31 g/mol.

Chromium (Cr) has an atomic mass of around 51.99 g/mol.

Oxygen (O) has an atomic mass of about 16.00 g/mol.

Now, we can calculate the molar mass of MgCrO4 by summing up the atomic masses of each element, considering the respective subscripts:

Molar mass = (Atomic mass of Mg) + (Atomic mass of Cr) + 4 × (Atomic mass of O)

Molar mass = (24.31 g/mol) + (51.99 g/mol) + 4 × (16.00 g/mol)

Molar mass = 24.31 g/mol + 51.99 g/mol + 64.00 g/mol

Molar mass ≈ 140.30 g/mol

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Answer:

Explanation:

The first step in order to solve this particular question is to make sure that the two reactions given in the question is balanced. Therefore, we have;

Reaction 1: C6H12O6 -----------------------------> 2 C2H5OH + 2 CO2.

Reaction 2: C6H12O6 ------------------------------> 2 C2H3CO2H + 2 H2O.

Next, we determine the number of moles of water and that of glucose. Recall that we are given from the question that the open flow reactor =  4000 kg of a 12% glucose-water solution flows in that is to say the percentage for water is [100% - 12% = 88%]. Also, the molar mass of water = H₂O = 18 kg/kmol and that for glucose =180 kg/kmol.

Number of moles of water = (4000 kg × 88%) ÷ 18 = 195.6 kmol.

Number of moles of glucose= (4000kg × 12%) ÷ 180 = 2.67 kmol.

Next thing to do is to determine the number of moles in the unreacted glucose . Therefore, the 90 kg of unreacted glucose ÷ 180kg/kmol = 0.5  kmol.

So, we have that During fermentation, 120 kg of carbon dioxide is produced.  Thus, the number of kmol = 120kg÷ 44kg/mol = 2.73kmol.

For reaction 1, we have 2 moles of CO₂ that is to say the extent of the reaction = 2.73kmol / 2 moles of CO₂ = 1.365 kmol.

For reaction 2, we have 2 moles of CO₂ that is to say the extent of the reaction = 2.67kmol - 0.5  kmol - 1.365kmol = 0.805 kmol.

For both reaction, the kmol for outflow of glucose = 2.73 kmol.

Also, 2 × 0.805 + 1.365 × 0 = 1.61kmol.

Hence, 195.6 kmol + 1.61 =197.21 kmol.

The mass of ethanol = 46.1 kg/kmol × 2.73 kmol = 125.853 kg.

The weight percent of ethanol has been 29.72%, and the percent mass of propionic acid has been 47.79%.

The balanced chemical equations of the reactions have been:

\(\rm C_6H_1_2O_6\;\rightarrow\;2\;C_2H_5OH\;+\;2\;CO_2\)

\(\rm C_6H_1_2O_6\;\rightarrow\;2\;C_2H_3CO_2H\;+\;2\;H_2O\)

The solution has consisted of 12% glucose.

Water in the solution = 88%

The mass of the solution = 4000 kg.

The moles of glucose = 12% of 4000 kg

Moles of glucose = \(\rm \dfrac{12}{100}\;\times\;4000\;\times\;\dfrac{1}{180\;g/mol}\)

Moles of glucose = 2.67 kmol.

Moles of water =  88 % of 4000 kg

Moles of water = \(\rm \dfrac{88}{100}\;\times\;4000\;\times\;\dfrac{1}{18\;g/mol}\)

Moles of water = 195.6 kmol.

The unreacted glucose in the mixture = 90 kg

Moles of unreacted glucose:

Moles = \(\rm \dfrac{weight}{molecular\;weight}\)

Moles of unreacted glucose = \(\rm \dfrac{90\;\times\;1000\;g}{180\;g/mol}\)

Moles of unreacted glucose = 0.5 kmol

The mass of carbon dioxide produced = 120 kg.

Moles of carbon dioxide produced = \(\rm \dfrac{120\;\times\;1000\;g}{44\;g/mol}\)

Moles of carbon dioxide produced = 2.73 kmol

Since 1 mole of glucose produces 2 moles of carbon dioxide.

2.73 kmol of carbon dioxide has been produced from 1.365 kmol of glucose.

The moles of ethanol produced by reaction 1 = 2 moles/ mole glucose.

The glucose present has been 2.67 kmol.

The ethanol produced = 5.34 kmol.

Moles of propionic acid produced = 5.34 kmol.

The mass of 5.34 kmol ethanol = Moles × molecular weight

The mass of ethanol produced = 5.34 × 1000 × 46.07g/mol

The mass of ethanol produced = 246.0138 kg.

The mass of propionic acid produced = 5.34  × 1000 × 74.08 g/mol

The mass of propionic acid produced = 395.5872 kg.

The mass of water produced = 5.34 × 1000 × 18 g/mol

The mass of water produced = 96.12 kg.

The remained glucose = 90 kg

The total mass in the reactor:

= Mass of glucose + water + propionic acid + ethanol

= 90 + 96.12 + 395.5872 + 246.0138 kg

= 827.721 kg.

% Mass of ethanol = \(\rm \dfrac{246.0138}{827.721}\;\times\;100\)

% Mass of ethanol = 29.72%

% Mass of Propionic acid = \(\rm \dfrac{395.5872}{827.721}\;\times\;100\)

% Mass of Propionic acid = 47.79 %.

The weight percent of ethanol has been 29.72%, and the percent mass of propionic acid has been 47.79%.

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Answer:

The forest products industry plays a major role in the Swedish economy, and accounts for between nine and 12 percent of Swedish industry's total employment, exports, sales and added value. It includes companies within the pulp and paper industry, as well as the wood-mechanical industry.

The zero order reaction's rate law is [A] = [4]. -K+ 1 Where [A] = Initial reactant concentration Order rate rate constant is K. At time t, [A] represents the reactant's concentration. At ty₂, [A] = [Ao] 2[A] - [A] = - K-

the procedure, event, or act of reacting. opposing or resisting a force, movement, or influence. Particularly: a propensity for an old, typically antiquated political or social structure or policy; a reaction to a given treatment, circumstance, or stimulus.

A physical reaction is one in which molecules rearrange themselves but do not undergo any chemical changes. Without a change in composition, physical responses can result in changes to the texture, size, temperature, and state.

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Answer:

168 ml

Explanation:

M1V1= M2V2

V1=M2V2/M1

V1= 0.195×15.5/18

V1 = 0.1679 L=168ml

Answer:

Explanation:

Given,

\(M_1 = 18M\\\\M_2 = 0.195\\\\V_1 = ?\\\\V_2 = 15.5\)

Therefore,

\(M_1 * V_1 = M_2 * V_2\\\\ V_1 = \frac{M_2*V_2}{M_1}\\\\V_1 = \frac{0.195*15.5}{18}\\\\V_1 = 0.168L \\\\V_1 = 168mL\)

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Divide the weight with 1,000 times the component or material's density to the conversion from grammes to litres. Sulfur hexafluoride in litres is produced from 11.0g of sulphur hexafluoride gas. 0.0753138778434317.

What volume of water is equivalent to one kilogramme. When pure water attains a maximum density of one kg/l at a temperature around 39.2 °F or 4 °C, it has a volume of 1 litre. 1 kilogramme of water is somewhat more than 1 litre at higher temperatures.

As there is one mole of SF in every six moles, there are now 0274 moles in total. After solving, we get the result of.164 mole of fluorine. Adding 1 mole will also have 6 moles od fluorine, giving us 0.0274 moles in s f, 6, corresponding to 6 times in 0.0274 mole of chlorine. Hence, in 4 grammes.

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Answer:

the electron will be drawn more toward the atom with the higher e lectronegativity resulting in a polar covalent bond.

Explanation:

When atoms of different elements share electrons through covalent bonding, the electron will be drawn more toward the atom with the higher e lectronegativity resulting in a polar covalent bond. When compared to ionic compounds, covalent compounds usually have a lower melting and boiling point, and have less of a tendency to dissolve in water.