Which kind of bond occurs when two atoms share an electron

Answer:

a. Rate = k×[A]

b. k = 0.213s⁻¹

Explanation:

a. When you are studying the kinetics of a reaction such as:

A + B → Products.

General rate law must be like:

Rate = k×[A]ᵃ[B]ᵇ

You must make experiments change initial concentrations of A and B trying to find k, a and b parameters.

If you see experiments 1 and 3, concentration of A is doubled and the Rate of the reaction is doubled to. That means a = 1

Rate = k×[A]¹[B]ᵇ

In experiment 1 and to the concentration of B change from 1.50M to 2.50M but rate maintains the same. That is only possible if b = 0. (The kinetics of the reaction is indepent to [B]

Rate = k×[A][B]⁰

b. Replacing with values of experiment 1 (You can do the same with experiment 3 obtaining the same) k is:

Rate = k×[A]

0.320M/s = k×[1.50M]

Answer:

municipal

Explanation:

I got state and federal wrong and after it told me it was municipal and if you where wondering I got the answers wrong on purpose to answer this one the fastest but at least I am going you the correct answer not like other people who just say a answer just because they want to but I actually got the right answer but yeah. BYEEEEEE

Answer:

municipal

Explanation:

did it and got it right

i hope you have you a great day

Answer:

C. Chemical change

Explanation:

A physical change is where something is changed but it doesnt affect the build up of the chemical. For example, if you broke sticks and threw them on the ground, that would be a physical change because the change is happening to the physical being of the object and not its chemical buildup. However, if you lit those sticks on fire, that would be considered a chemical change because you end up with two substances, ash and the remnants of the stick. A nuclear reaction would result in something blowing up so its not that. And a physical property is like what it looks like or how it smells. Hope I helped you!

Four peripheral hydrogen atoms and two singly-bonded nitrogen atoms make up the molecule of hydrazine.

Thus, It is a colourless, poisonous irritant and sensitizer in its anhydrous form, which harms the central nervous system and causes symptoms as severe as tumours and convulsions.

In addition to having a strong reducing agent that makes it highly explosive, hydrazine has a strong smell that is similar to that of ammonia.

Given this, it appears odd that over 100,000 metric tonnes of the substance are produced annually throughout the world. But hydrazine does have an impact on our daily activities. It can save our lives, give us food and clothing, keep us warm, and even transport us to the moon. It even has the ability to go back in time.

Thus, Four peripheral hydrogen atoms and two singly-bonded nitrogen atoms make up the molecule of hydrazine.

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Answer:

Explanation:

The skeleton equation for the complete combustion of propane (C3H8) is:

C3H8 + O2 -> CO2 + H2O

The balanced chemical equation for the complete combustion of propane (C3H8), including all states of matter, is:

2 C3H8 (g) + 9 O2 (g) -> 6 CO2 (g) + 8 H2O (g)

In this equation, the propane (C3H8) is in the gaseous state, and the oxygen (O2) and the products (CO2 and H2O) are also in the gaseous state. This equation shows that when propane (C3H8) undergoes complete combustion, it reacts with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O) as products. The coefficients in front of each chemical species indicate the number of moles of each substance that are involved in the reaction.

Most fuels, when burned with only a stoichiometric amount of air and under no heat loss conditions, produce flames in the temperature range of approximately 1600 to 2200 degrees Fahrenheit (870 to 1200 degrees Celsius). This temperature range is based on the chemical reactions that occur during combustion and the specific properties of the combustible material being burned.

It is important to note that in practice most combustion processes do not occur under conditions where heat is not lost. In real-life situations, heat is usually lost to the environment through conduction, convection, and radiation. This may result in lower flame temperatures than that indicated  above.

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Answer:

\(\boxed {\boxed {\sf 1.6 \times 10^{24} \ atoms \ Si}}\)

Explanation:

We are asked to find how many atoms are in a sample of 75 grams of silicon.

First, we must convert grams to moles using the molar mass. This is the mass of 1 mole of a substance. The molar mass is found on the Periodic Table because it is equal to the atomic mass, but the units are grams per mole instead of atomic mass units.

Look up the molar mass of silicon.

We convert using dimensional analysis, so we must set up a conversion factor.

\(\frac {28.085 \ g \ Si}{1 \ mol \ Si}\)

We are converting 75 grams of silicon to moles, so we multiply by this value.

\(75 \ g \ Si*\frac {28.085 \ g \ Si}{1 \ mol \ Si}\)

Flip the conversion factor so the units of grams of silicon cancel.

\(75 \ g \ Si*\frac {1 \ mol \ Si}{28.085 \ g \ Si}\)

\(75 *\frac {1 \ mol \ Si}{28.085 }\)

\(\frac {75}{28.085} \ mol \ Si\)

\(2.670464661 \ mol \ Si\)

Next, we convert moles to atoms using Avogadro's Number, or 6.022 ×10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this problem, the particles are atoms of silicon.

Set up another conversion factor, this time with Avogadro's Number.

\(\frac {6.022 \times 10^{23} \ atoms \ Si}{1 \ mol \ Si}\)

Multiply by the number of moles we found.

\(2.670464661 \ mol \ Si*\frac {6.022 \times 10^{23} \ atoms \ Si}{1 \ mol \ Si}\)

The units of moles of silicon cancel.

\(2.670464661 *\frac {6.022 \times 10^{23} \ atoms \ Si}{1 }\)

\(2.670464661 * {6.022 \times 10^{23} \ atoms \ Si}\)

\(1.60815382 \times 10^{24} \ atoms \ Si\)

The original value of grams (75) has 2 significant figures, so our answer must have the same. For the number we found, that is the tenths place. The 0 in the hundredths place tells us to leave the 6 in the tenths place.

\(1.6 \times 10^{24} \ atoms \ Si\)

Answer:

m=621.54g

Explanation:

number of moles= mass of H2O/molar mass of H2O

n=m/mm

make mass the subject of the formula

m=n*mm

find molar mass of water

H2O=1(16)+2(1)

mm= 18.0g/mol. n=34.53mol

hence, m=34.53mol*18.0g/mol

m=621.54g

Answer:

1. To find the pH of a 0.1 M solution of HNO₂, we first need to calculate the concentration of H+ ions in the solution using the dissociation constant K.

HNO₂ ⇌ H+ + NO₂-

K = [H+][NO₂-]/[HNO₂]

We know that the initial concentration of HNO₂ is 0.1 M, and that the dissociation constant K is 4.5×10^-4. Let x be the concentration of H+ ions in the solution.

K = [H+][NO₂-]/[HNO₂]

4.5×10^-4 = x(0.1-x)/0.1

Simplifying the equation gives us:

x^2 - 4.5×10^-5x + 4.5×10^-6 = 0

Using the quadratic formula, we get:

x = (4.5×10^-5 ± √(4.5×10^-5 - 4(1)(4.5×10^-6))) / 2(1)

x = 1.5×10^-3 or 3×10^-4

Since the concentration of H+ ions must be less than 0.1 M, we reject the larger value and take x = 3×10^-4 M.

To calculate the pH, we use the formula:

pH = -log[H+]

pH = -log(3×10^-4) = 3.52

2. To find the pH of a 0.05 M solution of NHẠOH, we first need to calculate the concentration of OH- ions in the solution using the dissociation constant K.

NHẠOH ⇌ NH₂- + OH-

K = [NH₂-][OH-]/[NHẠOH]

We know that the initial concentration of NHẠOH is 0.05 M, and that the dissociation constant K is 1.8×10^5. Let x be the concentration of OH- ions in the solution.

K = [NH₂-][OH-]/[NHẠOH]

1.8×10^5 = x(0.05-x)/0.05

Simplifying the equation gives us:

x^2 - 1.8×10^-3x + 9×10^-6 = 0

Using the quadratic formula, we get:

x = (1.8×10^-3 ± √(1.8×10^-3 - 4(1)(9×10^-6))) / 2(1)

x = 9.0×10^-3 or 1×10^-3

Since the concentration of OH- ions must be less than 0.05 M, we reject the larger value and take x = 1×10^-3 M.

To calculate the pOH, we use the formula:

pOH = -log[OH-]

pOH = -log(1×10^-3) = 3

To calculate the pH, we use the formula:

pH + pOH = 14

pH = 14 - pOH = 11

3. To find the pH of a 0.3 M solution of H₂S, we first need to calculate the concentration of H+ ions in the solution using the dissociation constant K.

H₂S ⇌ H+ + HS-

K = [H+][HS-]/[H₂S]

We know that the initial concentration of H₂S is 0.3 M, and that the dissociation constant K is 1.0×10^-7. Let

Explanation:

Answer:

moles reactant = 2, moles of products = 3

Explanation:

The reactants are on the left side of the equation. Although, energy (heat) is a reactant it will not factor into the calculation for moles. The coefficient is the number of moles for each substances. So for the reactant side NO would have 2 moles because the coefficient is 2. Using the same logic, NO on the product side will have 2 moles and O2 will be 1 mole. O2 has one mole because it is implied that you know anything multipled by one is the same number. So 1 mole of O2 is written as O2 and not 1 O2.

Hope this helps,

if not just leave a comment!

If iron has a density of 7.87g/cm³ and a mass of 3.729g, then the volume of iron is 0.474cm³

HOW TO CALCULATE VOLUME:

Volume (mL) = mass (g) ÷ density (g/mL)

Volume = 3.729 ÷ 7.87

Volume = 0.474cm³

Therefore, the volume of iron is 0.474cm³

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Answer:

TRUE

Explanation:

plz brian list

Answer:

True

Explanation:

Answer:

large object to small objects

Answer:

The answer is C or warm objects to cool objects

Explanation:

Heat is the thermal energy flows from one substance to another when the substances differ in temperature. When you drink hot chocolate the heat from the liquid will flow to your mouth.

Yes, because it describes action-reaction force pairs

A fruit fly with long wings can have either LL genotype or Ll genotype.

Explanation:

The allele for long wings (L) is dominant, meaning that it will express itself over the recessive allele for short wings (l). A fruit fly can inherit one of each allele from each parent, resulting in three possible genotypes: LL (homozygous dominant), Ll (heterozygous), or ll (homozygous recessive).

If a fruit fly has long wings, then it must have at least one L allele. It could have inherited one L allele from one parent and another L allele from the other parent (LL genotype), or it could have inherited one L allele from one parent and a l allele from the other parent (Ll genotype). Therefore, the likely genotype of a fruit fly with long wings is either LL or Ll.

Answer: its the first answer

Explanation:

The force of attraction between the objects would be 1.07824 x 10^-11 N if the masses of the objects were to be doubled and the distance between them remained constant.

A change in motion between two objects or systems is described by the physical quantity known as force. Any push or pull on an object that could cause it to accelerate or change shape is what is referred to as it.

The following formula determines the force of attraction between two masses:

F = G * (m1 * m2) / d²

where m1 and m2 are the masses of the two objects, d is the separation between them, F is the force of attraction, G is the gravitational constant.

In this instance, we are aware of the 3.4 x 10^-2 N force of attraction between the two masses. The new force of attraction can be estimated as follows if we double the masses of each object while maintaining their respective distances:

F' = G * (2m1 * 2m2) / d²

F' = G * (4m1m2) / d²

We can set F' equal to x since we're interested in learning about the new force of attraction:

x = G * (4m1m2) / d²

Now that the values have been substituted, we can find x:

x = G * (4m1m2) / d²

Now we can substitute the given values and solve for x:

x = (6.674 x 10^-11 N m²/ kg²) * (4 * (m1 * 2) * (m2 * 2)) / d²

x = (6.674 x 10^-11 N m² / kg²) * (16 * m1m2) / d²

x = (6.674 x 10^-11 N m² / kg²) * 16 * 3.4 x 10^-2 N * m² / kg²/ d²

x = 1.07824 x 10^-11 N.

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Answer:

I guess C.The external parts of the volcano are the summit, slope and magma chamber.

Hope this helps

Answer:

Option C.The external parts of the volcano are the summit, slope and magma chamber.

Explanation:

Because they didn't mention some of the other external parts.  

After the reaction is complete, no nitrogen and no hydrogen molecules remain, and 8.00 x 1014 molecules of NH3 are produced.

In the equation, nitrogen and hydrogen react at a high temperature, in the presence of a catalyst, to produce ammonia, according to the balanced chemical equation:N2(g)+3H2(g)⟶2NH3(g)The coefficients of each molecule suggest that one molecule of nitrogen reacts with three molecules of hydrogen to create two molecules of ammonia.

So, to determine how many molecules of ammonia are produced when four nitrogen and nine hydrogen molecules are present, we must first determine which of the two reactants is the limiting reactant.

To find the limiting reactant, the number of moles of each reactant present in the equation must be determined.

Calculations:
Nitrogen (N2) molecules = 4Hence, the number of moles of N2 = 4/6.02 x 1023 mol-1 = 6.64 x 10-24 mol
Hydrogen (H2) molecules = 9Hence, the number of moles of H2 = 9/6.02 x 1023 mol-1 = 1.50 x 10-23 mol

Now we have to calculate the number of moles of NH3 produced when the number of moles of nitrogen and hydrogen are known, i.e., mole ratio of N2 and H2 is 1:3.

The mole ratio of N2 to NH3 is 1:2; thus, for every 1 mole of N2 consumed, 2 moles of NH3 are produced.
The mole ratio of H2 to NH3 is 3:2; thus, for every 3 moles of H2 consumed, 2 moles of NH3 are produced.
From these mole ratios, it can be observed that the limiting reactant is nitrogen.

Calculation for NH3 production:
Nitrogen (N2) moles = 6.64 x 10-24 moles
The mole ratio of N2 to NH3 is 1:2; therefore, moles of NH3 produced is 2 × 6.64 × 10−24 = 1.33 × 10−23 moles.

Now, to determine how many molecules of NH3 are produced, we need to convert moles to molecules.
1 mole = 6.02 x 1023 molecules
Thus, 1.33 x 10-23 moles of NH3 = 8.00 x 1014 molecules of NH3 produced.

To find the amount of each reactant remaining after the reaction is complete, we must first determine how many moles of nitrogen are consumed, then how many moles of hydrogen are consumed, and then subtract these from the initial number of moles of each reactant.

The moles of nitrogen consumed = 4 moles × 1 mole/1 mole N2 × 2 mole NH3/1 mole N2 = 8 moles NH3
The moles of hydrogen consumed = 9 moles × 2 mole NH3/3 mole H2 × 2 mole NH3/1 mole N2 = 4 moles NH3
Thus, the moles of nitrogen remaining = 6.64 × 10−24 mol – 8 × 2/3 × 6.02 × 10^23 mol-1 = 5.06 × 10−24 mol
The moles of hydrogen remaining = 1.50 × 10−23 mol – 4 × 2/3 × 6.02 × 10^23 mol-1 = 8.77 × 10−24 mol

Finally, the number of molecules of each reactant remaining can be calculated as follows:
Number of N2 molecules remaining = 5.06 × 10−24 mol × 6.02 × 10^23 molecules/mol = 3.05 × 10−1 molecules ≈ 0 molecules
Number of H2 molecules remaining = 8.77 × 10−24 mol × 6.02 × 10^23 molecules/mol = 5.28 × 10−1 molecules ≈ 0 molecules.

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Answer: 13.5 minutes to run 3.2 km.

Explanation: To solve this problem, you need to convert the distance from kilometers to miles and the speed from miles per hour to kilometers per hour. 3.2 km is approximately 1.988 miles and 5.5 mph is approximately 8.851 kph. To find the time it takes to run 1.988 miles at 8.851 kph, you can use the formula time = distance ÷ speed. Plugging in the values, you get time = 1.988 miles ÷ 8.851 kph, which simplifies to approximately 0.225 hours or 13.5 minutes.

Therefore, it will take the student approximately 13.5 minutes to run 3.2 km.

The concentration of the acid is \(0.25 M\).

Titration is a laboratory technique used to determine the concentration of a substance in a solution by reacting it with a standardized solution of known concentration.

The titration formula can be given by,

(Volume of the Base) \(\times\) (Normality of the Base) = (Volume of the Acid) \(\times\) (Normality of the Acid)

\(\Rightarrow V_1N_1=V_2N_2\)

Given, the volume of the base (\(NaOH\)), \(V_1 =25 ml\).

The concentration of the base (\(NaOH\)), \(M_1=0.5 M\).

The equivalence of the base (\(NaOH\)) is \(1\).

Hence, the normality of the base (\(NaOH\)), \(N_1=\frac{0.5}{1}N=0.5N\).

Given, the volume of the acid (\(HCl\)), \(V_2 =50 ml\).

Let us assume that the normality of the acid (\(HCl\)) \(N_2\).

Substitute the values in the given formula of titration.

\((25\times0.5)=(50 \times N_2)\\\Rightarrow 12.5=50N_2\\\Rightarrow N_2=\frac{12.5}{50} N\\\Rightarrow N_2=0.25 N\)

Hence, the normality of the acid (\(HCl\)), \(N_2=0.25 N\).

The equivalence of the acid (\(HCl\)) is \(1\).

Therefore, the concentration of the acid, \(M_1=\frac{0.25}{1}=0.25 M\).

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The equilibrium constant for the system at this temperature is\(1.17 × 10^-31 mol^2/L^2\).

For the chemical equation:

N2(g) + O2(g) ⇌ 2NO(g)

The equilibrium mixture at a temperature of 1500 K is determined to contain 6.4 × 10^-3 mol/L of N2,\(1.7 × 10^-3\)mol/L of O2 and 1.1 × 10^-5 mol/L of NO.  First, we need to calculate the concentration of N2 and O2 required to produce

1.1 × 10^-5 mol/L of NO:

2NO(g) = N2(g) + O2(g)

Given that there are 1.1 × 10^-5 mol/L of NO, the number of moles of N2 and O2 are equal since the stoichiometric ratio is 1:1. Therefore:

\(1.1 × 10^-5 mol/L\) = [N2][O2]Kc = \(([NO]^2)/([N2][O2])Kc\)= \((1.1 × 10^-5 mol/L)^2/(6.4 × 10^-3 mol/L)(1.7 × 10^-3 mol/L)Kc\) =

1.17 × 10^-31 mol^2/L^2.

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In the reaction between Fe₂S₃(s) and HBr(aq), the sum of the coefficients for all reactants and products is 12.

If the reaction occurs between Fe₂S₃(s) and HBr(aq) and the equation is properly balanced with the smallest whole-number coefficients, what is the sum of the coefficients for all reactants and products?

Let's consider the unbalanced equation for the reaction between Fe₂S₃(s) and HBr(aq). This is a double displacement reaction.

Fe₂S₃(s) + HBr(aq) ⇒ FeBr₃(aq) + H₂S(g)

We will balance it using the trial and error method.

First, we will balance Fe atoms by multiplying FeBr₃ by 2.

Fe₂S₃(s) + HBr(aq) ⇒ 2 FeBr₃(aq) + H₂S(g)

Then, we will balance S atoms by multiplying H₂S by 3.

Fe₂S₃(s) + HBr(aq) ⇒ 2 FeBr₃(aq) + 3 H₂S(g)

Finally, we will get the balanced equation by multiplying HBr by 6.

Fe₂S₃(s) + 6 HBr(aq) ⇒ 2 FeBr₃(aq) + 3 H₂S(g)

The sum of the coefficients for all reactants and products is:

\(1 + 6 + 2 +3 = 12\)

In the reaction between Fe₂S₃(s) and HBr(aq), the sum of the coefficients for all reactants and products is 12.

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\(\text{Ammonia has been studied as an alternative "clean" fuel for internal combustion}\)

\(\text{engines, since its reaction with oxygen produces only nitrogen and water vapor,}\)

\(\text{and in the liquid form it is easily transported. An industrial chemist studying this}\)

\(\text{reaction fills a} \ \mathbf{100 \ L }\ \text{tank with} \ \mathbf{8.6 \ mol} \ \text{of ammonia gas and} \ \mathbf{28 \ mol} \ \ \text{of oxygen gas, }\)

\(\text{to be} \ \mathbf{2.6\ mol} \ .\ \text{Calculate the concentration equilibrium constant for the combustion of}\)

\(\text{ammonia at the final temperature of the mixture. Round your answer to 2 significant digits.}\)

Answer:

Explanation:

From the correct question above:

The reaction can be represented as:

\(\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }\)

From the above reaction; the ICE table can be represented as:

                    \(\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }\)

I (mol/L)     0.086            0.28                 0              0

C                   -4x                -3x               +2x           +6x

E                 0.086 - 4x     0.28 - 3x      +2x             +6x

At equilibrium;

The water vapor = \(\dfrac{2.6 \ mol}{100 \ L} = 6x\)

\(x = \dfrac{2.6}{100} \times \dfrac{1}{6}\)

\(x = 0.00433\)

\(\text{equilibrium constant} ({k_c}) = \dfrac{ [N_2]^2 [H_2O]^6 }{ [[NH_3]^4] [O_2]^3 }\)

\(\implies \dfrac{(2x)^2 (6x)^6}{(0.086-4x)^4\times (0.28-3x)^3} \\ \\\)

Replacing the value of x, we have:

\(K_c = \dfrac{4 \times 46,656 \times x^8}{(0.086-4x)^4\times (0.28 -3x)^3} \\ \\ K_c = \dfrac{4 \times 46656 \times (0.00433)^8}{(0.06868)^4(0.26701)^3} \\ \\ K_c = \mathbf{5.4446 \times 10^{-8}}\)

\(K_c = \mathbf{5.5 \times 10^{-8} \ to \ 2 \ significant \ figures}\)

Answer:

I dont either lol

Explanation:

Answer:

thete are 672 seeds in 14 kg

7.32 moles of Chromium is present in 4.41 × 10²⁴ atoms.

Number of moles = \(\frac{\text{Given number of atoms}}{\text{Avogadro's Number}}\)

Avogadro's number is the number of particles in one mole of substance. 6.022 × 10²³ is known as Avogadro's Constant / Avogadro's Number.

Avogadro's Number = 6.022 × 10²³

Now put the values in above formula we get

Number of moles = \(\frac{\text{Given number of atoms}}{\text{Avogadro's Number}}\)

                             = \(\frac{4.41 \times 10^{24}}{6.022 \times 10^{23}}\)

                             = 7.32 moles

Thus from the above conclusion we can say that 7.32 moles of Chromium is present in 4.41 × 10²⁴ atoms.

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Answer:

\(0.745~M\)

Explanation:

In this case, we have a dilution problem. So, we have to use the dilution equation:

\(C_1*V_1=C_2*V_2\)

Now, we have to identify the variables:

\(C_1~=~14.9~M\)

\(V_1~=~25~mL\)

\(C_2~=~?\)

\(V_2~=~0.5~L\)

Now, we have different units for the volume, so we have to do the conversion:

\(0.5~L\frac{1000~mL}{1~L}=~500~mL\)

Now we can plug the values into the equation:

\(C_2=\frac{14.9~M*25~mL}{500~mL}=0.745~M\)

I hope it helps!