A sample of a compound contains 41.33 g of carbon and 8.67 g of hydrogen. The molar mass of the
compound is 87.18 g/mol. What is the molecular formula for the compound?
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Wind direction temperature 20⁰C to 25⁰C is the temperature range at which the wind direction changes. The main answer is that the temperature range 20⁰C to 25⁰C is a relatively mild temperature range for wind direction changes, and wind direction changes can occur in this temperature range.

Wind direction is the direction from which the wind is blowing. Temperature affects wind direction in various ways. The main factor is the temperature gradient, which is the change in temperature over a given distance. When the temperature gradient is high, the wind tends to blow from areas of high pressure to areas of low pressure. Wind direction is also influenced by the Coriolis effect, which is the effect of the Earth's rotation on the direction of the wind. The Coriolis effect causes the wind to be deflected to the right in the Northern Hemisphere and to the left in the Southern Hemisphere. Wind direction can also be influenced by local topography, such as mountains and valleys, and by the presence of large bodies of water, such as lakes and oceans.

Wind direction temperature 20⁰C to 25⁰C is a relatively mild temperature range for wind direction changes, and wind direction changes can occur in this temperature range. Temperature affects wind direction in various ways, including through the temperature gradient, the Coriolis effect, and local topography and the presence of large bodies of water. Wind direction is an important factor in weather forecasting and can affect many aspects of human life, from agriculture to transportation to energy production.

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During the electrolysis of molten sodium chloride (NaCl), the substance produced at the cathode is metallic sodium (Na). Therefore, the correct answer is option E: Na metal.

During the electrolysis process, when an electric current is passed through molten NaCl, the NaCl dissociates into its ions: Na+ and Cl-. At the cathode, which is the negative electrode, the positively charged Na+ ions are attracted and undergo reduction. The reduction reaction at the cathode involves gaining electrons to form neutral sodium atoms, which then combine to form metallic sodium (Na). This is represented by the equation: 2Na+ + 2e- → 2Na.

The other options, such as O2 (option A), Cl2 (option B), NaOH (option C), and H2 (option D), are products that can be formed during different electrochemical processes, but they are not the products of the electrolysis of molten NaCl. In this particular case, the main product at the cathode is sodium metal.

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When naming ionic compounds, there are several rules that should be followed. Here are three important rules to remember:

1) Use the name of the cation first

2) Use the correct suffix for the anion

3) Use Roman numerals to indicate the charge of the cation

1) Use the name of the cation first: In an ionic compound, the name of the cation comes first, followed by the name of the anion. For example, in the compound NaCl, the cation is sodium and the anion is chloride.

2) Use the correct suffix for the anion: The suffix for the anion depends on its charge. For example, if the anion has a single negative charge, the suffix is "-ide". So, chloride becomes chloride ion (Cl-) and sulfide becomes sulfide ion (S2-).

3) Use Roman numerals to indicate the charge of the cation: If the cation can have multiple charges, the charge must be specified using a Roman numeral in parentheses. For example, iron (III)  chloride (FeCl3) indicates that the iron cation has a +3 charge. The Roman numeral is equal to the charge of the cation.

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The moles of hno3 are present if 0.180 molmol of ba(oh)2 was needed to neutralize the acid solution is 0.36 mole of HNO₃.

Neutralization or neutralisation is a chemical reaction in which acid and a base react quantitatively with each other. In a reaction in water, neutralization outcomes in there being no excess of hydrogen or hydroxide ions gift inside the answer.

Calculation:-

Valance factor of HNO₃ is = 1

valance factor of Ba(OH)₂ is = 2

mole of Ba(OH)₂ is = 0.180

therefore mole oh HNO₃ neede is = 2 * 0.180

                                                         =  0.36 mole of HNO₃

Neutralisation is a chemical response wherein an acid and a base react with each different quantitatively. it's also written as Neutralisation. The acid electricity of the reactant gives the pH of the neutralised solution.

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Answer:

A chemical change has occurred, with energy being given off.

Explanation:

The liquids mix and the stick gives off energy in light form

Knowing that the percent yield of this reaction was 72.0%, then 0.3681 moles I₂ was reacted with excess H₂.

Percent yield is defined as the ratio of the actual yield to the theoretical yield multiplied by 100.

Percent yield = (actual yield / theoretical yield) x 100

If a quantity of I₂ was reacted with excess H₂ to give 0.530 moles of HI, and the percent yield of this reaction was 72.0%, then the theoretical yield of the reaction should be 0.7361 moles of HI.

Percent yield = (actual yield / theoretical yield) x 100

72 = (0.530 / theoretical yield) x 100

theoretical yield = 0.7361 moles

Then, write a balanced equation for the reaction of I₂ and H₂ to produce HI.

I₂ + H₂ ⇒ 2 HI

From the balanced equation, 1 mole of I₂ requires 1 mole of H₂ to produce 2 moles of HI.

Theoretically, there should be 0.7361 moles of HI.

Let x = moles of I₂

1 mole of I₂ : 2 moles of HI = x mole of I₂ : 0.7361 moles of HI

x = 0.3681 moles I₂

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The molar mass of a 0.314-gram sample of gas having a volume of 1.6L at 287K and 0.92 atm is 0.503g/mol.

The temperature and pressure of a gas that behaves like an ideal gas is directly proportional to one another.

First, calculating molar mass :

The molar mass of a substance can be calculated by dividing the mass of the substance by its number of moles.

However, the number of moles can be calculated using the formula:

PV = nRT

From where;

P = pressure

V = volume

R = gas law constant

T = temperature

n = number of moles

0.92 × 1.6 = n × 0.0821 × 287

1.47 = 23.56n

n = 1.47/23.56

n = 0.624 moles

Molar mass = 0.314 g ÷ 0.624 mol

Molar mass = 0.503 g/mol

Therefore, the molar mass of a 0.314-gram sample of gas having a volume of 1.6L at 287K and 0.92 atm is 0.503g/mol.

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Answer:

water evaporation in a dish

water vapour condensation on a glass

Explanation:

because the is a more of heat

The total change in entropy of the system and the surroundings  277 J K−1 mol−1 .

The change in Gibbs energy can be calculated using the formula ΔG = ΔH - TΔS, where ΔG is the change in Gibbs energy, ΔH is the change in enthalpy, T is the temperature in Kelvin, and ΔS is the change in entropy. Plugging in the given values, we get:

ΔG = (−125 kJ mol−1) - (310 K)(−126 J K−1 mol−1)
ΔG = (−125 kJ mol−1) - (−0.126 kJ mol−1)(310 K)
ΔG = (−125 kJ mol−1) + (39.06 kJ mol−1)
ΔG = −85.94 kJ mol−1

The reaction is spontaneous if ΔG is negative. In this case, ΔG is −85.94 kJ mol−1, which is negative, so the reaction is spontaneous.

(The total change in entropy of the system and the surroundings can be calculated using the formula

ΔStotal = ΔSsystem + ΔSsurroundings.

We know that

ΔSsystem = −126 J K−1 mol−1,

and we can calculate ΔSsurroundings using the formula

ΔSsurroundings = −ΔH/T.

Plugging in the given values, we get:

ΔSsurroundings = −(−125 kJ mol−1)/(310 K)
ΔSsurroundings = 0.403 kJ K−1 mol−1
ΔSsurroundings = 403 J K−1 mol−1

So the total change in entropy is:

ΔStotal = (−126 J K−1 mol−1) + (403 J K−1 mol−1)
ΔStotal = 277 J K−1 mol−1

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Second Contributing Structure: Modify electron distribution with positive and negative formal charges.

The second and third contributing structures of the given molecules, along with the corresponding formal charges, are as follows:

Second Contributing Structure:

Draw the structure with modified electron distribution, considering one of the atoms to have a positive formal charge and another atom to have a negative formal charge.

Third Contributing Structure:

Draw the structure with another modified electron distribution, considering the positive and negative formal charges to be placed on different atoms compared to the second structure.

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\(C_3H_6\)  has a double bond in its carbon skeleton. This is a true statement.

Carbon skeleton refers to the chain of carbon atoms that make up an organic molecule. The presence or absence of double bonds in the carbon skeleton affects the properties of the molecule and how it interacts with other molecules. In \(C_3H_6\), there are three carbon atoms arranged in a linear chain, with each carbon atom forming single covalent bonds with two hydrogen atoms. The remaining valence electrons on each carbon atom form a double bond between the first and second carbon atoms.

This double bond is responsible for the unsaturated nature of the molecule. \(C_3H_6\)is an example of a simple alkene, also known as propene. Its carbon skeleton and double bond make it a versatile molecule that can be used in various applications, including the production of plastics, rubber, and other materials.

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Two possible sources of systematic (determinate) error in the experiment are; Incorrect calibration of volumetric glasswareIncorrect mass of NaCl

To calculate the concentration of NaCl in the resulting solution in mg/L NaCl, we can use the formula; Concentration (mg/L) = (Mass of solute ÷ Volume of solution in L) × 1000 g / 1 mg NaCl is present in the stock solution of 25 mL. So, the mass of NaCl in the solution would be;0.0842 g ÷ 25 mL = 0.00337 g/mL. Now, in the resulting solution, a 1.00 mL aliquot of the stock solution is transferred to a 50.00 mL volumetric flask and diluted to volume. Therefore, the volume of the resulting solution is 50.00 mL. We will substitute these values in the formula, Concentration (mg/L) = (0.00337 g/mL ÷ 50 mL) × 1000 g / 1 mg concentration (mg/L) = 67.4 mg/L. Therefore, the concentration of NaCl in the resulting solution in mg/L NaCl is 67.4 mg/L.7.  Concentration = 67.4 mg/LTolerance = 4.28 mg/LTotal concentration = 67.4 + 4.28 mg/L = 71.68 mg/LWe round off this value to one decimal place; Total concentration = 71.7 mg/LTherefore, the concentration of NaCl in the resulting solution using propagation of error through the calculation is 67.4+0.4 mg/L.8. Two possible sources of random (indeterminate) error in the experiment are; Errors in temperature measurement. Errors in measurement of water volume. Two possible sources of systematic (determinate) error in the experiment are; Incorrect calibration of volumetric glasswareIncorrect mass of NaCl.

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Answer:

A) The amount in grams of Li₃N produced is 1.243 g

B) N₂, is the limiting reagent

The mass of the non-limiting reagent, Li, remaining after the reaction is completed is 1.757 g

Explanation:

The given parameters are;

The mass of Li(s) = 2.5 grams

The mass of N₂ (g) = 0.5 grams

The chemical equation for the reaction can be presented as follows;

6 Li (s) + N₂ (g) → 2 Li₃N

Therefore, 6 moles of Li reacts with 1 mole of N₂  to produce 2 moles of Li₃N

The molar mass of Li = 6.941 g/mol

The molar mass of N₂ = 28.0134 g/mol

The number of moles of a reactant or product, n is given by the relation;

n = Mass of substance/(Molar mass of the substance)

For lithium, Li, n = 2.5/6.941 = 0.3602 moles

For Nitrogen gas, N₂, n = 0.5/28.0134  = 0.01785 moles

A) Given that 1 mole of  N₂ to produces 2 moles of Li₃N

0.01785  moles of  N₂ will produces 2×0.01785 = 0.0357 moles of Li₃N

The molar mass of Li₃N = 34.83 g/mol

The mass of Li₃N = 34.83 g/mol × 0.0357 moles = 1.243 g

B) 6  moles of Li reacts with 1 mole of N₂ to produce 2 moles of Li₃N

0.3602 moles will reacts with 1/6×0.3602 = 0.06003 mole of N₂

Therefore, N₂, is the limiting reagent and we have;

0.01785  moles of  N₂ will react with 6×0.01785 = 0.1071  moles of Li

The number of of moles of Li left = 0.3602 - 0.1071 =0.2531 moles

The mass of lithium left = 0.2531 moles × 6.941 g/mol = 1.757 g

The mass of lithium remaining after the reaction is completed = 1.757 g.

48.90% yield is produced in the given reaction

What is yield of a reaction?

The quantity of moles of a product created in relation to the amount of reactant consumed during a chemical reaction is measured in chemistry as yield, also known as reaction yield.

One of the most important aspects of organic and inorganic chemical synthesis that researchers must take into account is yield.  The terms "yield," "conversion," and "selectivity" are used in chemical reaction engineering to describe ratios of how much of a reactant was consumed (conversion), how much of the desired product was formed (yield), and how much of the undesired product was formed (selectivity), represented as X, Y, and S.

Hence, 48.90% yield is produced in the given reaction.

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The major product is 3-methyl-2-pentene and the minor product is 2-ethyl-1-butene.

The alkyl halide shown is 3-bromo-3-methyl pentane. We can see that it is a tertiary alkyl halide. The proportion of products formed during the reaction are determined by Markovnikov rule.

The two products of this reaction are; 3-methyl-2-pentene and 2-ethyl-1-butene. The major product is 3-methyl-2-pentene because it is formed from a more stable carbocation intermediate.  2-ethyl-1-butene is the minor product.

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Pipeline transportation is only available for specialist goods like liquids and gases.

Delicate, invisible particles that are constant motion and layer on top of one another make up liquids. The particles are held together by cohesive forces, yet they are not firmly adhered to one another as in a solid. Despite touching one another, the particles can still pass freely one another.

An liquid is still a sample of stuff that adopts the shape of the container it is stored in and develops a defined surface when gravity is present. The state, or situation, of substance having this attribute is also referred to as liquid.

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Answer:

gain

Explanation:

The rate constant (k) is approximately 0.00471 s⁻¹, and the reaction order (n) is approximately 1.64.

To determine the rate constant (k) and reaction order, we can use the relationship between the half-life and the rate constant for a first-order reaction. The half-life \((t_{1/2})\) of a first-order reaction is related to the rate constant (k) as follows:

\(t_{1/2\) = (0.693 / k)

Let's first calculate the rate constant (k) for each given half-life:

For the first half-life (\(t_{1/2\) = 147 s) and initial concentration (0.252 M):

147 s = (0.693 / k₁)

k1 = 0.693 / 147 s

k1 ≈ 0.00471 s⁻¹

For the second half-life (\(t_{1/2\) = 237 s) and initial concentration (0.156 M):

237 s = (0.693 / k2)

k₂ = 0.693 / 237 s

k₂ ≈ 0.00292 s⁻¹

Next, let's calculate the reaction order by comparing the rate constants for the two different concentrations. The reaction order (n) can be determined using the following equation:

k₁ / k₂ = (C₂ / C₁)\(^{(n-1)\)

Where C₁ and C₂ are the initial concentrations, and n is the reaction order.

Using the given concentrations (C₁ = 0.252 M and C₂ = 0.156 M), we have:

(k₁ / k₂) = (0.156 M / 0.252 M)\(^{(n-1)\)

(0.00471 s⁻¹) / (0.00292 s⁻¹) = (0.156 M / 0.252 M)\(^{(n-1)\)

Simplifying the equation:

1.61 ≈ (0.6190)\(^{(n-1)\)

To solve for n, we can take the natural logarithm (ln) of both sides:

ln(1.61) ≈ ln(0.6190)\(^{(n-1)\)

ln(1.61) ≈ (n-1) * ln(0.6190)

Using a calculator, we find:

n ≈ 1.64

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The concentration of the potassium dichromate used to make the new solution is 0.0763 M.

The total volume of the new solution is 379 + 448 = 827 mL.

The total amount of potassium ions in the new solution is 0.713 * 827 = 594.491 mol.

The amount of potassium bromide in the new solution is 0.12 * 379 = 45.48 mol. The amount of potassium dichromate in the solution is 594.491 - 45.48 = 549.011 mol.

The concentration of potassium dichromate in the solution is 549.011 / 448 = 0.0763 M.

The concentration of the potassium dichromate used to make the new solution is 0.0763 M.

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During this process, the hexanes in the flask will boil and turn into vapour at their boiling point of 68.73 °C (342 °K). This occurs because the heat being applied to the flask causes the temperature of the hexanes to increase, and when it reaches the boiling point, the liquid will begin to turn into a gas.

The vapour will then rise and escape from the flask. As the vapour exits the flask, it will cool and eventually condense back into a liquid. This process is known as "boiling."

Hexanes in the flask will boil and turn into vapour because of the heat being applied. Vapor rises and escapes from the flask, then condenses back into liquid. This is a boiling process.

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Answer:

hi nice to meet you, just be your self and ask your questions, its all good we are nice people

Explanation:

have a great day, please mark as brainliest please

Hello! Be ur self, ask questions and ask questions. Okibye

Answer:

1 × 10²⁴ atoms O

General Formulas and Concepts:

Chemistry - Atomic Structure

Explanation:

Step 1: Define

30 g O

Step 2: Identify Conversions

Avogadro's Number

Molar Mass of O - 16.00 g/mol

Step 3: Convert

\(30 \ g \ O(\frac{1 \ mol \ O}{16.00 \ g \ O} )(\frac{6.022 \cdot 10^{23} \ atoms \ O}{1 \ mol \ O} )\) = 1.12913 × 10²⁴ atoms O

Step 4: Check

We are given 1 sig fig. Follow sig fig rules and round.

1.12913 × 10²⁴ atoms O ≈ 1 × 10²⁴ atoms O

Answer: Concentration of the acid is 0.551 M.

Explanation:

Given: \(V_{1}\) = 71.1 mL,      \(M_{1}\) = 0.695 M

\(V_{2}\) = 89.7 mL,           \(M_{2}\) = ?

Formula used to calculate the concentration of acid is as follows.

\(M_{1}V_{1} = M_{2}V_{2}\)

Substitute the values into above formula as follows.

\(M_{1}V_{1} = M_{2}V_{2}\\0.695 \times 71.1 mL = M_{2} \times 89.7 mL\\M_{2} = 0.551 M\)

Thus, we can conclude that concentration of the acid is 0.551 M.

In order to neutralise 15.0 mL of 1.00 M KOH, we therefore require 15.0 mL of 0.50 M Sulfuric acid.

Strong acid and strong base reactions produce a salt that is neither acidic nor basic in nature, or neutral. For instance, sodium chloride and water are produced when HCl (hydrochloric acid), a powerful acid, combines with NaOH, a powerful base.

First, let's calculate the number of moles of potassium hydroxide in 15.0 mL of 1.00 M solution:

moles of Potassium hydroxide = concentration × volume

moles of Potassium hydroxide = 1.00 M × 15.0 mL / 1000 mL per L = 0.015 mol

moles of Sulfuric acid = moles of KOH / 2 = 0.015 mol / 2 = 0.0075 mol

moles of Sulfuric acid = concentration × volume

0.0075 mol = 0.50 M × volume

volume = 0.0075 mol / 0.50 M = 0.015 L

volume in mL = 0.015 L × 1000 mL per L = 15.0 mL

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Answer:

Why do atoms have no overall electrical charge?

Because atoms (and by extension matter in general) are usually electrically neutral, and have equal number of positively charged and negatively charged particles. The sign convention between electrons and protons is entirely arbitrary.

Explanation:

Your welcome!! Have a good day ^^

Answer:circular heliocentric orbits.

Explanation:

Answer:

b. positively

Explanation:

A single electron from the outermost shell of this element moves to the outermost shell of an atom with seven electrons.

When forming ions elements typically gain or lose the minimum number of electrons required to make a complete octet. For example, fluorine has 7 valence electrons, so it is most likely to gain an electron and form an ion of charge At the outermost energy level he states 8 electrons must be reached for an atom to be stable. Particles smaller than an atom.

Ionization is the process by which ions are formed by gaining or losing electrons from atoms or molecules. When an atom or molecule receives an electron it becomes a negatively charged anion and when it loses an electron it becomes a positively charged cation. Energy can be lost or gained in the formation of ions.

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The emitting element's atomic number and atomic mass number are both reduced by 2 and 4 respectively by alpha radiation. The atomic number of an element is increased by 1 by beta radiation, but the atomic mass number is unaffected. The atomic number or the atomic mass number are unaffected by gamma radiation.

An unstable atom will release energy on its own to move into a more stable state, which is radioactivity. The energy that is released by a radioactive atom is known as ionizing radiation. The term "radioactive isotope" refers to radioactive atoms of the same element with varying neutron counts.

Alpha radiation decreases the atomic number and atomic mass number of the emitting element by 2 and 4 correspondingly. Beta radiation raises an element's atomic number by 1, but it has no effect on its atomic mass number. Gamma radiation has no impact on the atomic number or atomic mass number.

The atomic or mass number are unaffected by the emission of gamma rays, however the nucleus will become more stable. The most significant change in atomic number and mass number is brought on by alpha radiation.

A table showing how each type of radiation affects the atomic number and mass number of an atom is attached in the image below.

Thus, The emitting element's atomic number and atomic mass number are both reduced by 2 and 4 respectively by alpha radiation.

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