Which is a characteristic of an atom?

Atoms are the largest unit of matter.

Atoms can be seen by the naked eye.

Atoms come in different varieties.

Atoms are created in the Earth’s core.

If the [OH-] concentration is 5.9 x 10^(-M), the pOH of the solution is approximately -4.77 and the solution is basic.

To calculate the pOH of a solution, we can use the formula:

pOH = -log[OH-]

Given that the [OH-] concentration is 5.9 x 10^(-M), we can substitute this value into the formula:

pOH = -log(5.9 x 10^(-M))

Calculating this expression, we find:

pOH = -log(5.9 x 10^(-M))

pOH ≈ -log(5.9) + (-log(10^(-M)))

Since log(10^(-M)) is equal to -M, the equation simplifies to:

pOH ≈ -log(5.9) - M

Now, we need the value of M (the exponent) to calculate the exact pOH value. It appears that the value of M is missing in the given information. However, assuming M is a positive value, we can continue the calculation.

If we consider M = 6, for instance, the equation becomes:

pOH ≈ -log(5.9) - 6

Now, we can evaluate the expression:

pOH ≈ 1.23 - 6

pOH ≈ -4.77

Therefore, if the [OH-] concentration is 5.9 x 10^(-M), the pOH of the solution is approximately -4.77.

To determine whether the solution is acidic, basic, or neutral, we can use the relationship between pH and pOH. The sum of the pH and pOH of a solution at 25°C is always equal to 14.

Since pOH = -4.77, the pH would be:

pH = 14 - pOH

pH ≈ 14 - (-4.77)

pH ≈ 18.77

A solution with a pH above 7 is considered basic. In this case, the calculated pH is greater than 7. Therefore, the solution is basic.

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The mass of hydrogen needed to produce 51.0 grams of ammonia is  ≈ 9.07 grams.

To determine the mass of hydrogen required to produce 51.0 grams of ammonia (NH3) using the Haber process, we need to calculate the stoichiometric ratio between hydrogen and ammonia.

From the balanced chemical equation:

3 H₂(g) + N₂(g) → 2 NH₃(g)

We can see that for every 3 moles of hydrogen (H₂), we obtain 2 moles of ammonia (NH₃).

First, we need to convert the given mass of ammonia (51.0 grams) to moles. The molar mass of NH₃ is 17.03 g/mol.

Number of moles of NH₃ = Mass / Molar mass

                     = 51.0 g / 17.03 g/mol

                     ≈ 2.995 moles

Next, using the stoichiometric ratio, we can calculate the moles of hydrogen required.

Moles of H₂ = (Moles of NH₃ × Coefficient of H₂) / Coefficient of NH₃

           = (2.995 moles × 3) / 2

           ≈ 4.493 moles

Finally, we can convert the moles of hydrogen to mass using the molar mass of hydrogen (2.02 g/mol).

Mass of H₂ = Moles × Molar mass

          = 4.493 moles × 2.02 g/mol

          ≈ 9.07 grams

Therefore, approximately 9.07 grams of hydrogen is needed to produce 51.0 grams of ammonia in the Haber process.

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D.) HF of the following molecules contain POLAR COVALENT bonds

A polar covalent link holds a hydrogen atom and a fluorine atom together in a hydrogen fluoride (HF) molecule.

The molecule that has a polar covalent link in it is c. H-F since fluorine and hydrogen atoms have different electronegativities, which causes the bond to become polarised.

A polar molecule is one that has a little positive charge on one end and a slight negative charge on the other. A polar molecule is a diatomic compound, such as HF, that has a polar covalent link.

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N2H4 was the allotted component that ran out first during the lunar module ascent phase.

Beginning with Apollo 11 in July 1969, which saw Neil Armstrong become the first person to set foot on the Moon, six missions eventually succeeded in putting people there. Apollo 13 was designed to land, but owing to a fault inside the spacecraft, it was only able to do a flyby. All nine crewed missions arrived back on Earth without incident.

Number of moles of N2H4 = 1200 kg / 32.0452 kg/mol = 37.46 mol

Number of moles of (CH3)2N2H2 = 1000 kg / 116.1596 kg/mol = 8.60 mol

Number of moles of N2O4 = 3687 kg / 92.0110 kg/mol = 40.06 mol

2 N2H4 + (CH3)2N2H2 + 3 N2O4 → 6 N2 + 2 CO2 + 8 H2O

2 mol N2H4 / 1 mol (CH3)2N2H2 = 17.20 mol N2H4

3 mol N2O4 / 1 mol (CH3)2N2H2 = 25.38 mol N2O4

The reaction will end after all of the available N2H4 has been used because we only have 37.46 mol of N2H4 available, which is less than the needed 17.20 mol. N2H4 is the limiting reagent, and as a result, the other reactants will be expelled.

N2H4 was the component allotted that was initially consumed during the ascent phase of the lunar module.

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Answer:

C. 6.99 moles

Explanation:

To convert molecules to moles we simply divide the given number of molecules by Avogadro's number, 6.022 x 10²³

Given number of molecules : 4.21 x 10^24

Number of moles : 4.21 x 10^24 / 6.022 x 10²³ = 6.99 moles ( rounded )

The answer is C

Answer:

true

Explanation:

np have a good day brainliest?

This is a perfect answer

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Answer:

2NH3(g) ----> N2(g) +3H2(g) ΔH= +92.4 kJ

Explanation:

Entropy increases with increase in the number of particles from left to right in a reaction. Hence, the reaction; 2NH3(g) ----> N2(g) +3H2(g) ΔH= +92.4 kJ is favoured by increase in entropy.

Similarly, the enthalpy change for the reaction is only +92.4 KJ. Hence for this reaction, both enthalpy and entropy changes drive the reaction in the same direction.

Answer:

1. pH is 4.66.

2. pH is 3.44.

3. pOH is 11.54.

Explanation:

1. Determination of the pH.

Concentration of Hydrogen ion, [H+] = 2.17×10^-5M

pH =..?

pH = –Log [H+]

pH = –Log 2.17×10^-5

pH = 4.66

2. Determination of the pH.

pOH = 10.56

pH =..?

pH + pOH = 14

pH + 10.56 = 14

Collect like terms:

pH = 14 – 10.56

pH = 3.44

3. Determination of the pOH.

Concentration of Hydrogen ion, [H+] = 3.48×10^-3M

pOH =..?

First, we shall determine the pH of the solution. This is illustrated below:

pH = –Log [H+]

pH = –Log 3.48×10^-3

pH = 2.46

Now, we can determine the pOH of the solution. This is illustrated below:

pH = 2.46

pOH =..?

pH + pOH = 14

2.46 + pOH = 14

Collect like terms

pOH = 14 – 2.46

pOH = 11.54

Answer:

Sodium has one valence electron. Valence electrons are electrons found in the outermost shell of an atom. The shell number representing the valence shell will differ depending on the atom in question. For sodium, which is in the 3rd row of the periodic table, the valence electrons will be found in the 3rd shell.

Consumers are organisms which depend on the producers or autotrophs for their food and nutrition.

Consumers are also known as heterotrophs, and they feed on the producers (autotrophs or self-feeders or plants) and herbivores (animals that eat plants) for food and energy. They don't produce their own food.

There are four types of consumers:

1) Primary – Herbivores are known as primary consumers and their source of food is plants or the first trophic level of the food chain. Example, Rabbits, Butterflies, Zooplanktons, etc.

2) Secondary – these consumers eat both plants and herbivores and are therefore also known as omnivores. Example, Ants, Crabs, Rats, Humans, etc.

3) Tertiary – these consumers eat primary and secondary consumers. Thus, they can be omnivores as well as carnivores. Example, Hawks, Snakes, Lions, etc.

4) Quaternary – these consumers prey on the tertiary consumers. Example, Polar bear, Alligator, etc.

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Answer:

The volume of an object can be calculated by multiplying its length, width and height. Further, depending on the shape of the object, its volume changes.

Explanation:

Hope this helps

Answer:

\(v_{rms}=866.32m/s\)

Explanation:

Hello,

In this case, since the rms speed of the molecules is computed by:

\(v_{rms}=\sqrt{\frac{3RT}{M} }\)

Whereas the absolute temperature is computed from the internal energy (by using the Cp of helium (3.1156 J/g*K) as shown below:

\(U=nCvT\\\\T=\frac{U}{nCv}=\frac{15kJ*\frac{1000J}{1kJ} }{10mol*\frac{4.00g}{1mol} *3.1156\frac{J}{g*K} } \\\\T=120.36K\)

Thereby, the rms speed results:

\(v_{rms}=\sqrt{\frac{3*8.314\frac{kg*m^2}{s^2*mol*K}*120.36K}{4.00\frac{g}{mol}*\frac{1kg}{1000} } } \\\\v_{rms}=866.32m/s\)

Regards.

Answer:

(A) Nickels and pennies

Answer: Climax Commuity.

Explanation: (hope it’s right! Good luck!)

Answer:

The correct answer is C, tertiary.

The following alcohol is a tertiary alcohol. Hence, option C is correct.

A primary alcohol is a compound in which a hydroxy group, ‒OH, is attached to a saturated carbon atom which has either three hydrogen atoms attached to it or only one other carbon atom and two hydrogen atoms attached to it.

Here, the carbon atom holding the OH group is attached directly to three alkyl groups.

Hence, the following alcohol is a tertiary alcohol.

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Answer:

Strong winds that blow for a long time over a great distance.

Explanation:

Generally, the biggest and most powerful wind-generated waves are produced by strong storms that blow for a sustained period over a large area. Huge and big waves, or swells, can travel over long distances. The size of the wave depends on wind speed, wind duration, and the area over which the wind is blowing. If The speed of the wind is more, it stays for along time and it covers a larger distance then the waves produced will be very powerful and large.

(a) The empirical formula of the compound is \(C_{29}H_ {36}O\)

(b) The molecular formula of the compound is approximately \(C_{8}H_{118}O_{3}\)

To determine the empirical formula of the organic compound, we need to find the simplest whole-number ratio of the elements present.

(a) The given percentages of carbon, hydrogen, and oxygen can be converted into moles by dividing them by their respective atomic masses:

Carbon: 0.624 g / 12.01 g/mol = 0.052 mol

Hydrogen: 0.065 g / 1.008 g/mol = 0.064 mol

Oxygen: 0.028 g / 16.00 g/mol = 0.0018 mol

Next, we divide each of the mole values by the smallest mole value (0.0018 mol in this case) to obtain the mole ratio:

Carbon: 0.052 mol / 0.0018 mol ≈ 29

Hydrogen: 0.064 mol / 0.0018 mol ≈ 36

Oxygen: 0.0018 mol / 0.0018 mol = 1

Therefore, the empirical formula of the compound is \(C_{29}H_ {36}O\)

(b) To find the molecular formula, we need the relative molecular mass of the compound, which is given as 1940 g/mol. The empirical formula mass can be calculated by summing the atomic masses in the empirical formula:

Empirical formula mass: (29 × 12.01 g/mol) + (36 × 1.008 g/mol) + (1 × 16.00 g/mol) = 588.94 g/mol

To find the multiplier, we divide the relative molecular mass by the empirical formula mass:

Multiplier: 1940 g/mol / 588.94 g/mol ≈ 3.29

Rounding to the nearest whole number, the molecular formula of the compound is approximately 3 times the empirical formula, resulting in \(C_{8}H_{118}O_{3}\).

In summary, the empirical formula of the compound is\(C_{29}H_ {36}O\), and the molecular formula is approximately \(C_{8}H_{118}O_{3}\).

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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:

1. the mass of H2CO3 produced is 1.675 grams.

2. the percentage yield of H2CO3 is 59.7%.

Explanation:

1. Using stoichiometry, we can find the moles of NaHCO3 and Na2CO3 produced in the reaction:

2 NaHCO3 -> Na2CO3 + H2CO3

Moles of NaHCO3 = 3.81 g / 84.01 g/mol = 0.04532 mol

Moles of Na2CO3 = 2.86 g / 105.99 g/mol = 0.02700 mol

Since the reaction produces an equal amount of Na2CO3 and H2CO3, we know that 0.02700 mol of H2CO3 was produced.

To find the mass of H2CO3 produced, we can use its molar mass:

Mass of H2CO3 = 0.02700 mol x 62.03 g/mol = 1.675 g

Therefore, the mass of H2CO3 produced is 1.675 grams.

2. The theoretical yield of H2CO3 is the amount that would be produced if the reaction went to completion and all of the NaHCO3 was converted to Na2CO3 and H2CO3. We can calculate the theoretical yield of H2CO3 by multiplying the moles of NaHCO3 used in the reaction by the molar mass of H2CO3:

Theoretical yield of H2CO3 = 0.04532 mol x 62.03 g/mol = 2.806 g

The percentage yield is calculated by dividing the actual yield (1.675 g) by the theoretical yield (2.806 g) and multiplying by 100:

Percentage yield of H2CO3 = (1.675 g / 2.806 g) x 100% = 59.7%

Therefore, the percentage yield of H2CO3 is 59.7%.

Answer:

              1.122 mol/L

Explanation:

Molarity = Moles ÷ Vol in L

First, convert volume from mL to L as,

713.1 mL ÷ 1000 = 0.7131 L

Putting values,

Molarity = 0.8 mol ÷ 0.7131 L

Molarity = 1.122 mol/L

Answer:

several hollow bones,endothermic

Explanation:

Answer:

i was looking for the same answer but i can't get one eather

Explanation:

Container 3 has the gas stored at the highest temperature.

Temperature is a measure of the average kinetic energy of the particles in a substance. In the given table, it is stated that container 3 has a large number of collisions with container walls, high average kinetic energy, and large number of particles with large spaces between them.

These properties indicate that the gas in container 3 has higher kinetic energy and more vigorous movement compared to the other containers.

Container 1 has a low number of collisions with container walls and a medium average kinetic energy. This suggests that the gas in container 1 has lower energy and less movement than the gas in container 3.

Container 2 has a large number of collisions with container walls, but it also has a small number of particles with little spaces between them. While the collisions may be frequent, the limited number of particles and the lack of space between them may result in lower overall kinetic energy compared to container 3.

Container 4 has few collisions with container walls, low average kinetic energy, and a small number of particles. These properties indicate that the gas in container 4 has the lowest energy and least movement among all the containers.

Container 3

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Answer:

Accuracy is the degree of closeness to true value.

Precision is the degree to which an instrument or process will repeat the same value.

Hope it helps....

Answer:

Accuracy and precision are alike only in the fact that they both refer to the quality of measurement, but they are very different indicators of measurement. Accuracy is the degree of closeness to true value. Precision is the degree to which an instrument or process will repeat the same value