Help please! I'll give brainliest and 5 stars if you show work!

Answer:

5. 0.566 g
6. A. 100 times more

Explanation:

5. The pH of a solution is defined as the negative logarithm (base 10) of the hydrogen ion concentration. For a solution with pH=2, the concentration of hydrogen ions is 10^-2 mol/L. Since HBr is a strong acid, it dissociates completely in water to produce H+ and Br- ions. Therefore, the concentration of HBr in the solution is also 10^-2 mol/L.

The molar mass of HBr is 80.91194 g/mol

So, in a 700 mL solution (0.7 L), there are

0.7 L * 10^-2 mol/L = 0.007 mol of HBr.

This corresponds to 0.007 mol * 80.91194 g/mol = 0.566 g of HBr dissolved in the solution.

6. The pH of a solution is defined as the negative logarithm (base 10) of the hydrogen ion concentration. This means that for each decrease in pH by 1 unit, the hydrogen ion concentration increases by a factor of 10. Since the difference in pH between the two solutions is 3 units (6-3=3), the hydrogen ion concentration in the solution with pH=3 is 10^3 = 100 times more than in the solution with pH=6.

We would need about 16 hydrogen atoms so that we can convert the compound to a saturated fat.

In animal products like meat and dairy, saturated fat is a form of dietary fat that is normally solid at room temperature. It is known as being "saturated" because each molecule of fat has the most hydrogen atoms possible, giving it a stable structure.

We can see this by counting the number of double bonds in the fat and there are eight of them so sixteen hydrogen atoms are needed for saturation.

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Answer:

A solvent front is the point on a chromatography paper or plate where the solvent has reached the end of the stationary phase and has migrated as far as it can go. It is the farthest point reached by the solvent in the chromatography process.

The ΔH°vap of water at the critical point is approximately 0.04614 kJ/mol.

To calculate the ΔH°vap (enthalpy of vaporization) of water at the critical point, we can use the Clausius-Clapeyron equation;

ln(P₂/P₁) = ΔH°vap/R [1/T₁ - 1/T₂]

where P₁ and T₁ are the pressure and temperature at which the enthalpy of vaporization is known (usually at standard conditions of 1 atm and 100 °C), P₂ and T₂ are the pressure and temperature at the critical point, ΔH°vap is the enthalpy of vaporization, and R is the gas constant (8.314 J/mol∙K).

Using the given values, we can plug them into the equation and solve for ΔH°vap;

ln(3.2 atm / 1 atm) = ΔH°vap / R [1/373 K - 1/275 K]

Simplifying;

ln(3.2) = ΔH°vap / R [0.0026819]

ΔH°vap / R = ln(3.2) / 0.0026819

ΔH°vap / R = 5.552

Multiplying both sides by R:

ΔH°vap = 5.552 x R

ΔH°vap = 5.552 x 8.314 J/mol∙K

ΔH°vap = 46.14 J/mol

Converting to kJ/mol;

ΔH°vap = 0.04614 kJ/mol

Therefore, the ΔH°vap of water is 0.04614 kJ/mol.

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5.221 grams of Al are required to react with 352 mL of 1.65 M HCl solution.

Molarity (M) is defined as the moles of solute per liter of the solution.

Balanced chemical equation is : 2Al + 6HCl → 2AlCl₃ + 3H₂

From the equation, we can see that 2 moles of Al react with 6 moles of HCl to produce 2 moles of AlCl₃ and 3 moles of H₂.

As moles of HCl = Molarity × Volume

moles of HCl = 1.65 mol/L × 0.352 L

moles of HCl = 0.58128 mol

and moles of Al = (2/6) × moles of HCl

moles of Al = (1/3) × 0.58128 mol

moles of Al = 0.19376 mol

mass of Al = moles of Al × molar mass of Al

mass of Al = 0.19376 mol × 26.98 g/mol

mass of Al = 5.221 g

So, 5.221 grams of Al are required to react with 352 mL of 1.65 M HCl solution.

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The mass of sodium sulfate [tex](Na_2SO_4)[/tex] formed by the complete reaction of 137.3 grams of sodium hydroxide [tex](NaOH)[/tex] is 486.74 grams.

The balanced chemical equation for the reaction between sodium hydroxide [tex](NaOH)[/tex] and sulfuric acid [tex](H_2SO_4)[/tex] is:

[tex]NaOH[/tex] + [tex](H_2SO_4)[/tex] → [tex]Na_2SO_4[/tex] + [tex]2H_2O[/tex]

From the balanced equation, we can see that 1 mole of [tex]NaOH[/tex] reacts with 1 mole of [tex](H_2SO_4)[/tex] to produce 1 mole of [tex]Na_2SO_4[/tex] and 2 moles of water [tex](H_2O).[/tex]

The molar mass of [tex]NaOH[/tex] is 40.00 g/mol, which means that 137.3 grams of [tex]NaOH[/tex] is equal to 137.3 g / 40.00 g/mol = 3.4325 moles of [tex]NaOH[/tex].

Since 1 mole of [tex]NaOH[/tex] reacts with 1 mole of [tex](H_2SO_4)[/tex] to produce 1 mole of [tex](Na_2SO_4)[/tex] , we can say that 3.4325 moles of [tex]NaOH[/tex] will react with 3.4325 moles of [tex]H_2SO_4[/tex] to produce 3.4325 moles of [tex]Na_2SO_4[/tex].

The molar mass of [tex](Na_2SO_4)[/tex] is 142.04 g/mol, which means that 1 mole of [tex](Na_2SO_4)[/tex] has a mass of 142.04 g.

Therefore, 3.4325 moles of [tex](Na_2SO_4)[/tex] has a mass of 3.4325 moles x 142.04 g/mol = 486.74 grams.

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1. The long-haired cat in the P generation is a purebred. This is because it has two copies of the recessive allele (hh) responsible for long hair.

2. The short-haired cat in the P generation is a hybrid. We know this because the offspring include both short-haired (Hh) and long-haired (hh) cats, indicating that the short-haired parent must have one dominant (H) and one recessive (h) allele (Hh).

3. If the short-haired cat in the P generation were purebred (HH), all offspring would have short hair, as they would inherit one dominant allele (H) from the short-haired parent and one recessive allele (h) from the long-haired parent, resulting in Hh offspring.

4. The black horse is a hybrid. Since the cross between a black horse (B...) and a brown horse (bb) produced a brown foal (bb), the black horse must carry one dominant allele (B) and one recessive allele (b) - making it a hybrid (Bb).

5. To determine whether a guinea pig with a smooth coat (S...) is a hybrid or a purebred, perform a test cross by mating it with a guinea pig with a rough coat (ss). If all offspring have smooth coats (Ss), the smooth-coated guinea pig is likely purebred (SS). If any offspring have a rough coat (ss), the smooth-coated guinea pig is a hybrid (Ss).

A dominant allele is a variant of a gene that expresses its trait even when only one copy is present in an individual's genotype. In other words, it masks the effect of another variant (allele) of the same gene when they are together.

A recessive allele is a variant of a gene that only expresses its trait when two copies are present in an individual's genotype. The trait associated with the recessive allele is "masked" by the presence of a dominant allele, and it will only be expressed if both copies of the gene are recessive.

The above answer is based on the question below;

In a test cross, the organism with the trait controlled by a dominant allele is crossed with an organism with a trait controlled by a recessive allele. If all offspring have the trait controlled by the dominant allele, then the parent is probably a purebred. If any offspring has the recessive strait, then the dominant parent is a hybrid.

1. Is the long-haired cat in the P generation a hybrid or a purebred? Explain your answer.

2. Is the short-haired cat in the P generation a hybrid or a purebred? Explain your answer.

3. If the short-haired cat in the P generation were purebred, what would you expect the offspring to look like?

4. In horses, the allele for a black coat (B) is dominant over the allele for a brown coat (b). A cross between a black horse and a brown horse produces a brown foal. Is the black horse a hybrid or a purebred? Explain.

5. In guinea pigs, the allele for a smooth coat (S) is dominant over the allele for a rough coat (s). Explain how you could find out whether a

guinea pig with a smooth coat is a hybrid or a purebred.

H= Short hair

h = Long hair

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Answer:

2.000 moles.

Explanation:

To solve this question, we need to use the formula:

n = V / VM

where n is the number of moles, V is the volume of water in milliliters, and VM is the molar volume of water in milliliters per mole. The molar volume of water at standard temperature and pressure (STP) is 18.02 mL/mol. Plugging in the given value, we get:

n = 36.030 mL / 18.02 mL/mol n = 2.000 moles

The answer is 2.000 moles.

Answer:

649.090

Explanation:

The S-P difference (sec) is used to calculate the distance (km) between an earthquake epicenter and a seismic station, while the magnitude (M) is a measure of the energy released during the earthquake.

These parameters are important for understanding the severity and impact of an earthquake, as well as for predicting future seismic activity.

The S-P difference (sec) refers to the time difference between the arrival of the primary (P) waves and the secondary (S) waves at a seismic station. This time difference is used to calculate the distance (km) between the earthquake epicenter and the seismic station, using the equation: distance (km) = S-P difference (sec) x 8 km/sec. This calculation assumes that the waves travel at a constant speed through the Earth's interior.
The magnitude (M) of an earthquake is a measure of the energy released during the earthquake, and is usually determined using a seismometer. The magnitude scale is logarithmic, meaning that each increase of one unit represents a tenfold increase in seismic energy. For example, an earthquake with a magnitude of 5.0 is ten times more powerful than one with a magnitude of 4.0, and 100 times more powerful than one with a magnitude of 3.0.

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Answer:

There are 1000 milliliters (ml) in one liter. Therefore, the teacher has a total of 6 x 1000 = 6000 ml of sodium nitrate solution.

Explanation:

To divide the solution evenly among the 24 students, we need to divide the total volume of the solution by the number of students:

6000 ml ÷ 24 students = 250 ml per student

Therefore, each student will receive 250 milliliters of sodium nitrate solution.

Answer:

Answer- 0.25ml

Explanation:

So there are 24 students and 6 liters of Solution.So to evenly distribute

Just divide 6 by 24(6÷24/)... So the answer will be 0.25

Answer:

Explanation:

Coal deposits have been found in Antarctica, particularly in the Transantarctic Mountains where they are interbedded in sedimentary rocks of the flat-lying Beacon Supergroup. The presence of coal in Antarctica supports Wegener's continental drift hypothesis because it suggests that Antarctica was once part of a larger landmass that had a warm climate suitable for the formation of coal. Coal is formed from ancient plant matter that has been compressed and heated over millions of years. The presence of coal in Antarctica suggests that the continent was once located closer to the equator and had a climate that supported lush vegetation.

Answer:

Explanation:

We can use the formula:

q = mcΔT

where q is the heat absorbed, m is the mass, c is the specific heat, and ΔT is the change in temperature.

Given:
specific heat of iron (c) = 0.449 J/g°C
initial temperature (T1) = 24.0°C
final temperature (T2) = 82.1°C
heat absorbed (q) = 948.0 J

Substituting the given values into the formula, we get:

q = mcΔT
948.0 J = m(0.449 J/g°C)(82.1°C - 24.0°C)
948.0 J = m(0.449 J/g°C)(58.1°C)
m = 948.0 J ÷ (0.449 J/g°C × 58.1°C)
m = 33.1 g

Therefore, the mass of the iron piece is 33.1 g (to three significant figures)

the oxidation number of copper in copper oxide is 2...

Answer: +2

Explanation: Copper has a +2 oxidation number in CuO.

This is due to the fact that oxygen has an oxidation number of 2, and the entire chemical has a neutral charge. Consequently, the following equation can be used to determine copper's oxidation number:

(+2) + (-2) = 0

In order to counteract the -2 oxidation number of oxygen in CuO, copper must have an oxidation number of +2.

Charles's law of gases states that the density of an ideal gas is inversely proportional to its temperature at constant pressure.

The equation is as follows;

Va/Ta = Vb/Tb

Where;

0.67/362 = 1.12/Tb

0.00185Tb = 1.12

Tb = 605.41K

This temperature in °C is 605.41 - 273 = 332°C

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If a skimmer and a gull eat the same fish and the skimmer is more successful at catching the fish over time, it is likely that the skimmer population would increase, while the gull population may decrease.

The skimmer's success in catching the fish would give it an advantage in obtaining the necessary nutrients for survival and reproduction. As a result, the skimmer population would likely grow over time as more individuals are able to survive and reproduce due to the abundance of food.

On the other hand, the gull population may decrease due to the competition with the skimmer for the same food source. If the skimmer population grows significantly, it may lead to a reduction in the availability of fish for the gulls to feed on. Over time, this could result in a decline in the gull population due to reduced food availability.

However, it is important to note that the impact on the bird populations may depend on various factors such as the size of the populations, availability of other food sources, and environmental factors. Therefore, the outcome of this scenario cannot be predicted with certainty and would require further analysis and investigation.

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A silver bar with the mass of the 300 grams is heated from the 30 °C to 55 °C. The amount heat does the silver bar absorb in the joules is 1762.5 J.

The mass of the silver bar = 300 g

The initial temperature = 30 °C

The final temperature = 55 °C

The heat energy is expressed as :

Q = mc ΔT

Where,

The m is mass of the silver bar = 300 g

The c is the specific heat capacity = 0.235 J/g °C

The ΔT is the change in the temperature = final temperature - initial temperature

The ΔT is the change in the temperature = 55 °C - 30 °C

The ΔT is the change in the temperature = 25 °C

The heat energy, Q = 300 × 0.235 × 25

The heat energy, Q = 1762.5 J

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The best solution to suppress the dissolution of MgCO3 is option D 0.200 M Na2CO3

We need to add an ion or compound that will react with MgCO3 and form a precipitate, thus removing Mg2+ and CO32- ions from the solution.

Therefore, Option D, 0.200 M Na2CO3, contains CO32- ions that can react with Mg2+ ions to form MgCO3 precipitate. This would effectively suppress the dissolution of MgCO3 by removing Mg2+ and CO32- ions from the solution.

Therefore, option D is the best solution to suppress the dissolution of MgCO3.

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Hydrazine, reacts with the oxygen to form the nitrogen gas and the water. The percent yield of the reaction is 3.18 %.

The balanced reaction is :

N₂H₄  + O₂  --->  N₂ + 2H₂O

The mass of the N₂H₄  = 3.55 g

The moles of N₂H₄ = mass / molar mass

The moles of N₂H₄ = 3.55 / 32

The moles of N₂H₄ = 0.110 mol

The theoretical yield = 0.110 mol × 28 g/mol

The theoretical yield = 3.08 g

The gas equation is :

P V = n R T

n = P V / R T

n = (1 × 0.850 ) / ( 0.0823 ×295 )

n = 0.0035 mol

The actual yield = 0.0035 × 28

The actual yield = 0.098 g

The percent yield = ( 0.098 / 3.08 ) × 100 %

The percent yield = 3.18 %.

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A 6.50-g sample of copper metal at 25.0 °C is heated by the addition of 145 J of energy. The final temperature of the copper is 83.7 °C. The specific heat capacity of copper is 0.38 J/g-K.

The correct answer choice is "83.7"

To solve this problem, we can use the equation:

q = mcΔT

where q is the amount of energy absorbed by the copper, m is the mass of the copper, c is the specific heat capacity of copper, and ΔT is the change in temperature of the copper.

Rearranging this equation to solve for ΔT, we get:

ΔT = q / (mc)

Substituting the given values, we get:

ΔT = 145 J / (6.50 g x 0.38 J/g-K)

ΔT = 58.7 K

Therefore, the final temperature of the copper is:

25.0 °C + 58.7 °C = 83.7 °C

So the correct option is 83.7.

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The term that would best describe the solution formed if 80g KBr dissolved in 100g water is unsaturated solution.

A saturated solution is a solution with solute that dissolves until it is unable to dissolve anymore, leaving the undissolved substances at the bottom.

On the other hand, an unsaturated solution is that solution that is capable of dissolving more of a solute at the same temperature.

According to this question, 80 grams of KBr were dissolved in 100 grams of water at 35 degrees Celsius. This means that the solution is unsaturated because it can still dissolve more KBr.

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The value of the equilibrium constant, K, for the reaction is 0.0049

The following data were obtained from the quesion:

The equilibrium constant for the reaction can be obtain as illustrated below:

Equilibrium constant = [B₂][AC] / [AB₂C]

Equilibrium constant = (0.007 × 0.0118) / 0.0168

Equilibrium constant = 0.0049

Thus, from the above calculation, the equilibrium constant, K for the reaction is 0.0049

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Answer: They have eight outer electrons.

Explanation: They don't need any more electrons to be added to them and can't give out any electrons to other groups. They have a complete outer shell.

The amount of heat required to vaporize 181.20 g of ethanol would be 170.1 kJ.

Using the formula:

Q = n ∆Hvap

where:

Moles of 181.20 g of ethanol = 181.20 g / 46.07 g/mol = 3.933 mol

Substituting the values:

Q = 3.933 mol x 43.3 kJ/mol = 170.1 kJ

In other words, the amount of heat required to vaporize 181.20 g of ethanol is 170.1 kJ.

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1. To solve for the grams of H2 needed, we need to use the given ΔH value to calculate the amount of moles of N2 that reacted. From the balanced chemical equation, we know that for every 3 moles of H2 that reacts, 1 mole of N2 reacts. Therefore, we can use the mole ratio to convert the moles of N2 to moles of H2 and then use the molar mass of H2 to convert to grams.

First, we need to calculate the moles of N2 that reacted to produce 150.9kJ of heat:
ΔH = -92.40 kJ/mol N2
150.9 kJ = (1 mol N2 / -92.40 kJ) x (-150.9 kJ)
mol N2 = 1.63 mol

Using the mole ratio from the balanced chemical equation:

1 mol N2 : 3 mol H2

We can calculate the moles of H2 needed:3 mol H2 = 1 mol N2

3 mol H2 = 1.63 mol N2
mol H2 = 0.543 mol

Finally, we can convert moles of H2 to grams:

mol H2 = 0.543 mol
molar mass of H2 = 2.02 g/mol
grams of H2 = (0.543 mol) x (2.02 g/mol)
grams of H2 = 1.10 g

Therefore, 1.10 grams of H2 are needed to involve 150.9kJ of heat.

2. To solve for the moles of NH3 produced, we can use the same mole ratio from the balanced chemical equation:

1 mol N2 : 2 mol NH3

From the moles of N2 that reacted calculated in part 1, we can calculate the moles of NH3 produced:
1 mol N2 = 2 mol NH3
1 mol N2 = 1.63 mol N2
mol NH3 = (2 mol NH3 / 1 mol N2) x (1.63 mol N2)
mol NH3 = 3.26 mol

Therefore, 3.26 moles of NH3 were produced in the process.

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The mass (in grams) of aluminum, Al that must react with iodine to form AlI₃, given that -836.0 KJ of heat is relaesd is 149.0 g

The mass of aluminum required to react with iodine to produce AlI₃ can be obtain as shown below:

2Al(s) + 3I₂(s) → 2AlI₃(s) ∆H = -302.9 KJ

From the balanced equation above,

When -302.9 KJ of heat energy is released, 54 g of aluminum, Al reacted.

Therefore,

When -836.0 KJ of heat energy will be release = (-836.0KJ × 54 g) / -302.9 KJ = 149.0 g of aluminum, Al will react.

Thus, from the above calculation, we can conclude that the mass of aluminum, Al required is 149.0 g

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If in a reaction between vinegar and antacid tablets, the antacid is the limiting reagent. cm of gas. At constant pressure and temperature, three tablets produce 600 cm³ . The  volume that four tablets will produce is: C. 800 cm³.

Since the antacid is the limiting reagent, the amount of gas produced will be directly proportional to the number of tablets used.

We know that three tablets produced 600 cm³ of gas. Therefore, we can set up a proportion:

3 tablets produce 600 cm³ of gas

4 tablets produce x cm³ of gas

To solve for x, we can use cross-multiplication:

3 tablets × x cm³ of gas = 4 tablets × 600 cm³ of gas

3x = 2400

x = 800 cm³

Therefore the answer is C. 800 cm³.

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The height of fall of the ice is determined as 313.25 m.

The total heat energy of the ice during the fall is calculated as follows;

Q = ml + mcΔT

where;

Q = 334000 x 0.4  +  0.4 x 4200 x (10)

Q = 150,400 J

The energy converted into potential energy is calculated as;

3%K.E = 150,400 J

0.03K.E = 150,400 J

K.E = 5,013,333.33 J

¹/₂mv² = 5,013,333.33 J

v = √(2 x 5,013,333.33)/(0.4)

v = 5,006.67 m/s

The height of fall is calculated as;

h = √2gh

h = √(2 x 5,006.67 x 9.8)

h = 313.25 m

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The decay of Xenon-146 follows an exponential decay model, where the amount of Xenon-146 remaining after time t is given by:

N(t) = N0 * (1/2)^(t/T)

where N0 is the initial amount of Xenon-146, T is the half-life of Xenon-146, and t is the time that has passed.

We are given that N0 = 416 g, N(t) = 13 g, and T = 37 hours. We can solve for t as follows:

N(t) = N0 * (1/2)^(t/T)

13 = 416 * (1/2)^(t/37)

Taking the natural logarithm of both sides:

ln(13) = ln(416) + (t/37) * ln(1/2)

Solving for t:

t = 37 * [ln(13/416) / ln(1/2)]

t ≈ 111.2 hours

Therefore, the Xenon-146 has been decaying for approximately 111.2 hours.