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8. How many molecules of aspartame (C₁4H₁N₂O5) are in 745 grams of C14H18N₂O5?

9. What is the mass of 5 x 10^25 molecules of Fe₂(SO3)3?

10. Calculate the mass of 6.3 moles of Ba3(PO4)2

Explanation:

Foremost among the threats is fishing bycatch—marine animals incidentally caught and killed in fishing operations.

This is a serious crisis, not only for these charismatic animals, but also for the health of our oceans, where cetaceans play an irreplaceable role as apex predators.

In order to make a bulb light up, certain conditions need to be met. Firstly, there must be a closed circuit, meaning that there is a continuous flow of current through the circuit. This can be achieved by connecting the bulb to a power source, such as a battery or a generator, and completing the circuit with wires.

Secondly, the bulb must be connected properly to the circuit. This involves connecting the positive and negative terminals of the bulb to the corresponding terminals of the power source or the wires.

Finally, the bulb must have the correct voltage and wattage rating. Using a bulb with a higher or lower rating than the power source can cause the bulb to either not light up or burn out quickly.

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The final temperature of the water is 32.3°C.

The final temperature of the mixture is 53.3°C.

The heat lost by the copper is 4.19 kJ.

To determine the final temperature of the water, we can use the principle of conservation of energy. The total heat gained by the cooler water is equal to the total heat lost by the hotter water. The heat gained or lost by a substance can be calculated using the formula:

Q = mcΔT

where Q is the heat gained or lost, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature.

Let's assume that the final temperature of the water is T. Then, the heat lost by the hotter water is:

Q1 = mcΔT = (100.0 g) x (4.184 J/g°C) x (73.7°C - T)

The heat gained by the cooler water is:

Q2 = mcΔT = (100.0 g) x (4.184 J/g°C) x (T - 27.8°C)

Since the total heat gained is equal to the total heat lost, we can equate Q1 and Q2:

(100.0 g) x (4.184 J/g°C) x (73.7°C - T) = (100.0 g) x (4.184 J/g°C) x (T - 27.8°C)

Simplifying the equation:

3134.4 - 41.84T = 418.4T - 11702.72

459.24T = 14837.12

T = 32.3°C

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Percent of water of crystallisation in MgSO4⋅7H2O=2467×18×100=51. 2 %.

Divide by 100 the mass of the hydrate and the volume of water lost. The amount of water in one mole of the hydrate must be multiplied by 100 after being divided by the hydrate's molar mass in order to determine the theoretical (actual) percent hydration (percent water) using the formula.

Hence, this ratio might be represented by one magnesium sulfate formula unit to seven water molecules. Since MgSO47H2O is the chemical formula for magnesium sulfate hydrate, the value of x must be seven. The value of x in the chemical formula for magnesium sulfate hydrate is thus seven.

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The positive difference between the

pressure in the balloon and the pressure of the troposphere is 0.048 kPa .

The balloon is being compressed by atmospheric force. The balloon is trying to inflate owing to gas pressure. The atmospheric pressure (outside pressure) in a fully inflated balloon is lower than the internal gas pressure.

An inflated balloon has a higher internal pressure than its surrounding air. Because the contracting force produced by the rubber's surface tension balances the expanding force caused by the pressure differential, the balloon remains stable.

Because to the air pressure on the walls, a balloon's surface feels tighter when it is inflated. The internal air pressure gives the balloon its shape. The balloon will therefore be more pressurized after being inflated.

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3.28 x 10³ mol of water has a mass of 59.1156 kg. This is the accurate answer to the given question.

We can use the molar mass of water, which is 18.02 g/mol, to calculate the mass of 3.28 x 10³ moles of H2O. The following equation can be used to translate the quantity of moles into grams:

mass is determined by multiplying the number of moles by the molar mass.

Inputting the values provided yields:

18.02 g/mol times 3.28 x 10 3 moles of mass equals 59,115.6 g.

Lastly, by dividing by 1000, we can change the mass from grams to kilograms:

mass is 59,115.6 grams per kilogram (59.1156 kg).

As a result, 3.28 x 10³ moles of H2O have a mass of 59.1156 kg.

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The name of the formula BF2 is : C.) Boron difluoride. The number of atoms of each element in a compound is known by chemical formula.

Chemical formula is an expression that shows elements in a compound and relative proportions of those elements. If only one atom of specific type is present, then no subscript is used. For atoms that have two or more of a specific type of atom present, then a subscript is written after the symbol for that atom.

Chemical formula contains the symbols of atoms of the elements present in the compound and also how many there are for each element in the form of subscripts.

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The pH of the solution is 1.8. Option C is the right answer

First, we need to calculate the moles of LiOH and HNO3:

Moles of LiOH = 0.1 mol/L × 0.025 L = 0.0025 mol

Moles of HNO3 = 0.1 mol/L × 0.035 L = 0.0035 mol

Next, we need to determine which of the reactants will be limiting. LiOH and HNO3 react in a 1:1 stoichiometric ratio, so whichever reactant is present in the smaller amount will be limiting. In this case, LiOH is limiting because it has fewer moles than HNO3.

The balanced chemical equation for the reaction between LiOH and HNO3 is:

LiOH + HNO3 → LiNO3 + H2O

Since LiOH is limiting, all of it will react with HNO3. This means that all of the moles of HNO3 will be consumed, and there will be some excess LiOH remaining.

Moles of LiOH remaining = Moles of LiOH - Moles of HNO3 = 0.0025 mol - 0.0035 mol = -0.0010 mol

Note that the result is negative, which means that all of the HNO3 has been consumed, and there is still some LiOH remaining.

The concentration of LiOH in the final solution is:

[LiOH] = Moles of LiOH remaining / Total volume of solution = -0.0010 mol / (0.025 L + 0.035 L) = -0.010 mol/L

Again, the result is negative, which is not physically meaningful. This indicates that our assumption that LiOH is limiting is incorrect.

To correct this, we can assume that HNO3 is limiting and repeat the calculations. In this case, all of the HNO3 will react with LiOH, and there will be some excess HNO3 remaining.

Moles of HNO3 remaining = Moles of HNO3 - Moles of LiOH = 0.0035 mol - 0.0025 mol = 0.0010 mol

The concentration of HNO3 in the final solution is:

[HNO3] = Moles of HNO3 remaining / Total volume of solution = 0.0010 mol / (0.025 L + 0.035 L) = 0.01 mol/L

The balanced chemical equation for the reaction between HNO3 and LiOH is:

HNO3 + LiOH → LiNO3 + H2O

The stoichiometry of this reaction tells us that one mole of HNO3 reacts with one mole of LiOH to produce one mole of water. Therefore, the moles of water produced in the reaction are equal to the moles of HNO3 remaining:

Moles of water = 0.0010 mol

The volume of the final solution is:

Volume = 0.025 L + 0.035 L = 0.06 L

The molarity of the water in the final solution is:

Molarity of water = Moles of water / Volume of solution = 0.0010 mol / 0.06 L = 0.0167 mol/L

The pH of water is 7.0, and the pH of a solution is related to the concentration of H+ ions in the solution. Since water is neutral and does not contribute any H+ ions, the pH of the final solution is:

pH = 7.0 - log[H+] = 7.0 - log(0.0167) ≈ 1.8

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Determine the pH of the solution when 25ml of 0.1mLiOH is added to 35ml of 0.1m nitric acid

(A) 0.23

(B) 0.47

(C) 1.8

(D) 7.0

Factors that make the higher yield of crop possible are as follows: enough water for irrigation. access of organic manure of high quality . Fertilizers are a factor in the higher yields of high-cost farming.

Fertilizer is used to ensure good vegetative growth (leaves, branches and flowers), giving rise to healthy plants. They are artificially made but have a rapid response. Overuse of fertilisers can cause negative effects like loose soil fertility. There's a risk with any bond that the issuing company might not be able to meet its obligations.

However, the risks of default are typically higher for companies that issue high-yield bonds. Interest rate . If the octet is expanded with preferred number of bonds, formal charge = 0. The S-atom will gain two electrons in order to complete its octet.

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Answer: Its the bottom one

Enzymes speed up chemical reactions by lowering the activation energy needed for the reaction. The correct option is d.

Enzymes are biological molecules, typically proteins, that act as catalysts for chemical reactions. Catalysts are substances that can speed up a chemical reaction without being consumed in the process.

Enzymes work by lowering the activation energy needed for a chemical reaction to occur. Activation energy is the energy required to break the bonds of the reactants and initiate the chemical reaction. By lowering the activation energy, enzymes make it easier for reactants to come together and form products.

Enzymes achieve this by binding to the reactants, or substrates, and bringing them into close proximity to each other. This allows the substrates to interact more easily and lowers the energy barrier for the reaction to occur. Enzymes also stabilize the transition state of the reaction, further reducing the activation energy needed for the reaction to proceed.

Therefore, the correct answer is d) they lower the activation energy needed for the reaction.

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Answer:

the mass of Na2SO4 produced is 27.6 g.

Explanation:

determine the balanced chemical equation for the reaction between NaCl and H₂SO4, and then use stoichiometry to calculate the mass of Na2SO4 produced.

The balanced chemical equation for the reaction is:

2 NaCl + H₂SO4 → Na2SO4 + 2 HCl

This equation tells us that 2 moles of NaCl react with 1 mole of H₂SO4 to produce 1 mole of Na2SO4.

First, we need to determine the number of moles of NaCl and H₂SO4 we have:

moles of NaCl = mass / molar mass = 48.0 g / 58.44 g/mol = 0.821 mol

moles of H₂SO4 = mass / molar mass = 19.0 g / 98.08 g/mol = 0.194 mol

Since H₂SO4 is the limiting reactant (it is present in less amount than required for complete reaction), we need to use the number of moles of H₂SO4 to calculate the number of moles of Na2SO4 produced:

moles of Na2SO4 = moles of H₂SO4 / 1 x (1 mole Na2SO4 / 1 mole H₂SO4) = 0.194 mol x (1/1) = 0.194 mol

Finally, we can use the molar mass of Na2SO4 to calculate the mass of Na2SO4 produced:

mass of Na2SO4 = moles of Na2SO4 x molar mass = 0.194 mol x 142.04 g/mol = 27.6 g

Therefore, the mass of Na2SO4 produced is 27.6 g.

The particles in suspension can be separated from heterogenous mixtures by passing the mixture through a filter. Option 1.

The particles in suspension can be separated from heterogeneous mixtures by passing the mixture through a filter.

A suspension is a heterogeneous mixture in which the particles are large enough to settle out over time and can be separated by physical means such as filtration.

Other options such as solution, colloid, and pure substances cannot be separated using a filter.

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Answer:

11.2823 grams of NH3 are produced
Explanation:

No. of moles for H2 = 22.3/22.4 = 0.9955 moles

By calculating number of moles produced of NH3, by using ratios:

N2 + 3H2 → 2NH3
1    :   3    :    2
? : 0.9955:  ?

part value = 0.9955/3 = 0.33183
No. of moles for NH3 = 2 * 0.33183 = 0.6637 moles

mass of produced NH3 (Molar mass = 17 g/mol) = 0.6637*17 = 11.2823 grams

a. Sodium hydroxide(aq) + ammonium sulfate(aq) will react to form solid ammonium hydroxide and sodium sulfate(aq). b. Niobium(V) sulfate(aq) + barium nitrate(aq) will react to form solid barium sulfate(s) and niobium(V) nitrate(aq).

Predicting whether a reaction will occur in aqueous solution is important in understanding chemical reactions and their potential outcomes. It allows chemists to determine what products will form and whether a reaction is feasible or not based on the solubility of the reactants and products.

To predict whether a reaction will occur in aqueous solution, one can use solubility rules to determine whether the products of the reaction will form an insoluble precipitate or remain in solution. If a precipitate forms, it indicates that a reaction has occurred, whereas if all the products remain in solution, the reaction will not occur. Additionally, one can use other chemical principles, such as redox reactions and acid-base reactions, to predict whether a reaction will occur.

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Yes, the diamond cubic structure is close-packed. In a close-packed structure, you can find a plane of atoms where all the atoms touch. To understand why the diamond cubic structure is close-packed, follow these steps:

1. Identify the structure: The diamond cubic structure consists of two interpenetrating face-centered cubics (fcc) lattices with a relative shift of 1/4 along the space diagonal.

2. Locate the planes: In the diamond cubic structure, there are several planes where atoms touch each other, forming a hexagonal close-packed (hcp) arrangement. For instance, the (111) plane in one of the fcc sublattices is close-packed.

3. Examine neighboring atoms: In these close-packed planes, the atoms are arranged so that each atom touches its neighboring atoms. This close arrangement of atoms maximizes the packing efficiency.

So, based on the presence of close-packed planes and the neighboring atoms touching each other, the diamond cubic structure is considered a close-packed structure.

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The percentage of NaCl in the impure sample is 76.25%.

The given weight of an impure sample of table salt is 0.8421 g, and the weight of AgCl formed is 2.044 g. The percentage of NaCl in the impure sample can be determined as follows:

Calculations: Gram molecular weight of NaCl = 23 + 35.5 = 58.5 g

Number of moles of AgCl = weight of AgCl / gram molecular weight of AgCl= 2.044 / 143.5 = 0.0142 mol

The equation of the reaction between NaCl and AgNO3 is given as follows:

NaCl + AgNO3 → NaNO3 + AgCl

It can be seen from the above equation that 1 mole of NaCl gives 1 mole of AgCl.So, the number of moles of NaCl present in the sample is 0.0142 mol.

Percentage of NaCl in the sample = number of moles of NaCl present in the sample / number of moles of the impure sample × molar mass of NaCl × 100= (0.0142 / (0.8421 / 58.5)) × 100= 76.25%

Therefore, the percentage of NaCl in the impure sample is 76.25%.

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The pressure inside the aerosol can at 845°C is 33.8 atm. This is why it is extremely dangerous to incinerate aerosol cans, as they can explode when exposed to high temperatures.

To solve this problem, we can use the combined gas law, which relates the pressure, volume, and temperature of a gas:

(P1 x V1) / T1 = (P2 x V2) / T2

where P1, V1, and T1 are the initial pressure, volume, and temperature of the gas, and P2, V2, and T2 are the final pressure, volume, and temperature.

We are given the following:

Initial pressure P1 = 3.00 atm

Initial temperature T1 = 25°C = 298 K

We need to find the final pressure P2 at T2 = 845°C = 1118 K.

Since the can is sealed, volumes V1 and V2 remain the same.

Plugging these values into the combined gas law, we get:

(P1 x V1) / T1 = (P2 x V2) / T2

(3.00 atm x V1) / 298 K = (P2 x V1) / 1118 K

Simplifying this equation, we get the following:

P2 = (3.00 atm x V1 x 1118 K) / (298 K x V1)

P2 = 33.8 atm

Therefore, the pressure inside the aerosol can at 845°C is 33.8 atm, which is significantly higher than at room temperature. This demonstrates why it is extremely dangerous to incinerate aerosol cans, as they can explode when exposed to high temperatures.

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the volume of a 1.80 M solution of potassium chloride that is required to prepare 0.100 L of a 0.600 M solution is 0.0333 L, or 33.3 mL.

To determine the volume of a 1.80 M solution of potassium chloride that is required to prepare 0.100 L of a 0.600 M solution, we can use the formula:

M1V1 = M2V2

where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume.

Let's use this formula to solve the problem:

M1 = 1.80 M (since we are starting with a 1.80 M solution)

V1 = ? (what we are trying to find)

M2 = 0.600 M (since we want to prepare a 0.600 M solution)

V2 = 0.100 L (since we want to prepare 0.100 L of the final solution)

Substituting these values into the formula, we get:

1.80 M x V1 = 0.600 M x 0.100 L

Simplifying, we get:

V1 = (0.600 M x 0.100 L) / 1.80 M

V1 = 0.0333 L, or 33.3 mL

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The valence electrons of oxygen are; 2s2 2p4

The valence electrons of nitrogen are 2s2 2p3

This is how we can be able to know the number of valence electrons.

Valence electrons are the electrons in the outermost shell of an atom. They are the electrons involved in chemical reactions and bonding with other atoms. In the case of nitrogen, its electron configuration is 1s^2 2s^2 2p^3, which means it has two electrons in the first shell, two electrons in the second shell, and three electrons in the outermost p orbital, giving it a total of 5 valence electrons.

Oxygen has an electron configuration of 1s^2 2s^2 2p^4, which means it has two electrons in the first shell, two electrons in the second shell, and four electrons in the outermost p orbital, giving it a total of 6 valence electrons.

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Answer:

33.67g

Explanation:

To calculate the mass of carbon monoxide, we can use the ideal gas law, which relates the pressure, volume, number of moles, and temperature of a gas:

PV = nRT

Where:

P = pressure

V = volume

n = number of moles

R = ideal gas constant

T = temperature

At STP (standard temperature and pressure), the pressure is 1 atm and the temperature is 273.15 K. The ideal gas constant is R = 0.08206 L atm/(mol K).

To find the number of moles of carbon monoxide in 29.3 L at STP, we can rearrange the ideal gas law to solve for n:

n = PV/(RT)

n = (1 atm)(29.3 L)/(0.08206 L atm/(mol K) * 273.15 K)

n = 1.203 mol

The molar mass of carbon monoxide (CO) is 28.01 g/mol. Therefore, the mass of 1.203 mol of CO is:

mass = n * molar mass

mass = 1.203 mol * 28.01 g/mol

mass = 33.67 g

Therefore, the mass of 29.3 L of carbon monoxide at STP is approximately 33.67 grams.

Answer:- NaBr and Cl2 are reactants .

The atoms/molecules/ions on RHS of the reaction which react to give products are called reactants.

So here Sodium bromide and chlorine are the reactants.

Option A is the correct choice.

Additional information:-

Law of conservation of Mass :-

According to this law , the mass of reactants is equal to the mass of the products . Take a chemical reaction ,

C + O2 ==> CO2

Now suppose that if 12g of carbon reacts with 16g of oxygen to form carbon dioxide, then what will be the mass of carbon dioxide? So by law of conservation of Mass we can say that mass of carbon dioxide will be equal to mass of the reactants that will be 12g + 16g = 28g

With this I am leaving a sample problem for you to solve,

Question: Consider the following reaction .

2C + O2 ==> 2CO

If the mass of carbon taken is 24g and the amount of carbon monoxide formed is 56g , find the amount of oxygen used in the reaction.

Answer:D. Yields

Explanation:

Answer:

0.9299 x [tex]10^-7[/tex] moles of [tex]OH^-[/tex]

Explanation:

[tex]OH^-[/tex]: 5.6 x 10^(15)  (brainly doesn't allow exponents greater than 9)

This number of particles can also be represented as 5.6e15.

'e' in this case means 'times 10 to the n' where n is the number that follows e.

Use stoichiometry to convert between different units.

[tex]\frac{5.6e15 particles}{1 sample} *\frac{1 mole}{6.022 e22 particles}= 0.9299 e -7 moles[/tex]

So in that sample of particles, there are 0.9299 x [tex]10^-7[/tex] moles of [tex]OH^-[/tex].

Answer: The vapor pressure of pure water at 100°C is 760mm of Hg.

Explanation: We know tat 100 degree Celsius the water will convert into gas phase.  At boiling point the vapor pressure of any liquid is 1 atmosphere and if we boil the water at 100° Celsius  it means 1 atmosphere . So, 1 atmosphere = 760mm of Hg.

Answer:

First, we need to find the number of moles of HCl used in the titration:

moles HCl = M x V = 2.0 mol/L x 1.50 L = 3.00 mol

Since zinc hydroxide and HCl react in a 1:2 molar ratio according to the balanced chemical equation:

Zn(OH)2 + 2 HCl → ZnCl2 + 2 H2O

we know that twice as many moles of HCl are required to react with each mole of Zn(OH)2. Therefore, the number of moles of zinc hydroxide in the 5.00 kg sludge can be calculated as follows:

moles Zn(OH)2 = 1/2 x moles HCl = 1/2 x 3.00 mol = 1.50 mol

Finally, we can use the molar mass of Zn(OH)2 to convert the number of moles to mass:

mass Zn(OH)2 = moles Zn(OH)2 x molar mass Zn(OH)2

= 1.50 mol x 99.39 g/mol

= 149.1 g

Therefore, the mass of zinc contained in the 5.00 kg sludge is 149.1 grams.

Explanation:

The formula for the first Compound I is likely A2B, or O2N.

To show that the data are in accordance with the law of multiple proportions, we need to calculate the ratios of the masses of element B that combine with a fixed mass of element A in each compound. If these ratios are in simple whole-number ratios, then the law of multiple proportions is upheld.

For Compound I:

Mass of A (23.3 g) / Mass of B (3.00 g) = 7.77

For Compound II:

Mass of A (7.00 g) / Mass of B (4.50 g) = 1.56

These ratios are not equal, indicating that the two compounds have different ratios of A and B. However, we can simplify the ratio for Compound II by dividing both masses by the mass of B:

Mass of A (7.00 g / 4.50 g) = 1.56

This suggests that Compound II has a 2:3 ratio of A to B.

Using this information, we can determine the formula of Compound II. Let x be the atomic mass of A and y be the atomic mass of B:

(2x) + (3y) = 4.50 g

Solving for y gives us y = 1.00 g/mol. Since the formula of Compound II is AB2, we know that the atomic mass of A is 2 g/mol. Therefore, A is likely oxygen (atomic mass 16 g/mol).

For Compound I, we know that A is oxygen (atomic mass 16 g/mol). Using the same approach as above, we can determine that B has an atomic mass of 14 g/mol. Therefore, the formula for Compound I is likely A2B, or O2N.

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Answer:  

Generally true.

Explanation:

As you go down the columns of the Periodic Table, the radius of the atom increases.

As you go across the Periodic Table (left to right), the radius of the atoms decrease.

Therefore, the second element on the table, Helium, has the smallest radius and element 87, Francium has the largest radius.

There are a few exception scattered throughout the table, such as the lanthanides and actinides, but generally speaking the radius increases as you travel from right to left and top to bottom.

Final volume of the gas would be 850.5 mL when the pressure is increased from 787.0 mmHg to 897.0 mmHg while keeping the temperature constant.

Boyle's Law explains that when temperature of given mass of confined gas is constant, then the product of its pressure and volume is also constant.

Given, P1 = 787.0 mmHg ; V1 = 970.0 mL; P2 = 897.0 mmHg

V2 = ?

As P1V1 = P2V2

V2 = P1V1/ P2

V2 = (787.0 mmHg x 970.0 mL) / 897.0 mmHg

V2 = 850.5 mL

Therefore, the final volume of the gas would be 850.5 mL when the pressure is increased from 787.0 mmHg to 897.0 mmHg while keeping the temperature constant.

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The birth rate is  28.2 births per year

The number of births per 1,000 persons in a particular population over a given time period is commonly used to determine the birth rate.

The following is the formula that can be used to calculate the birth rate:

(Number of births / Population) x 1,000 is the birth rate.

The birth rate can be obtained from;

Birth rate = (Number of births / Population) x 1,000

We know that the birth rate can be obtained from;

7.34 x 10^4 / 2.6 x 10^6  * 1000

= 28.2 births per year

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Law of conservation of mass was crucial to progression of chemistry, as it helped scientists understand that substances did not disappear as result of reaction.

Law of conservation of mass explains that in chemical reaction mass is neither created nor destroyed.

To ensure Conservation of Mass is followed when mixing vinegar and baking soda in a bottle, you should make sure to do the following:

Weigh the bottle and its contents before adding vinegar and baking soda. This will give the initial mass of system.

Carefully measure the amounts of vinegar and baking soda you add to bottle.

Quickly and securely cap the bottle after adding baking soda to avoid any gas from escaping.

Weigh the bottle and its contents again after reaction has completed and the gas has stopped escaping. This will give the final mass of the system.

Compare the initial and final masses of system. They should be equal within experimental error.

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