3. Each container holds one litre of water. Write the least amount of salt needed to make each solution saturated. Then draw a picture of the saturated solution in the circle. answer of this.
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The wavelength of one of the spectral lines for strontium chloride SrCl₂ is approximately 460.7 nanometers (nm).

When strontium chloride SrCl₂ is heated, it emits a characteristic red color, which indicates that it produces spectral lines in the red part of the visible spectrum. The most intense spectral line for SrCl₂ is at approximately 460.7 nm, which corresponds to the transition from the 5² electronic configuration to the 4d state.

This transition is responsible for the red color observed when strontium chloride is introduced to a flame. The wavelength of a spectral line is related to the energy of the transition and is given by

λ = hc ÷ E

where λ is the wavelength, h is Planck's constant, c is the speed of light, and E is the energy of the transition. In the case of SrCl₂, the energy of the transition from 5s² to 4d is approximately 2.69 eV, which corresponds to a wavelength of 460.7 nm.

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The complete question is:

What is the wavelength of one of the spectral lines for strontium chloride SrCl₂?

Total, 127.8 moles of the steam (water vapor) will be produced.

Iron(III) nitrate nonahydrate, or Fe(NO₃)₃·9H₂O, is a compound that contains iron (III) cations (Fe³⁺) and nitrate anions (NO₃⁻) combined with nine water molecules (H₂O) per formula unit. The nine water molecules are referred to as "nonahydrate," which indicates that there are nine water molecules associated with each formula unit of Fe(NO₃)₃. This compound is also commonly known as ferric nitrate nonahydrate.

Balanced equation for the given reaction is;

Fe(NO₃)₃·9H₂O(s) → Fe(NO₃)₃(s) + 9H₂O(g)

According to the equation, one mole of Fe(NO₃)₃·9H₂O produces 9 moles of water vapor (H₂O) when it undergoes the reaction. Therefore, if there are initially 14.2 moles of Fe(NO₃)₃·9H₂O, the number of moles of water vapor (H₂O) produced will be;

14.2 moles of Fe(NO₃)₃·9H₂O × 9 moles of H₂O per 1 mole of Fe(NO₃)₃·9H₂O = 127.8 moles of H₂O

Therefore, 127.8 moles of water vapor will be produced.

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Gas legislation

PV = nRT

where P is atmospheric pressure

Volume (measured in litres)

The number of moles is n.

R = gas constant (0.0821 L atm/(mol K))

Temperature (in Kelvin) equals T.

We need to rearrange the ideal gas law to solve for n:

n = PV / RT

To use this equation, we need to know the pressure, temperature, and volume of the carbon dioxide.

0.002355 moles of carbon dioxide are equal to 235.5 litres of the petrol. As a result, 0.002355 moles of carbon dioxide are equivalent to 0.053 litres.

This is because 1 mole of carbon dioxide is equal to 22.4 litres. This indicates that there are 0.002355 moles of carbon dioxide for every 0.053 litres of carbon dioxide.

Therefore, there are 0.002355 moles of carbon dioxide in 235.5 litres of carbon dioxide.

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The color of the object appears as yellow/orange.

The visible light spectrum spans a wavelength range of 400 to 700 nanometers (nm), with shorter wavelengths appearing as blue or violet and longer wavelengths as red.

This object is absorbing light with wavelengths between 430 and 400 nm, indicating that it is absorbing light that is visible in the blue and violet spectrum.

We know that this color that we see is actually a complementary color to blue/violet from the color wheel.

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The mass of iron (III) nitrate needed to prepare a 125 mL solution of 0.250 M is approximately 7.56 grams.

To calculate the mass of iron (III) nitrate needed to prepare a 125 mL solution of 0.250 M, we can use the formula:

mass (in grams) = volume (in liters) x concentration (in moles/liter) x molar mass (in grams/mole)

First, let's convert the volume from milliliters (mL) to liters (L):

125 mL = 0.125 L

Next, we can use the given concentration of 0.250 M to calculate the number of moles of Fe(NO3)3 needed:

moles = concentration x volume

moles = 0.250 mol/L x 0.125 L

moles = 0.03125 mol

Finally, we can use the molar mass of Fe(NO3)3 to convert from moles to grams:

mass = moles x molar mass

mass = 0.03125 mol x 241.86 g/mol

mass ≈ 7.56 g

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Answer:H = 5

Explanation: You would get this answer if you divide by 2 then multiple by 7

The concentration of the sugar in the soft drink in mass percent is 8.64%

First, we shall determine the mass of the solution. Details below:

Mass of solution = Mass of sugar + mass of water

Mass of solution = 33 + 349

Mass of solution = 382 grams

Finally, we shall determine the mass percent of the sugar in the solution. Details below:

Percentage = (mass of sugar / mass of solution) × 100

Percentage of sugar = (33 / 382) × 100

Percentage of sugar = 8.64%

Thus, we can conclude that the mass percent of the sugar is 8.64%

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All the balanced chemical equations obey the law of conservation of mass. The numbers which are used to balance the chemical equation are called the coefficients. So here the mass of zinc is 55 g.

According to the law of conservation of mass, the mass can neither be created nor be destroyed but it can be converted from one form to another. The reactants appear on the left hand side and the products appear on the right hand side.

The amount of products is equal to the amount of reactants according to the law of conservation of mass.

25 + 125 = 95 + Zn

Zn = 55 g

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There are different methods to calculate the concentration of a solution. Mass percentage is one among them. Mass percentage is mainly used to calculate the concentration of a binary solution. Here mass percent of water is 10.13.

Mass percentage of a particular component in a solution is equal to mass in grams of that component present per 100 g of the solution. For example, a 5% aqueous solution of urea means 5g of urea in 100 g of its aqueous solution.

Mass percentage = Mass of the component / Total mass of solution × 100

Mass of water = 306.0 - 275.0 = 31

% Mass = 31 / 306.0 × 100 = 10.13%

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The heat of fusion of water is 79.9 cal/g. If a 7.2 g piece of ice melts in 105 g of water at 34.3 deg C in an insulated bottle, 35071.6 °C is the  final temperature of the water.

The physical concept of temperature indicates in numerical form how hot or cold something is. A thermometer is used to determine temperature. Thermometers are calibrated using a variety of temperature scales, which historically defined distinct reference points or thermometric substances.

The most popular scales were the Celsius scale, sometimes known as centigrade, with the unit symbol °C, the scale of Fahrenheit (°F), or the Kelvin scale (K), with the latter being mostly used for scientific purposes.

Δ T = T(initial) - T(final)

T(final)= m × c × q - T(initial)

T(final)= 79.9 x 4.184 x 105 - 30.0

T(final)= 35071.6 °C

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Answer:

a. The chemical equation for the reaction is:

2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

b. To find the mass of required sulfuric acid, we need to use stoichiometry. We can start by finding the number of moles of aluminum used in the reaction:

Molar mass of Al = 27 g/mol

Number of moles of Al = 5.4 g / 27 g/mol = 0.2 mol

According to the balanced equation, 3 moles of H₂SO₄ are required to react with 2 moles of Al. Therefore, the number of moles of H₂SO₄ required is:

Number of moles of H₂SO₄ = 3/2 x 0.2 mol = 0.3 mol

Molar mass of H₂SO₄ = 2 x 1 g/mol + 32 g/mol + 4 x 16 g/mol = 98 g/mol

Mass of H₂SO₄ required = 0.3 mol x 98 g/mol = 29.4 g

Therefore, 29.4 g of sulfuric acid is required to react with 5.4 g of aluminum.

c. To find the volume of hydrogen gas obtained, we need to use the ideal gas law:

PV = nRT

where P is the pressure of the gas, V is its volume, n is the number of moles of the gas, R is the universal gas constant (0.0821 L atm/mol K), and T is the temperature in Kelvin.

We can start by finding the number of moles of hydrogen gas produced in the reaction. According to the balanced equation, 3 moles of H₂ are produced for every 2 moles of Al. Therefore, the number of moles of H₂ produced is:

Number of moles of H₂ = 3/2 x 0.2 mol = 0.3 mol

Assuming the reaction occurs at standard temperature and pressure (STP), which is 0°C (273 K) and 1 atm, we can use the molar volume of a gas at STP, which is 22.4 L/mol. Therefore:

V = nRT/P = 0.3 mol x 0.0821 L atm/mol K x 273 K / 1 atm = 6.58 L

Therefore, the volume of hydrogen gas produced at STP is 6.58 L.

Explanation:

The Tafel slope for Mechanism 2 is the sum of the slopes of both steps, presuming that each step in Mechanism 2 is RCS::

b2 = b2_1 + b2_2 = 0.1184 + 0.1184 = 0.2368 V

To identify the rate-controlling step (RCS) utilizing the Tafel slope, we initially need to calculate the Tafel slope for each suggested mechanism when the electron transfer coefficient (alpha, α) equals 0.5.

The target reaction involves Cu oxidation.

Mechanism 1:

Cu -> Cu²⁺ + 2e⁻

Mechanism 2:

Cu -> Cu⁺ + e⁻

Cu⁺ -> Cu²⁺ + e⁻

The Tafel slope (b) can be computed with the following formula:

b = (2.303 * R * T) / (α * n * F)

Where:

R signifies the gas constant (8.314 J/mol K)

T represents the temperature in Kelvin (let's assume 298 K, standard room temperature)

α denotes the electron transfer coefficient (0.5)

n is the number of electrons exchanged in the RCS

F is the Faraday constant (96,485 C/mol)

For Mechanism 1, n = 2 (since 2 electrons are exchanged in the rate-controlling step):

b1 = (2.303 * 8.314 * 298) / (0.5 * 2 * 96,485) = 0.0592 V

For Mechanism 2, we must examine both steps. Let's initially evaluate the Tafel slope for each step.

Step 1 (n = 1):

b2_1 = (2.303 * 8.314 * 298) / (0.5 * 1 * 96,485) = 0.1184 V

Step 2 (n = 1):

b2_2 = (2.303 * 8.314 * 298) / (0.5 * 1 * 96,485) = 0.1184 V

The Tafel slope for Mechanism 2 is the sum of the slopes of both steps, presuming that each step in Mechanism 2 is RCS::

b2 = b2_1 + b2_2 = 0.1184 + 0.1184 = 0.2368 V

Having obtained the Tafel slopes for both mechanisms, we can now compare them to the experimental Tafel slope to ascertain which mechanism is more likely the RCS.

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The oxidation state of oxygen in magnesium pyrophosphate (Mg2P2O7) is +5.

o calculate the oxidation state of oxygen in magnesium pyrophosphate (Mg2P2O7), we first need to know the oxidation states of the other atoms in the molecule.

The magnesium ion (Mg2+) has a fixed oxidation state of +2 in this compound, and the phosphate ion (PO43-) has an overall oxidation state of -3.

We can set up an equation to solve for the oxidation state of oxygen (O) in the pyrophosphate ion:

2(+2) + 2(O) + 7(-2) = 0

Simplifying the equation gives:

4 + 2O - 14 = 0

2O = 10

O = +5

Therefore, the oxidation state of oxygen in magnesium pyrophosphate (Mg2P2O7) is +5.

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The oxidation number of oxygen in magnesium pyrophosphate is -2.

Oxidation number refers to the hypothetical charge of an atom within a molecule.

For monoatomic ions, the oxidation number always has the same value as the net charge corresponding to the ion.

0 = +2(2) + 5(2) + X(7)

0 = 14 + 7x

-14 = 7x

x = -2

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For alcohol, ether, aldehyde, ketone, carboxylic acid, ester, amine and amide, the condensed structural formulas of six carbon members of the series are shown below.

The condensed structural formula for an alcohol with six carbon atoms is C6H13OH

The condensed structural formula for an ether with six carbon atoms is C6H14O

The condensed structural formula for an aldehyde with six carbon atoms is C6H10O

The condensed structural formula for a ketone with six carbon atoms is C6H10O

The condensed structural formula for a carboxylic acid with six carbon atoms is C6H10O2

The condensed structural formula for an ester with six carbon atoms is C6H12O2

The condensed structural formula for an amine with six carbon atoms is C6H15N

The condensed structural formula for an amide with six carbon atoms is C6H13NO

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The number of atoms remaining after 4 half-lives can be calculated using the formula: N = N0 /[tex]2^4[/tex] . Therefore, after 4 half-lives, approximately 56,250 atoms of the radioactive substance remain.

Radioactive decay is the process by which a nucleus of an atom loses energy by emitting ionizing radiation. The rate of decay of a radioactive substance is measured by its half-life, which is the time it takes for half of the radioactive atoms in a sample to decay.

N = N0 /[tex]2^n[/tex]

where: N0 = initial number of atoms N = final number of atoms

Substituting the given values,

N = 900,000 / [tex]2^4[/tex]

N = 56,250

Rounding to the nearest whole number,

N ≈ 56,250 atoms

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To prepare a buffer solution of pH 9.00, we need to use the Henderson-Hasselbalch equation:

Where [A-]/[HA] is the ratio of the concentrations of the conjugate base and acid of the buffer, respectively. Since we are given the pH and the overall concentration of the buffer, we can solve for the ratio [A-]/[HA]:

Next, we can use the definition of the concentration of a solution to find the concentration of NH4Cl needed to make a 0.125 mol-L^-1 buffer solution:

0.125 mol-L^-1 = [NH4Cl] + [NaOH]

Since the NaOH solution is 1.00 mol-L^-1, we can assume that the contribution of NaOH to the total concentration of the buffer is negligible, and so:

Finally, we can use the molar mass of NH4Cl to find the mass of NH4Cl needed to prepare 1 L of the buffer solution:

So we need to use 6.63 g of NH4Cl and enough volume of 1.00 mol-L^-1 NaOH solution to make a total volume of 1 L. To find the volume of NaOH solution needed, we can use the definition of molarity:

Rearranging this equation, we get:

Since we are adding NaOH solution to NH4Cl to make a total volume of 1 L, the molality of NaOH solution is also 0.125 mol-L^-1. Therefore:

Substituting the values we know:

So we need to use 6.63 g of NH4Cl and 56 mL of 1.00 mol-L^-1 NaOH solution to prepare 1 L of a buffer solution of pH 9.00.

The number of moles of the gas by the ideal gas law is 0.18 moles.

The behavior of an ideal gas, a hypothetical gas made up of randomly moving particles with little volume and no intermolecular interactions, is described by the ideal gas law.

Although intermolecular interactions and non-zero particle volume prevent gases from always behaving in an ideal manner, the ideal gas law is nevertheless a good approximation for many gases under some circumstances.

We know that;

PV = nRT

We have ;

P = 1.2 atm

V = 3.4 L

T = 10 + 273 = 283 K

n = ?

n = PV/RT

n = 1.2 * 3.4/0.082 * 283

n =4.08 /23.2

n = 0.18 moles

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Answer: I got you fam

Explanation:

A limiting reactant is a reactant in a chemical reaction that limits the amount of product created.

So for example if there are elements X and Y reacting to create product XY, once say element X runs out, the reaction stops, even though there is still more of the reactant Y. So there is 0 g of element X remaining, and maybe 2 g left of element Y. X is the limiting reactant since it limits the amount of product made.

Theoretical yield is the maximum amount of product that could be made in an experiment. This occurs if a reaction is 100% effective (and experimentally, this doesn't usually happen, which is why it is called theoretical).

An example of a composition which fulfills the qualifications of being both a base and a weak electrolyte is ammonia (NH3).

A base is any constituent which voluntarily receives protons (H+) in an associated chemical reaction while an electrolyte denotes any material that can conduct electricity through liquids or in melted state.

Upon dissolution in water, it is apt to accept a proton from such and thus create the acidic ion known as ammonium (NH4+). Nonetheless, due to its scarce dissociation into hydroxide (OH-) and ammonium ions, it is deemed a weak electrolyte.

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5380 mL of hydrogen chloride gas will produce 25.0 g of  BaCl₂ at STP.

To solve this problem, we need to use stoichiometry to determine the amount of hydrogen chloride gas produced from the given amount of BaCl₂.

First, we need to balance the chemical equation:

BaCl₂(s) + H₂SO₄(aq) → BaSO₄(aq) + 2HCl(g)

According to the balanced equation, 1 mole of BaCl₂ produces 2 moles of HCl.

The molar mass of BaCl₂ is 208.23 g/mol.

So, the number of moles of BaCl₂ in 25.0 g is:

n(BaCl₂) = mass ÷ molar mass = 25.0 g ÷ 208.23 g/mol = 0.120 mol

From the balanced equation, 1 mole of BaCl₂ produces 2 moles of HCl. Therefore, the number of moles of HCl produced is:

n(HCl) = 2 × n(BaCl₂) = 2 × 0.120 mol = 0.240 mol

At STP (standard temperature and pressure), 1 mole of any gas occupies 22.4 L.

Therefore, the volume of HCl gas produced at STP is:

V(HCl) = n(HCl) × 22.4 L/mol = 0.240 mol × 22.4 L/mol = 5.38 L

However, the question asks for the volume of hydrogen chloride gas in mL, so we need to convert the answer to mL:

1 L = 1000 mL

Therefore, V(HCl) = 5.38 L × 1000 mL/L = 5380 mL

So, 25.0 g of BaCl₂ at STP will produce 5380 mL of hydrogen chloride gas.

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The reaction between calcium nitrate (Ca(NO₃)₂) and sodium oxalate (Na₂C₂O₄) can be represented by the following chemical equation:

Ca(NO₃)₂ + Na₂C₂O₄ → CaC₂O₄ + 2NaNO₃

This is a double displacement reaction, where the calcium ion (Ca²⁺) from calcium nitrate and the oxalate ion (C₂O₄²⁻) from sodium oxalate switch places to form calcium oxalate (CaC₂O₄) and sodium nitrate (NaNO₃). The balanced chemical equation shows that one mole of calcium nitrate reacts with one mole of sodium oxalate to form one mole of calcium oxalate and two moles of sodium nitrate.

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The percentage composition of Hydrogen (H) in water (H₂O) is 11.1%

Percentage composition is finding the amount of individual elements that made up a compound.

To calculate, we use the formula:

% composition = [tex]\frac{Atomic Mass}{Molar Mass} x 100[/tex]

(a) Water: H₂O

To calculate the percentage composition of elements in water, we need to find the molar mass of water, which is:

Molar mass of H₂O = (2 × atomic mass of H) + (1 × atomic mass of O)

Molar mass of H₂O = (2 × 1.0 g/mol) + (1 × 16.00 g/mol)

Molar mass of H₂O = 18.0 g/mol

%Hydrogen = 2/18 x100 = 11.11

%Oxygen: 16/18 x 100 = 88.89%

b. Glucose: C₆H₁₂O₆

Molar mass of C₆H₁₂O₆ =  6 (12) + 12(1) + 6(16)

                                       = 72 + 12 + 96

                                       = 180g/mol

Now, we can calculate the percentage composition of elements in glucose:

% Carbon: 72/180 = 40%

% Hydrogen: 12/180 = 7%

Percentage composition of oxygen: (6 × atomic mass of O) / molar mass of C6H12O6 × 100%

Percentage composition of oxygen: (6 × 16.00 g/mol) / 180.18 g/mol × 100%

Percentage composition of oxygen: 53.29%

Therefore, the percentage composition of elements in glucose is 39.99% carbon, 6.72% hydrogen, and 53.29% oxygen.

c. Calcium nitrate: Ca(NO3)2

To calculate the percentage composition of elements in calcium nitrate, we need to find the molar mass of calcium nitrate, which is:

Molar mass of Ca(NO3)2 = atomic mass of Ca + (2 × atomic mass of N) + (6 × atomic mass of O)

Molar mass of Ca(NO3)2 = 40.08 g/mol + (2 × 14.01 g/mol) + (6 × 16.00 g/mol)

Molar mass of Ca(NO3)2 = 164.09 g/mol

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Answer:

To find the mass of sodium bicarbonate (NaHCO3) in a sample of Alka Seltzer that weighs 0.17 grams, we need to know the percentage of NaHCO3 in Alka Seltzer. This information is usually provided on the label of the product. Once we know the percentage of NaHCO3, we can use the following formula to calculate the mass of NaHCO3 in the sample:

mass of NaHCO3 = sample mass (in grams) x % NaHCO3 / 100

For example, if the percentage of NaHCO3 in Alka Seltzer is 50%, then the mass of NaHCO3 in a 0.17 gram sample would be:

mass of NaHCO3 = 0.17 x 50 / 100 = 0.085 grams

Therefore, the mass of sodium bicarbonate in the student's 0.17 gram sample of Alka Seltzer depends on the percentage of NaHCO3 in the Alka Seltzer, which should be provided on the product label.

Explanation:

The mass of the ammonia that is required is  258 g.

The quantitative correlations between the reactants and products in a chemical reaction are the focus of the chemistry subfield known as stoichiometry.

We have to know that;

1 mole of CO2 produces 1 mole of urea

2.2 moles of CO2 produces 2.2 urea

Given that the number of moles of urea = 455 g/60 g/mol

= 7.58 moles

Now;

2 moles of NH3 produces 1 mole of urea

x moles of NH3 produces 7.58 moles of urea

x = 7.58 * 2/1

= 15.16 moles

Mass of the ammonia =  15.16 moles * 17 g/mol

= 258 g

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Answer:

Explanation:

molar mass Cr(NO3)3 = 238 g/mol

Convert 325 ml to liters:  325 mls x 1 L / 1000 mls = 0.325 L

Convert 10.0 g to moles:  10.0 g x 1 mol / 238 g = 0.0420 moles

Molarity = moles/liters = 0.0420 moles / 0.325 L = 0.129 M (3 sig. figs.)

The reaction between sodium carbonate and hydrochloric acid

Na2CO3 + 2HCl = 2NaCl + CO2 + H2O

For reaction A

Mass of sodium bicarbonate used = (Mas of evaporating dish + watch glas + sodium bicarbonate) - (Mas of evaporating dish + watch glas)

= 46.582 - 46.263

= 0.319 g

Mass of sodium chloride = (mas of evaporating dish + watch glas + sodium chloride) - (Mas of evaporating dish + watch glas)

= 46.473 - 46.263

= 0.210 g

Moles of sodium bicarbonate (NaHCO3) used = mas/molecular weight

= (0.319 g) / (84 g/mol)

= 0.00380 mol

Moles of sodium chloride (NaCl) used = mas/molecular weight

= (0.210 g) / (58.44 g/mol)

= 0.00359 mol

Mol ratio of NaHCO3 : NaCl = 0.00380 : 0.00359

Divide by 0.00359

Simple mol ratio

NaHCO3 : NaCl = 1.06 : 1

After rounding

Mol ratio of NaHCO3 : NaCl = 1 : 1

Moles of NaHCO3 = moles of NaCl = 0.00359 mol

Theoretical yield of NaCl = moles x molecular weight

= 0.00359 mol x 58.44 g/mol

= 0.210 g

the percent yield of sodium chloride

= actual yield x 100 / theoretical yield

= 0.210*100/0.210

= 100%

the reaction between sodium bicarbonate and hydrochloric acid

NaHCO3 + HCl = NaCl + CO2 + H2O

For reaction B

Mass of sodium carbonate used = (Mas of evaporating dish + watch glas + sodium carbonate) - (Mas of evaporating dish + watch glas)

= 51.677 - 51.368

= 0.309 g

Mass of sodium chloride = (mas of evaporating dish + watch glas + sodium chloride) - (Mas of evaporating dish + watch glas)

= 51.671 - 51.368

= 0.303 g

Moles of sodium carbonate (Na2CO3) used = mas/molecular weight

= (0.309 g) / (106 g/mol)

= 0.00292 mol

Moles of sodium chloride (NaCl) used = mas/molecular weight

= (0.303 g) / (58.44 g/mol)

= 0.00518 mol

Mol ratio of

Na2CO3 : NaCl = 0.00292 : 0.00518

Divide by 0.00292

Simple mol ratio

Na2CO3 : NaCl = 1 : 1.78

After rounding

Mol ratio of Na2CO3 : NaCl = 1 : 2

Moles of NaCl = 2 x moles of Na2CO3

= 2 x 0.00292 = 0.00584 mol

Theoretical yield of NaCl = moles x molecular weight

= 0.00584 mol x 58.44 g/mol

= 0.341 g

the percent yield of sodium chloride

= actual yield x 100 / theoretical yield

= 0.303*100/0.341

= 88.86%

the reaction between sodium carbonate and hydrochloric acid

Na2CO3 + 2HCl = 2NaCl + CO2 + H2O

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