If your lawn is 21.0 ft wide and 20.0 ft long, and each square foot of lawn accumulates 1350 new snow flakes every minute. How much snow (in kilograms) accumulates on your lawn per hour? Assume an average snow flake has a mass of 1.60 mg.

Answer : The molarity of vitamin c in organic juice is, [tex]2.97\times 10^{-3}mole/L[/tex]

Solution : Given,

Mass of ascorbic acid (solute) = 124 mg = 0.124 g        [tex](1mg=0.001g)[/tex]

Volume of juice = 236.6 ml

Molar mass of ascorbic acid = 176.12 g/mole

Formula used :

[tex]Molarity=\frac{\text{Moles of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution}}[/tex]

Now put all the given values in this formula, we get the molarity of vitamin c in organic juice.

[tex]Molarity=\frac{0.124g\times 1000}{176.12g/mole\times 236.6ml}=2.97\times 10^{-3}mole/L[/tex]

Therefore, the molarity of vitamin c in organic juice is, [tex]2.97\times 10^{-3}mole/L[/tex]

Final answer:

The molarity of vitamin C in orange juice, where one cup contains 124 mg of ascorbic acid and is equal to 236.6 mL, is approximately 0.00297 M.

Explanation:

To calculate the molarity of vitamin C in orange juice, we can use the formula for molarity:

Molarity = moles of solute / liters of solution

First, we need to find the number of moles of ascorbic acid in one cup of orange juice.

1 cup of orange juice contains 124 mg of ascorbic acid. Since the molar mass of ascorbic acid, C6H8O6, is 176.12 g/mol, we convert the mass from mg to grams and then to moles:

124 mg * (1 g / 1000 mg) * (1 mol / 176.12 g) = 0.000704 mol

Next, we convert the volume from milliliters to liters:

236.6 mL * (1 L / 1000 mL) = 0.2366 L

Now, we can calculate the molarity:

Molarity = 0.000704 mol / 0.2366 L ≈ 0.00297 M

Thus, the molarity of vitamin C in the orange juice is approximately 0.00297 M.

Answer:

A. Tools are appropriate for the children in your setting and meet professional requirements for quality.

Explanation:

This is correct! C::

The room is approximately 0.0212 hectometers long.

To convert the length of the room from meters to hectometers, we need to know that 1 hectometer (hm) is equal to 100 meters (m). Therefore, to convert meters to hectometers, we divide the length in meters by 100.

Given that the room is 212 meters long, we perform the following calculation:

[tex]\[ \text{Length in hectometers} = \frac{\text{Length in meters}}{100} \][/tex]

[tex]\[ \text{Length in hectometers} = \frac{212 \text{ m}}{100} \][/tex]

[tex]\[ \text{Length in hectometers} = 2.12 \text{ hm} \][/tex]

However, to express this length in a more standard form, we can write 2.12 hectometers as 212/10000 hectometers, which simplifies to 212 hm divided by 100, or 2.12 hm. To be more precise, we can express this as a decimal:

[tex]\[ \text{Length in hectometers} = 0.0212 \text{ hm} \][/tex]

Thus, the room is approximately 0.0212 hectometers long.

The solubility of nitrogen in a diver's blood at a pressure of 2.85 atm and a temperature of 25.0 ∘c is 1.78 x 10^-3 moles per litre, according to Henry's law.

In this context, we will use Henry's law to calculate the solubility of N2 in a diver's blood. According to Henry's law, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. The equation is: C = P * k, where C is the solubility of the gas in the liquid, P is the partial pressure of the gas, and k is Henry's law constant.

In the given situation, the total pressure is 2.85 atm (which is the sum of the partial pressures of all gases), and the value of Henry's law constant (k) for N2 is 6.26×10−4mol/(l⋅atm). So, the solubility of nitrogen can be calculated by multiplying the pressure by Henry's law constant: C = 2.85 atm * 6.26×10−4mol/(l⋅atm) = 1.78 x 10^-3 mol/L.

Therefore, at a pressure of 2.85 atm and a temperature of 25.0 ∘c, the solubility of nitrogen in a diver's blood is 1.78 x 10^-3 moles per litre.

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After a balanced neutralization reaction between Sr(OH)2 and HNO3, the unused Sr(OH)2 determines the solution is basic.

To address this question, let's first write down the balanced equation for the reaction between strontium hydroxide (Sr(OH)2) and nitric acid (HNO3) which is:

Sr(OH)2 (aq) + 2HNO3 (aq) → Sr(NO3)2 (aq) + 2H2O (l)

Now let's calculate the moles of HNO3 added:

Volume of HNO3 = 55.0 mL = 0.055 L

Concentration of HNO3 = 0.200 M

Moles of HNO3 = Volume × Concentration = 0.055 L × 0.200 M = 0.011 mol

Next, calculate the moles of Sr(OH)2:

Mass of Sr(OH)2 = 15.0 g

Molar mass of Sr(OH)2 = 121.63 g/mol (approximately)

Moles of Sr(OH)2 = Mass ÷ Molar mass = 15.0 g ÷ 121.63 g/mol = 0.123 mol

According to the stoichiometry of the balanced equation, we need twice as many moles of HNO3 to react completely with Sr(OH)2. In this scenario, we have excess Sr(OH)2 (0.123 mol) compared to HNO3 (0.011 mol). Hence, all of the HNO3 will react, leaving some Sr(OH)2 unreacted.

After reaction, moles of Sr(OH)2 remaining = 0.123 mol - (0.011 mol × 1/2) = 0.1175 mol

The concentration of remaining Sr(OH)2 can be calculated by assuming the final volume is the sum of the volumes of the solutions mixed, which is an approximation for dilute solutions.

Since all the HNO3 has reacted, the resulting solution will be basic due to the excess Sr(OH)2 remaining.

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(a) The balanced equation for the reaction between the solutes is: [tex]\text{Sr(OH)}_2(aq) + 2\text{HNO}_3(aq) \rightarrow \text{Sr(NO}_3\text{)}_2(aq) + 2\text{H}_2\text{O}(l)[/tex]. (b) The concentrations of ions are: Sr²⁺ is 2.136 M, NO₃⁻ is 0.2 M, and OH⁻ is 4.272 M. (c) Resultant solution is basic in nature.

To solve the given problem, let's follow the steps one by one:

(a) Write a balanced equation for the reaction that occurs between the solutes.

Strontium hydroxide Sr(OH)₂ reacts with nitric acid HNO₃ to form strontium nitrate Sr(NO₃)₂ and water:

[tex]\text{Sr(OH)}_2(aq) + 2\text{HNO}_3(aq) \rightarrow \text{Sr(NO}_3\text{)}_2(aq) + 2\text{H}_2\text{O}(l)[/tex]

(b) Calculate the concentration of each ion remaining in solution.

1. Determine the moles of each reactant:

- Moles of Sr(OH)₂:

[tex]\text{Molar mass of Sr(OH)}_2 = 87.62 \, (\text{Sr}) + 2 \times 16.00 \, (\text{O}) + 2 \times 1.01 \, (\text{H}) = 121.64 \, \text{g/mol} \\\\\text{Moles of Sr(OH)}_2 = \frac{15.0 \, \text{g}}{121.64 \, \text{g/mol}} \approx 0.123 \, \text{mol}[/tex]

- Moles of HNO₃:

[tex]\text{Molarity of HNO}_3 = 0.200 \, \text{M} \\\\\text{Volume of HNO}_3 = 55.0 \, \text{mL} = 0.055 \, \text{L} \\\\\text{Moles of HNO}_3 = 0.200 \, \text{mol/L} \times 0.055 \, \text{L} = 0.011 \, \text{mol}[/tex]

2. Determine the limiting reactant:

According to the balanced equation, 1 mole of Sr(OH)₂ reacts with 2 moles of HNO₃. Therefore, the reaction requires:

[tex]\text{Moles of HNO}_3 \text{ required} = 0.123 \, \text{mol Sr(OH)}_2 \times 2 = 0.246 \, \text{mol}[/tex]

Since we only have 0.011 moles of HNO₃, it is the limiting reactant.

3. Calculate the remaining moles of Sr(OH)₂:

[tex]\text{Moles of Sr(OH)}_2 \text{ reacted} = \frac{0.011 \, \text{mol HNO}_3}{2} = 0.0055 \, \text{mol} \\\\\text{Remaining moles of Sr(OH)}_2 = 0.123 \, \text{mol} - 0.0055 \, \text{mol} = 0.1175 \, \text{mol}[/tex]

4. Determine the concentrations of ions in the solution:

- Volume of the final solution = volume of HNO₃ + volume of water from Sr(OH)₂ dissolution (approximately equal to volume of water added):

Assuming the solution volume remains approximately 55.0 mL + a negligible volume from Sr(OH)₂, the final volume is roughly  0.055 L.

- Concentration of Sr²⁺ ions:

Since only 0.0055 moles of Sr(OH)₂ reacted to form Sr(NO₃)₂, 0.1175 moles of Sr(OH)₂ remain, giving the concentration of Sr²⁺:

[tex]\text{Concentration of Sr}^{2+} = \frac{0.1175 \, \text{mol}}{0.055 \, \text{L}} \approx 2.136 \, \text{M}[/tex]

- Concentration of NO₃⁻ ions:

All HNO₃ dissociates, producing:

[tex]\text{Concentration of NO}_3^{-} = \frac{0.011 \, \text{mol}}{0.055 \, \text{L}} = 0.2 \, \text{M}[/tex]

- Concentration of OH⁻ ions:

From Sr(OH)₂, OH⁻ concentration:

[tex]\text{OH}^{-} \text{ from Sr(OH)}_2: \text{2 moles OH}^{-}\text{ per mole of Sr(OH)}_2 \\\\\text{Concentration of OH}^{-} = 2 \times \frac{0.1175 \, \text{mol}}{0.055 \, \text{L}} \approx 4.272 \, \text{M}[/tex]

(c) Is the resultant solution acidic or basic?

To determine if the solution is acidic or basic, we compare the concentrations of H⁺ and OH⁻. Since HNO₃ (a strong acid) was completely neutralized and excess Sr(OH)₂ (a strong base) remains, the solution will be basic due to the presence of significant OH⁻ ions.

Summary

- Balanced equation:

[tex]\text{Sr(OH)}_2(aq) + 2\text{HNO}_3(aq) \rightarrow \text{Sr(NO}_3\text{)}_2(aq) + 2\text{H}_2\text{O}(l)[/tex]

- Concentrations of ions in solution:

[tex]\text{Sr}^{2+}: 2.136 \, \text{M} \\\\ \text{NO}_3^{-}: 0.2 \, \text{M} \\\\ \text{OH}^{-}: 4.272 \, \text{M} \\\\[/tex]

- Resultant solution is basic due to the presence of excess OH⁻ ions.

Answer:

The answer is B.

Explanation:

The term that name the water in a chemical reaction that water  decomposes is.

  reactant ( answer D)

Explanation

Reactant is a substance  that take part in and undergoes change  during a reaction.   In other  words reactant  are consumed to form product.

water  is   reactant when it decomposes because  it is consumed  to form  hydrogen and oxygen.

The decomposition  of water  is as below  equation,

2H2O→ 2 H2 + O2

The molar mass of element X is determined in g/mol by taking into account the mole ratio in the X₂O₃ compound. The mass of X (1.0000 g) is first divided by the moles of X, which is two-thirds of the moles of Oxygen, calculated by dividing the mass of Oxygen in X₂O₃ by its molar mass.

To calculate the molar mass of element X, the mole ratio in the X₂O₃compound needs to be considered. This mole ratio suggests that for every 1 mole of X, 1.5 moles of Oxygen is used given that from the formula of X₂O₃hat would be 2 moles of X for every 3 moles of oxygen (2:3).

The mass of Oxygen in the compound can be calculated by subtracting the mass of X from the total compound weight. Given that 1.1xxx grams of X₂O₃ was obtained from 1.0000 grams of X, the mass of Oxygen in the compound would be 1.1xxx g - 1.0000 g giving a mass of Oxygen in the compound as 0.1xxx grams.

We know that the molar mass of Oxygen (O) is 16.00 g/mol. So, the number of moles of Oxygen in the compound can be calculated as (0.1xxx g)/(16.00 g/mol) which gives the number of moles for Oxygen. Given the mole ratio from the X₂O₃ formula, the number of moles of X will be two-thirds of the moles of Oxygen.

Finally, the molar mass of X can be calculated by dividing the mass of X (1.0000 g) by the number of moles of X calculated. This will give the answer for the molar mass of X in g/mol.

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Answer:

C

Explanation:

Answer is: the reducing agent is C₂H₅OH.

Balanced chemical reaction:

16H⁺ + 2Cr₂O₇²⁻ + C₂H₅OH → 4Cr³⁺ + 11H₂O + 2CO₂.

Oxidation half reaction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O/ ×2.

2Cr₂O₇²⁻ + 28H⁺ + 12e⁻ → 4Cr³⁺ + 14H₂O.

Reduction half reaction: C₂H₅OH + 3H₂O → 2CO₂ + 12H⁺ + 12e⁻.

Net reaction: 2Cr₂O₇²⁻ + 28H⁺ + C₂H₅OH + 3H₂O → 4Cr³⁺ + 14H₂O + 2CO₂ + 12H⁺.

Reducing agent is element or compound who loose electrons in chemical reaction. Ethanol (C₂H₅OH) lost electrons and it is oxidized to carbon dioxide (CO₂).

The reducing agent in the reaction is [tex]\boxed{\text{C}_{2}\text{H}_{5}\text{OH}}[/tex].

Further Explanation:

Redox reaction:

Redox is a term that is used collectively for the reduction-oxidation reaction. It is a type of chemical reaction in which the oxidation states of atoms are changed. In this reaction, both reduction and oxidation are carried out simultaneously. Such reactions are characterized by the transfer of electrons between the species involved in the reaction.

The process of gain of electrons or the decrease in the oxidation state of the atom is called reduction while that of loss of electrons or the increase in the oxidation number is known as oxidation. In redox reactions, one species lose electrons and the other species gain electrons. The species that lose electrons and itself gets oxidized is called as a reductant or reducing agent. The species that gains electrons and gets reduced is known as oxidant or oxidizing agent. The presence of redox pair or redox couple is a must for the redox reaction.

The general representation of a redox reaction is,

[tex]\text{X}+\text{Y}\rightarrow\text{X}^{+}+\text{Y}^{-}[/tex]

The oxidation half-reaction can be written as:

[tex]\text{X}\rightarrow\text{X}^{+}+e^{-}[/tex]  

The reduction half-reaction can be written as:

[tex]\text{Y}+e^{-}\rightarrow\text{Y}^{-}[/tex]

Here, X is getting oxidized and its oxidation state changes from  to +1 whereas B is getting reduced and its oxidation state changes from 0 to -1. Hence, X acts as the reducing agent, whereas Y is an oxidizing agent.

Rules to calculate the oxidation states of elements:

1. The oxidation number of free element is always zero.

2. The oxidation number of oxygen is generally taken as -2, except for peroxides.

3. The oxidation state of hydrogen is normally taken as +1.

4. The sum of oxidation numbers of all the elements present in a neutral compound is zero.

5. The oxidation numbers of group 1 and group 2 elements are +1 and +2 respectively.

The oxidation state of O is -2.

The expression to calculate the oxidation number in [tex]\text{Cr}_{2}\text{O}_{7}^{2-}[/tex] is:

[tex]\left[2(\text{Oxidation state of Cr})+7(\text{oxidation state of O})\right]=-2[/tex]     ...... (1)

Rearrange equation (1) for oxidation number of Cr

[tex]2(\text{Oxidation number of Cr})=\left[(-2)-7(\text{oxidation number of O})\right][/tex]                  …… (2)

Substitute -2 for oxidation state of O in equation (2).

[tex]\begin{aligned}\text{Oxidation state of Cr}&=\dfrac{[(-2)-7(-2)]}{2}\\&=\dfrac{[-2+14]}{2}\\&=+6\end{aligned}[/tex]

The charge on Cr is +3 so its oxidation state is also +3. The oxidation state of Cr changes from +6 to +3. This shows it decreases during the reaction and therefore Cr is an oxidizing agent.

The oxidation state of O is -2 and the oxidation state of H is +1.

The expression to calculate the oxidation in [tex]\text{C}_{2}\text{H}_{5}\text{OH}[/tex] is:

[tex]\left[2(\text{oxidation state of C})+1(oxidation state of O)+6(oxidation state of H)\right]=0[/tex]   ...... (3)

Rearrange equation (3) for oxidation state of C

[tex]2(\text{oxidation state of C})=\left[-1(oxidation state of O)-6(oxidation state of H)\right][/tex]       ...... (4)

Substitute -2 for oxidation state of O and +1 for oxidation state of H in equation (4).

[tex]\begin{aligned}\text{Oxidation state of C}&=\dfrac{[-1(-2)-6(+1)]}{2}\\&=\dfrac{[+2-6]}{2}\\&=-2\end{aligned}[/tex]

The oxidation state of O is -2.

The expression to calculate the oxidation state in [tex]\text{CO}_{2}[/tex] is:

[tex]\left[(\text{oxidation state of C})+2(oxidation state of O)\right]=0[/tex]   ...... (5)

Rearrange equation (5) for oxidation state of C

[tex](\text{oxidation state of C})=\left[-2(oxidation state of O)\right][/tex]       ...... (6)

Substitute -2 for oxidation state of O in equation (6).

[tex]\begin{aligned}\text{Oxidation state of C}&=[-2(-2)]\\&=+4\end{aligned}[/tex]

The oxidation state of C changes from -2 to +4. This shows it increases during the reaction and therefore [tex]\textbf{C}_{2}\textbf{H}_{5}\textbf{OH}[/tex] acts as a reducing agent.

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Answer details:

Grade: High School

Subject: Chemistry

Chapter: Redox reactions

Keywords: redox reaction, C2H5OH, Cr, C, CO2, Cr2O72-, oxidation, reduction, reductant, oxidant, reducing agent, oxidizing agent, electrons, redox pair, redox couple, oxidation state, oxidized, reduced, simultaneously.

Answer:

Species

Explanation:

The acronym, DKPCOFGS; Gives you an explanation of the subgroups highest - lowest.

Did

King

Phil

Come

Over

For

Good

Spagettii

Spagettii is species in the acronym

Enjoy your test :D

-Snooky

Answer:      c. 373 K

Explanation:

The percentage by mass of calcium carbonate present in rock is [tex]\boxed{{\mathbf{77}}{\mathbf{.62 \% }}}[/tex].

Further Explanation:

First, we have to find the excess number of moles of HCl acid that are neutralized by NaOH.

The number of moles of NaOH in 11.56 ml of 1.010 M NaOH solution is calculated as follows:

[tex]\begin{aligned}{\text{Number of moles of NaOH}}\left({{\text{mol}}}\right)&={\text{Concentration }}\left( {{\text{mol/L}}}\right) \times{\text{Volume }}\left({\text{L}} \right)\\&= 1.010{\text{ mol/L}}\left({11.56{\text{ ml}}\times \frac{{1{\text{ L}}}}{{1000{\text{ml}}}}}\right)\\&=0.011676{\text{ mol}}\\\end{aligned}[/tex]

The balanced chemical reaction of NaOH and HCl is as follows:

[tex]{\text{NaOH}}\left({aq}\right)+{\text{HCl}}\left({aq}\right)\to{\text{NaCl}}\left({aq} \right) +{{\text{H}}_{\text{2}}}{\text{O}}\left( l \right)[/tex]

Since NaOH and HCl are reacted in 1:1 ratio, therefore, the excess number of moles of HCl are equal to number of moles of NaOH that is 0.011676 mol.

Now, we have to find how many moles of HCl initially reacted with limestone.

The initial number of moles of HCl in 30.00 ml of 1.035 M HCl solution is calculated as follows:

[tex]\begin{aligned}{\text{Number of moles of HCl}}\left( {{\text{mol}}}\right)&={\text{Concentration }}\left( {{\text{mol/L}}}\right) \times {\text{Volume }}\left({\text{L}} \right)\\&= 1.035{\text{ mol/L}} \times \left( {30.00{\text{ ml}}\times \frac{{1{\text{ L}}}}{{1000\,{\text{ml}}}}}\right)\\&=0.03105{\text{ mol}}\\\end{aligned}[/tex]

Therefore, the number of moles of HCl initially reacted with limestone is calculated as follows:

[tex]\begin{aligned}{\text{Number of moles of HCl reacted with CaC}}{{\text{O}}_3}&=\left({0.03105{\text{ mol}} - 0.011676{\text{ mol}}}\right)\\&={\text{0}}{\text{.019374 mol of HCl}}\\\end{aligned}[/tex]

Therefore, the number of moles of HCl initially reacted with limestone is 0.019374 mol.

The balanced chemical equation for the reaction of limestone [tex]\left( {{\text{CaC}}{{\text{O}}_{\text{3}}}}\right)[/tex] with HCl is as follows:

[tex]{\text{CaC}}{{\text{O}}_3}\left( s \right) + 2{\text{HCl}}\left( {aq} \right)\to{\text{C}}{{\text{O}}_{\text{2}}}\left( g \right) + {{\text{H}}_{\text{2}}}{\text{O}}\left( l \right) + {\text{CaC}}{{\text{l}}_2}\left( {aq} \right)[/tex]

The balanced chemical equation shows that 1 mole of [tex]{\text{CaC}}{{\text{O}}_3}[/tex] reacted with 2 moles of HCl to neutralize the reaction completely, therefore, the number of moles of  [tex]{\text{CaC}}{{\text{O}}_3}[/tex] neutralized by 0.019374 mol of HCl are calculated as follows:

[tex]\begin{aligned}{\text{Amount of CaC}}{{\text{O}}_3}\left( {{\text{mol}}}\right)&= \left( {{\text{0}}{\text{.019374 mol of HCl}}}\right)\left({\frac{{1{\text{ mol CaC}}{{\text{O}}_3}}}{{{\text{2 mol HCl}}}}}\right)\\&=0.009687{\text{ mol CaC}}{{\text{O}}_3}\\\end{aligned}[/tex]

The molar mass of [tex]{\text{CaC}}{{\text{O}}_3}[/tex]  is 100.0 g/mol.

Mass of 0.009687 mol of [tex]{\text{CaC}}{{\text{O}}_3}[/tex] is calculated as follows:

[tex]\begin{aligned}{\text{Mass}}\left( {\text{g}}\right)&={\text{Number of moles}} \times{\text{Molarmass}}\left({{\text{g/mol}}}\right)\\&=0.009687{\text{mol}}\times{\text{100}}{\text{.0 g/mol}}\\&=0.9687{\text{g}}\\\end{aligned}[/tex]

The percentage by mass can be calculated as follows:

[tex]\begin{aligned}{\text{Percent by mass}}\left( \%\right)&=\frac{{{\text{Mass of CaC}}{{\text{O}}_3}}}{{{\text{Mass of lime stone}}}}\times 100\\&=\frac{{0.9687{\text{ g}}}}{{1.248{\text{ g}}}}\times 100\\&=77.62{\text{ }}\%\\\end{aligned}[/tex]

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Mole concept

Keywords: Percentage by mass, calcium carbonate in rock, number of moles of HCl, excess number of moles, CaCO3, balance chemical equation, limestone.

Final answer:

To calculate the percent by mass of calcium carbonate in the limestone, we calculate the moles of HCl that reacted with the limestone, subtract the moles neutralized by NaOH, convert this to grams of CaCO₃, and divide by the sample mass.

Explanation:

To calculate the percent by mass of calcium carbonate in the limestone rock, we need to find out how much of the HCl was used to react with the calcium carbonate and not neutralized by NaOH, and then convert this amount to grams of CaCO3.

We start by calculating the number of moles of NaOH that reacted with the excess HCl:

Number of moles of NaOH = Volume (L) × Molarity (M)

Number of moles of NaOH = 0.01156 L × 1.010 M = 0.0116776 mol

Since the reaction between NaOH and HCl is 1:1, the moles of HCl that were neutralized by NaOH are also 0.0116776 mol.

Now we can calculate the moles of HCl that reacted with the CaCO3:

Total moles of HCl initially = Volume (L) × Molarity (M)

Total moles of HCl = 0.03000 L × 1.035 M = 0.03105 mol

Moles of HCl that reacted with CaCO₃ = Total moles of HCl - Moles of HCl neutralized by NaOH

Moles of HCl that reacted with CaCO₃ = 0.03105 mol - 0.0116776 mol = 0.0193724 mol

The reaction between CaCO₃ and HCl is also 1:1:

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

Moles of CaCO₃ = Moles of HCl that reacted with CaCO₃ = 0.0193724 mol

To find the mass of CaCO₃ we multiply the moles of CaCO₃ by the molar mass of CaCO₃ (100.09 g/mol):

Mass of CaCO₃ = Moles of CaCO₃ × Molar Mass of CaCO₃

Mass of CaCO₃ = 0.0193724 mol × 100.09 g/mol = 1.93936 g

Finally, we calculate the percent by mass of CaCO₃ in the rock:

Percent by mass of CaCO₃ = (Mass of CaCO₃ / Mass of Rock Sample) × 100%

Percent by mass of CaCO₃ = (1.93936 g / 1.248 g) × 100% = 155.413%

However, a percent mass over 100% indicates an error in the calculation as it's not possible to have more calcium carbonate than the total mass of the rock. It likely means that we must take into account other substances in the limestone that might react with HCl or a calculation error.

297440787 tons of carbon dioxide is produced by the plant after a combustion reaction taking place.

Reaction which takes place in presence of air is combustion reaction.When carbon burns in air it produces carbon dioxide .

C+O₂[tex]\rightarrow[/tex]CO₂

80.7% mass of carbon means 80.7/100×12=9.684 g

12 g C is present in 44 g carbon dioxide

∴9.684 g of C  is present in 9.684×44/12=35.508 g

As 9.6854 g carbon/coal produces 35.508 g carbon dioxide ,

∴35.508×8376726=297440787 tonnes of carbon dioxide.

Combustion reactions are reactions which take place at high temperatures and are exothermic and produce products which are oxidized.In combustion reactions ,chemical equilibrium is difficult to achieve. There are 3 types of combustion:

1) rapid combustion

2) explosive combustion

3) spontaneous combustion

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Answer: The weight of the mouse on the Moon is 0.036 N

Explanation:

Weight is defined as the force exerted by the body on any surface. It is also defined as the product of mass of the body multiplied by the acceleration due to gravity.

Mathematically,

[tex]W=mg[/tex]

where,

W = weight of the mouse

m = mass of the mouse = [tex]22g=0.022kg[/tex]

g = acceleration due to gravity on moon = [tex]1.625m/s^2[/tex]

Putting values in above equation, we get:

[tex]W=0.022kg\times 1.625 m/s^2=0.036N[/tex]

Hence, the weight of the mouse on the Moon is 0.036 N

In the given question, Insulin, adrenaline, and estrogen are examples of hormones. Each hormone has a specific function and plays an important role in maintaining homeostasis in the body.

Insulin is a hormone made up of 51 amino acids and is released by the beta cells of pancreas. It helps in regulating blood glucose level.

Adrenaline is also known as epinephrine. It is produced by the adrenal glands and is involved in the body's "fight or flight" response to stress.

Estrogen is a female sex hormone produced by the ovaries and is involved in the development of female secondary sexual characteristics, such as breast development and the menstrual cycle.

Therefore, examples of hormones are insulin, adrenaline, and estrogen, which are chemical messengers that regulate many physiological processes in the body.

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