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CaCO3 + HNO3 --> Ca(NO3)2 + <H2CO3>

<H2O + CO2>

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Lakes have been acidified by acid rain hno3 and h2so4 can be neutralized by a process called liming in which limestone caco3 is added to the acudified water what mass of limestone in kg would?

To calculate the mass of limestone (CaCO3) needed to neutralize HNO3 and H2SO4 in acidified water, you can use stoichiometry. First, determine the amount of acid in moles. Then, use the balanced chemical equation to find the moles of CaCO3 needed to neutralize the acid. Finally, convert the moles of CaCO3 to mass in kilograms using its molar mass.


Calcium carbonate plus nitric acid makes water and what else?

CaCO3 + 2 HNO3 =&gt; H2O + CO2 + Ca(NO3)2


What mass of limestone in kg would be required to completely neutralize a 15.4 billion-liter lake that is 1.9105 M in H2SO4 and 8.7106 M in HNO3?

To calculate the amount of limestone needed, first determine the moles of H2SO4 and HNO3 in the lake. Then, use the stoichiometry of the neutralization reaction between limestone (CaCO3) and the acids to calculate the moles of limestone required to neutralize the acids. Finally, convert the moles of limestone to kilograms using the molar mass of CaCO3.


What happens when H2O is added to CaCO3?

When H2O is added to CaCO3 (calcium carbonate), it can cause a chemical reaction where calcium carbonate may dissolve to form calcium ions (Ca2+) and carbonate ions (CO3 2-). This can result in the formation of a solution containing calcium ions and carbonate ions.


How do you prepare 0.01 M from 65 percent HNO3?

The molar mass of HNO3 is 63 gram/mol and of water is 18 gram/mol. Lets assume that the density of the HNO3 + water mixture is equal to the density of water at room temperature is 1000 gram/L. 0.01 M is equal to 0.01 mol HNO3 per liter HNO3 + water. Multiplying the molar mass of HNO3 by the molarity shows this is equal to 0.63 gram HNO3 per liter HNO3 + water. Using the density of water shows that this is equal to 0.000063 gram HNO3 per gram water. This is therefore 0.0065 percent by weight. (The assumption that the density of the mixture is approximately equal to that of pure water seems justified because the amount of HNO3 in the mixture is very low) Assuming we start with an initial 65 percent by weight of HNO3 (10.3 M using the same type of calculation), water will need to be added to dilute to 0.0065 percent (0.01M). To find the required amount of water that is added a total and component mass balance can be used. The total mass balance is given by: Mt = M0 + Ma Here, M0 is the initial amount of water + HNO3, Ma is the added amount of (pure) water and Mt is the total mass of the water + HNO3 after adding more water. The component mass balance over HNO3 is given by: xtMt = x0M0 + xaMa= x0M0 Where xt is the final weight fraction of HNO3 in water, x0 is the initial weight fraction in water and xa is the weight fraction of HNO3 in pure water (which is logically equal to zero). Lets say we have a starting mass M0 = 100 gram HNO3 + water with the initial weight fraction x0 = 0.65 gram HNO3 per gram and we want a final weight fraction xt = 0.000065, rearranging for the total weight after having added water to reach this dilution gives: Mt = x0M0/xt = 0.65x100/0.000065 = 1000000 gram HNO3 + water The amount of water which needs to be added then simply follows from: Ma = Mt - M0 = 1000000 - 100 = 999900 gram water

Related Questions

Lakes have been acidified by acid rain hno3 and h2so4 can be neutralized by a process called liming in which limestone caco3 is added to the acudified water what mass of limestone in kg would?

To calculate the mass of limestone (CaCO3) needed to neutralize HNO3 and H2SO4 in acidified water, you can use stoichiometry. First, determine the amount of acid in moles. Then, use the balanced chemical equation to find the moles of CaCO3 needed to neutralize the acid. Finally, convert the moles of CaCO3 to mass in kilograms using its molar mass.


What is thr Molecular equation of calcium carbonate and nitric acid?

The correct molecular equation of, calcium carbonate-CaCO3 nitric acid-HNO3


What is the balanced symbol equation for calcium carbonate plus nitric acid?

CaCO3 + HNO3. NO3^- and CO3^-2 are both polyatomic ions.


How much 2.5 molarity of HNO3 is required to react the CaCO3?

This depends on the mass of calcium carbonate.


Calcium carbonate plus nitric acid makes water and what else?

CaCO3 + 2 HNO3 =&gt; H2O + CO2 + Ca(NO3)2


When CaCo3 is added to Hcl what is formed?

Calcium chloride, Water, and Carbon Dioxide. Here is the BALANCED reaction equation. CaCO3(s) + 2HCl(aq) = CaCl2(aq) + H2O(l) + CO2(g). NB When writing chemical formulae, please use the correct symbols. CaCo3 means one calcium atom and three(3) COBALT atoms. whereas I think you mean 'CaCO3' , which means 1 x (Ca)Calcium, 1 x (C)Carbon and 3 x (O)Oxygen. Similarly, Hcl means nothing. , whereas it should be written as 1 x (H)Hydrogen and 1 x (Cl)Chlorine. Note the use of CAPITAL letters for single letter elemental symbols and as the first letter for two letter elemental symbols. Please see the Periodic Table for the correct form of writing elemental symbols.


What acid is added to aluminium to make salt?

Some examples are: HNO3, HF, HCl, H2SO4.


What mass of limestone in kg would be required to completely neutralize a 15.4 billion-liter lake that is 1.9105 M in H2SO4 and 8.7106 M in HNO3?

To calculate the amount of limestone needed, first determine the moles of H2SO4 and HNO3 in the lake. Then, use the stoichiometry of the neutralization reaction between limestone (CaCO3) and the acids to calculate the moles of limestone required to neutralize the acids. Finally, convert the moles of limestone to kilograms using the molar mass of CaCO3.


What happens when H2O is added to CaCO3?

When H2O is added to CaCO3 (calcium carbonate), it can cause a chemical reaction where calcium carbonate may dissolve to form calcium ions (Ca2+) and carbonate ions (CO3 2-). This can result in the formation of a solution containing calcium ions and carbonate ions.


How do you prepare 0.01 M from 65 percent HNO3?

The molar mass of HNO3 is 63 gram/mol and of water is 18 gram/mol. Lets assume that the density of the HNO3 + water mixture is equal to the density of water at room temperature is 1000 gram/L. 0.01 M is equal to 0.01 mol HNO3 per liter HNO3 + water. Multiplying the molar mass of HNO3 by the molarity shows this is equal to 0.63 gram HNO3 per liter HNO3 + water. Using the density of water shows that this is equal to 0.000063 gram HNO3 per gram water. This is therefore 0.0065 percent by weight. (The assumption that the density of the mixture is approximately equal to that of pure water seems justified because the amount of HNO3 in the mixture is very low) Assuming we start with an initial 65 percent by weight of HNO3 (10.3 M using the same type of calculation), water will need to be added to dilute to 0.0065 percent (0.01M). To find the required amount of water that is added a total and component mass balance can be used. The total mass balance is given by: Mt = M0 + Ma Here, M0 is the initial amount of water + HNO3, Ma is the added amount of (pure) water and Mt is the total mass of the water + HNO3 after adding more water. The component mass balance over HNO3 is given by: xtMt = x0M0 + xaMa= x0M0 Where xt is the final weight fraction of HNO3 in water, x0 is the initial weight fraction in water and xa is the weight fraction of HNO3 in pure water (which is logically equal to zero). Lets say we have a starting mass M0 = 100 gram HNO3 + water with the initial weight fraction x0 = 0.65 gram HNO3 per gram and we want a final weight fraction xt = 0.000065, rearranging for the total weight after having added water to reach this dilution gives: Mt = x0M0/xt = 0.65x100/0.000065 = 1000000 gram HNO3 + water The amount of water which needs to be added then simply follows from: Ma = Mt - M0 = 1000000 - 100 = 999900 gram water


What happens whenCaCO3 is mixed with H2O?

CaCO3 + 2H2O ==&gt; Ca(OH)2 + CO2 + H2O


What is the reaction of HNO3 when added to galactose?

When nitric acid (HNO3) is added to galactose, it may lead to the oxidation of galactose to form galactonic acid or mucic acid. This reaction is typically carried out under controlled conditions to avoid unwanted side reactions or degradation of the sugar molecule.