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How do you standardize 0.5 m hcl?
To standardize 0.5 M HCl, you would typically titrate it using a primary standard solution such as sodium carbonate (Na2CO3). By titrating a known volume of the HCl with the sodium carbonate solution and using the mole ratio between the two, you can calculate the exact concentration of the HCl solution. This process ensures that the concentration of the 0.5 M HCl is accurate for future use in experiments.
How many moles of HCL in 50ml of 1.0 M HCL?
50ml = .05L of HCL 1.0 M = 1mol / 1L of HCL simply multiply - .05 by 1.0, and get your answer!
What volume of a 1.50 M HCl solution should you use to prepare 2.00 L of a 0.100 M HCl solution?
To prepare a 0.100 M HCl solution from a 1.50 M HCl solution, you need to use the dilution formula, which is M1V1 = M2V2. You would need to use (V_1 = \frac{M_2V_2}{M_1}) to calculate the volume needed. Plugging in the values, you would need to use ( V_1 = \frac{0.100 M \times 2.00 L}{1.50 M} = 0.133 L or 133 mL) of the 1.50 M HCl solution.
How many moles of HCl are in 50 ml of 4.0 M HCl?
To find the moles of HCl, first calculate the millimoles of HCl in 50 mL: 4.0 mol/L * 50 mL = 200 mmol. Then convert millimoles to moles by dividing by 1000: 200 mmol / 1000 = 0.2 moles of HCl. Therefore, there are 0.2 moles of HCl in 50 mL of 4.0 M HCl.
What is the molarity of an hcl solution if 7ml hcl solution is titrated with 27.6ml of 0.170m of naoh solution?
The balanced chemical equation for the reaction is: HCl + NaOH -> NaCl + H2O. From the equation, it is a 1:1 mole ratio reaction. Therefore, the moles of HCl can be calculated from the volume and concentration of NaOH used in the titration. Then, use the moles of HCl and the volume of HCl solution used to calculate the molarity of the HCl solution.
How do you standardize 0.5 m hcl?
To standardize 0.5 M HCl, you would typically titrate it using a primary standard solution such as sodium carbonate (Na2CO3). By titrating a known volume of the HCl with the sodium carbonate solution and using the mole ratio between the two, you can calculate the exact concentration of the HCl solution. This process ensures that the concentration of the 0.5 M HCl is accurate for future use in experiments.
How many moles of HCL in 50ml of 1.0 M HCL?
50ml = .05L of HCL 1.0 M = 1mol / 1L of HCL simply multiply - .05 by 1.0, and get your answer!
What volume of a 1.50 M HCl solution should you use to prepare 2.00 L of a 0.100 M HCl solution?
To prepare a 0.100 M HCl solution from a 1.50 M HCl solution, you need to use the dilution formula, which is M1V1 = M2V2. You would need to use (V_1 = \frac{M_2V_2}{M_1}) to calculate the volume needed. Plugging in the values, you would need to use ( V_1 = \frac{0.100 M \times 2.00 L}{1.50 M} = 0.133 L or 133 mL) of the 1.50 M HCl solution.
How many moles of HCl are in 50 ml of 4.0 M HCl?
To find the moles of HCl, first calculate the millimoles of HCl in 50 mL: 4.0 mol/L * 50 mL = 200 mmol. Then convert millimoles to moles by dividing by 1000: 200 mmol / 1000 = 0.2 moles of HCl. Therefore, there are 0.2 moles of HCl in 50 mL of 4.0 M HCl.
What is the molarity of an hcl solution if 7ml hcl solution is titrated with 27.6ml of 0.170m of naoh solution?
The balanced chemical equation for the reaction is: HCl + NaOH -> NaCl + H2O. From the equation, it is a 1:1 mole ratio reaction. Therefore, the moles of HCl can be calculated from the volume and concentration of NaOH used in the titration. Then, use the moles of HCl and the volume of HCl solution used to calculate the molarity of the HCl solution.
What volume of 0.100 M HCl is required to completely react with ml of 0.161 M Na2CO3?
To solve this problem, we need to use the balanced chemical equation between HCl and Na2CO3. From the equation, we can see that it is a 1:2 ratio for HCl to Na2CO3. Therefore, we need twice the volume of 0.161 M Na2CO3 to react completely with HCl. Calculate the volume of HCl required by multiplying the volume of Na2CO3 by 2.
Calculate the molarity of HCl of pH 5.7?
To calculate the molarity of a solution from its pH, use the formula: pH = -log[H+]. In this case, pH 5.7 corresponds to [H+] = 1 x 10^-5.7 M. Given that HCl is a strong acid and dissociates completely in water, the molarity of HCl is also 1 x 10^-5.7 M.
65 ml of hcl is titrated with a solution of 0.15 m koh it requires 44 ml of koh to reach the equivalence point what is the concentration m of the hcl solution?
The balanced chemical equation for the reaction is HCl + KOH -> KCl + H2O. From the reaction, we see that the moles of HCl are equal to the moles of KOH at the equivalence point. Using the equation n = M x V where n is the number of moles, M is the molarity, and V is the volume in liters, we can calculate the concentration of the HCl solution to be 0.15 M.
What mass of caco3 is required to react with 100 ml of 2 m hcl solution?
To find the mass of CaCO3 required to react with 100 mL of 2 M HCl, you need to first calculate the number of moles of HCl using its molarity and volume. Then, use the balanced chemical equation to determine the mole ratio between HCl and CaCO3, allowing you to calculate the mass of CaCO3 needed.
Is 1 m HCl is more reactive than 4m HCl?
1 m HCl is not more reactive than 4m HCl, but 4m HCl is more concentrated.
What mass of calcium carbonate in grams can be dissolved by 3.9g of HCl?
To solve this, you have to be aware that this is a acid-base reaction, and that HCl is a gas, that usually is not applied in this form.However: Hydrochloric acid reacts with the calcium salt of carbonic acid, to form calcium chloride, water and (volatile) carbon dioxide.Thus, you must first calculate the moles (n) of H+ contained in 3.9g of HCl. 1 mole of HCl contains 1 mole of H+. So you can calculate:M(HCl) = M(H) + M(Cl) = 36.45g/molm(HCl) = 3.90gn(HCl) = m(HCl) / M(HCl) = 0.11molNext, you must calculate the moles of carbonate that can be dissolved.Using the following formula:CO32- + 2 H+ ↔ H2O + CO2↑you can see, that you need 2 moles of H+ for 1 mole of CO32-.Subsequently, you have to calculate the molar mass of calcium carbonate:M(CaCO3) = 40.08g/mol + 12.01g/mol + 3*16.00g/mol = 100.09g/molAnd finally, you can calculate the mass of calcium carbonate you can dissolve using 0.11mol HCL:m(CaCO3)= M(CaCO3) * [½ * n(HCl)] = 5.35gFrom the equations above, considering the molarities, we can draw a more dense formula that allows us to neglect the absolute molarities, so we only have to use the relative molarities. The equation can also be used to check if we calculated correctly)m(A)/(2*M(A)) = m(B)/M(B)We transpose to calculate m(B):m(B)= (½*m(A)/M(A))*M(B)and when we insert the values:m(CaCO3) = (0.50*(3.90g/36.45g/mol))*100.09g/mol = 5.35gAnd next time, you'll be able to do this by yourself ;)
How much HCL exists in 247 mL of 0.43 M solution of HCL?
To determine the amount of HCl in the solution, you would first calculate the number of moles of HCl present using the formula Molarity = moles/volume. Then, you would multiply the number of moles by the molar mass of HCl (36.46 g/mol) to find the mass.
