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How many m in 146cm?
1.46 m 100cm = 1 m
How long is 1 feet in meters?
A search on Google for "1 meter in feet" gives 3.2808399 feet.Direct Conversion Formula 1 m*1 ft0.3048 m=3.280839895 ft
How many feet is 2.5 meters?
82.02 ft 1 m = 3.28 ft 1 ft = 0.3048 m
How many inches are in 1 m?
1 m = 1 / .0254 inches = 39.3701 inches
20 meters is how many feet?
6.096 m Algebraic Steps / Dimensional Analysis Formula 20 ft*12 in 1 ft*2.54 cm 1 in*1 m 100 cm=6.096 m Direct Conversion Formula 20 ft*0.3048 m 1 ft=6.096 m
Calculate 0.5 M HCL?
0.5 M HCl means the concentration of hydrochloric acid is 0.5 moles per liter of solution. To calculate this, you need to use the formula: moles = molarity x volume (in liters). So, if you have, for example, 1 liter of 0.5 M HCl solution, you would have 0.5 moles of HCl in that solution.
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.
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.
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 can 1 M of HCl be prepared from 32 percent HCl concentration?
percentage = well, mostly weight percentage is the most commonly given percentage density = the density of a sol'n is constant, for HCl, the value is 1.18 g/ mL now assuming you have 34% of HCl in the sol'n in preparing 1M HCl, calculate the mass of HCl Needed. assuming there are 100g of the whole solution, therefore there will be 34g of HCl: percent by weight = (34% x 100g) / 100g = 34g now to get the mass, we need dimensional analysis. 34 g HCl x 1 mol HCl x 1 L sol'n x 1.18 g HCl x 1 mL = FINAL ANSWER 36.45 g HCl 1 mol HCl 1 mL 1 L sol'n you just get the final answer.... percentage = well, mostly weight percentage is the most commonly given percentage density = the density of a sol'n is constant, for HCl, the value is 1.18 g/ mL now assuming you have 34% of HCl in the sol'n in preparing 1M HCl, calculate the mass of HCl Needed. assuming there are 100g of the whole solution, therefore there will be 34g of HCl: percent by weight = (34% x 100g) / 100g = 34g now to get the mass, we need dimensional analysis. 34 g HCl x 1 mol HCl x 1 L sol'n x 1.18 g HCl x 1 mL = FINAL ANSWER 36.45 g HCl 1 mol HCl 1 mL 1 L sol'n you just get the final answer....
Why you use 85 ml Conc HCl for 1 M HCl?
To prepare 1 M hydrochloric acid (HCl) solution, you usually need to dilute concentrated HCl solution (~35-37% HCl) with water in a specific ratio. The specific volume of concentrated HCl needed depends on its concentration and the final volume of the 1 M solution you want to make. In this case, 85 ml of concentrated HCl is likely to be the volume needed to make 1 L of 1 M 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.
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 1 M NaHCO3 solution does it take to fully react with 5 ml of a 1 M solution of HCl?
The answer is 5 m L sodium bicarbonate, 1 M solution.
What is the pH of a 0.1 mole per liter HCl solution?
The pH of a 0.1 M HCl solution is 1. HCl is a strong acid that completely dissociates in water to form H+ ions. Thus, the concentration of H+ ions in a 0.1 M HCl solution is 0.1 M, resulting in a pH of 1.
How do you prepare a 0.0001 M soulution HCL using 0.1 M of HCl?
you need to make 1000 times dilutions, this could be done in multi-steps: transfer 1 ml 0.1 M HCl into 100 ml volumetric flask and complete volume with water --------(1) from solution (1) transfer 2 ml into 20 ml volumetric falsk and complete volume with water, this is 0.0001 M HCL.
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.
