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If 3.5 L of a 5.0 M SrCl2 solution is diluted to 45 L what is the molarity of the diluted solution?
To find the molarity of the diluted solution, first calculate the number of moles of SrCl2 in the original solution: [ \text{moles} = \text{Molarity} \times \text{Volume} = 5.0 , \text{M} \times 3.5 , \text{L} = 17.5 , \text{moles} ] Next, divide the moles by the new volume: [ \text{Molarity} = \frac{\text{moles}}{\text{Volume}} = \frac{17.5 , \text{moles}}{45 , \text{L}} \approx 0.389 , \text{M} ] Thus, the molarity of the diluted solution is approximately 0.39 M.
If a gas has a volume of 1 L at a pressure of 270 kPa what volume will it have when the pressure is increased to 540 kPa Assume the temperature and number of particles are constant. Show your work. Ma?
To solve for the new volume when the pressure is increased, we can use Boyle's Law, which states that ( P_1 V_1 = P_2 V_2 ). Given: ( P_1 = 270 , \text{kPa} ) ( V_1 = 1 , \text{L} ) ( P_2 = 540 , \text{kPa} ) Plugging in the values: [ 270 , \text{kPa} \times 1 , \text{L} = 540 , \text{kPa} \times V_2 ] Solving for ( V_2 ): [ V_2 = \frac{270 , \text{kPa} \times 1 , \text{L}}{540 , \text{kPa}} = 0.5 , \text{L} ] Thus, the new volume will be 0.5 L.
What is the pressure exerted by 2.26 mol of gas in a 2.92 L container at 32C?
To find the pressure exerted by the gas, we can use the Ideal Gas Law, which is ( PV = nRT ). Here, ( n = 2.26 ) mol, ( R = 0.0821 , \text{L} \cdot \text{atm} / (\text{mol} \cdot \text{K}) ), and ( T = 32°C = 305 , \text{K} ). Plugging in the values, we get ( P = \frac{nRT}{V} = \frac{(2.26 , \text{mol}) \cdot (0.0821 , \text{L} \cdot \text{atm} / (\text{mol} \cdot \text{K})) \cdot (305 , \text{K})}{2.92 , \text{L}} \approx 18.1 , \text{atm} ).
What mass of calcium chloride is needed to prepare 2.85 L of a 1.56 M solution?
To calculate the mass of calcium chloride (CaCl₂) required for a 1.56 M solution, first use the formula: [ \text{mass (g)} = \text{molarity (mol/L)} \times \text{volume (L)} \times \text{molar mass (g/mol)} ] The molar mass of CaCl₂ is approximately 110.98 g/mol. Therefore, for 2.85 L of a 1.56 M solution: [ \text{mass} = 1.56 , \text{mol/L} \times 2.85 , \text{L} \times 110.98 , \text{g/mol} \approx 49.3 , \text{g} ] Thus, about 49.3 grams of calcium chloride is needed.
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To properly format an APA in-text citation, you need to include the author's last name and the publication year in parentheses within the text of your paper. For example, (Smith, 2019).
What is the pressure of 5.0 mol nitrogen (N2) gas in a 2.0 L container at 268 K?
To find the pressure of 5.0 mol of nitrogen gas in a 2.0 L container at 268 K, we can use the Ideal Gas Law, which is expressed as ( PV = nRT ). Here, ( n = 5.0 ) mol, ( R = 0.0821 , \text{L} \cdot \text{atm} / (\text{mol} \cdot \text{K}) ), and ( T = 268 , \text{K} ). Plugging in the values: [ P = \frac{nRT}{V} = \frac{(5.0 , \text{mol})(0.0821 , \text{L} \cdot \text{atm} / (\text{mol} \cdot \text{K}))(268 , \text{K})}{2.0 , \text{L}} \approx 55.2 , \text{atm}. ] Thus, the pressure is approximately 55.2 atm.
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Who exerts more pressure on the ground a 500n woman standing in high heels with a total area of 0.005 m cube or the same woman standing in work boots with a total area of 0.05 m cube calculate both?
To calculate the pressure exerted on the ground, we use the formula ( \text{Pressure} = \frac{\text{Force}}{\text{Area}} ). For the woman in high heels: [ \text{Pressure} = \frac{500 , \text{N}}{0.005 , \text{m}^2} = 100,000 , \text{Pa} ] For the woman in work boots: [ \text{Pressure} = \frac{500 , \text{N}}{0.05 , \text{m}^2} = 10,000 , \text{Pa} ] Thus, the woman in high heels exerts significantly more pressure on the ground than when she is in work boots.
How many grams of LiCl are required to make 2.0 of 0.65?
To calculate the grams of LiCl needed to make a 2.0 L solution at a concentration of 0.65 M, use the formula: [ \text{grams of solute} = \text{molarity} \times \text{volume (L)} \times \text{molar mass} ] The molar mass of LiCl is approximately 42.39 g/mol. Thus: [ \text{grams of LiCl} = 0.65 , \text{mol/L} \times 2.0 , \text{L} \times 42.39 , \text{g/mol} \approx 55.85 , \text{g} ] Therefore, approximately 55.85 grams of LiCl are required.
How many grams of solute are needed to prepare 2.5L of a 3.0M of AlNO33 solution?
To find the grams of solute needed, use the formula: [ \text{grams} = \text{molarity (M)} \times \text{volume (L)} \times \text{molar mass (g/mol)}. ] The molar mass of Al(NO₃)₃ is approximately 213.00 g/mol. For a 3.0 M solution in 2.5 L: [ \text{grams} = 3.0 , \text{mol/L} \times 2.5 , \text{L} \times 213.00 , \text{g/mol} = 1597.5 , \text{g}. ] Thus, you would need approximately 1597.5 grams of Al(NO₃)₃.
