The correct answer is option B, that is, add 1.46 grams of NaCl to 250 milliliters of H₂O.
First there is a need to find the moles of NaCl in 250 ml of 0.10 M NaCl,
Moles of NaCl = molarity × volume = 0.10 M × (250/1000L) = 0.025 mol
The corresponding mass of NaCl is,
Mass of NaCl = moles × molar mass = 0.025 mol × 58.5 g/mol = 1.46 g
Thus, there is a need to dissolve 1.46 grams of NaCl solid into 50 ml of H₂O and dilute to 250 ml.
How many grams Nitrogen in 1 mol? The answer is 14.0067. We assume you are converting between grams Nitrogen and mole. You can view more details on each measurement unit: molecular weight of Nitrogen or mol The molecular formula for Nitrogen is N. The SI base unit for amount of substance is the mole. 1 grams Nitrogen is equal to 0.071394404106606 mole. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between grams Nitrogen and mole. Type in your own numbers in the form to convert the units!
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
density 15.4 grams per 1.2³ cm³ ≅ 8.9 grams per cm³
<span>Find the metal that has a density of approximately 8.9 g/cm³</span>
Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ
Explanation:
The balanced chemical reaction is,
The expression for enthalpy change is,
![\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactants}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Bn%5Ctimes%20H_f_%7Bproducts%7D%5D-%5Bn%5Ctimes%20H_f_%7Breactants%7D%5D)
Putting the values we get :
![\Delta H=[8\times H_f_{CO_2}+10\times H_f_{H_2O}]-[2\times H_f_{C_4H_{10}+13\times H_f_{O_2}}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B8%5Ctimes%20H_f_%7BCO_2%7D%2B10%5Ctimes%20H_f_%7BH_2O%7D%5D-%5B2%5Ctimes%20H_f_%7BC_4H_%7B10%7D%2B13%5Ctimes%20H_f_%7BO_2%7D%7D%5D)
![\Delta H=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.7)+(13\times 0)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%288%5Ctimes%20-393.5%29%2B%2810%5Ctimes%20-241.82%29%5D-%5B%282%5Ctimes%20-125.7%29%2B%2813%5Ctimes%200%29%5D)
2 moles of butane releases heat = 5314.8 kJ
1 mole of butane release heat = 
Thus enthalpy of combustion per mole of butane is -2657.4 kJ
