Answer:
liquid
Explanation:
the state of matter is liquid
Answer:
identical conditions, separate samples of O2 and an unknown gas were allowed to effuse through identical membranes simultaneously. After a certain amount of time, it was found that 6.23 mL of O2 had passed through the membrane, but only 3.85 mL of of the unknown gas had passed through. What is the molar mass of the unknown gas
identical conditions, separate samples of O2 and an unknown gas were allowed to effuse through identical membranes simultaneously. After a certain amount of time, it was found that 6.23 mL of O2 had passed through the membrane, but only 3.85 mL of of the unknown gas had passed through. What is the molar mass of the unknown gas
Explanation:
identical conditions, separate samples of O2 and an unknown gas were allowed to effuse through identical membranes simultaneously. After a certain amount of time, it was found that 6.23 mL of O2 had passed through the membrane, but only 3.85 mL of of the unknown gas had passed through. What is the molar mass of the unknown gas
The balanced equation for the given titration reaction can be written as
![2 HNO_{3}(aq)+ Ca(OH)_{2}(aq)\rightarrow Ca(NO_{3})_{2}+ 2 H_{2}O](https://tex.z-dn.net/?f=%202%20HNO_%7B3%7D%28aq%29%2B%20Ca%28OH%29_%7B2%7D%28aq%29%5Crightarrow%20Ca%28NO_%7B3%7D%29_%7B2%7D%2B%202%20H_%7B2%7DO%20)
Step 1 : Find moles of Ca(OH)₂
Volume of Ca(OH)₂ = 20 mL
Concentration of Ca(OH)₂ = 5.00 x 10⁻³ M
Moles of Ca(OH)₂ = ![20 mL * \frac{1L}{1000 mL} * \frac{5.00*10^{-3}mol}{L}](https://tex.z-dn.net/?f=%2020%20mL%20%2A%20%5Cfrac%7B1L%7D%7B1000%20mL%7D%20%2A%20%5Cfrac%7B5.00%2A10%5E%7B-3%7Dmol%7D%7BL%7D%20%20%20)
Moles of Ca(OH)₂ = 1.00 x 10⁻⁴
Step 2 : Find moles of HNO3 using mole ratio from balanced equation
The mole ratio of HNO₃ and Ca(OH)₂ is 2 : 1
Let us use this as a conversion factor to find moles of HNO₃.
Moles of HNO₃ = ![1 * 10^{-4} mol (Ca(OH)_{2} ) * \frac{2mol(HNO_{3})}{1mol(Ca(OH)_{2})}](https://tex.z-dn.net/?f=%201%20%2A%2010%5E%7B-4%7D%20mol%20%28Ca%28OH%29_%7B2%7D%20%29%20%2A%20%5Cfrac%7B2mol%28HNO_%7B3%7D%29%7D%7B1mol%28Ca%28OH%29_%7B2%7D%29%7D%20%20%20%20)
Moles of HNO₃ = 2.00 x 10⁻⁴
Step 3 : Find concentration of HNO3
Concentration of HNO3 can be calculated as
![[HNO_{3} ] = \frac{mol (HNO_{3})}{Volume (L)}](https://tex.z-dn.net/?f=%20%5BHNO_%7B3%7D%20%5D%20%3D%20%5Cfrac%7Bmol%20%28HNO_%7B3%7D%29%7D%7BVolume%20%28L%29%7D%20%20%20)
The concentration of HNO₃ is given as 5.00 x 10⁻³
Moles of HNO3 = 2.00 x 10⁻⁴
Let us plug in these values in above equation.
![5 * 10^{-3} = \frac{2.00*10^{-4}}{Volume (L)}](https://tex.z-dn.net/?f=%205%20%2A%2010%5E%7B-3%7D%20%3D%20%5Cfrac%7B2.00%2A10%5E%7B-4%7D%7D%7BVolume%20%28L%29%7D%20%20%20)
![Volume = \frac{2.00*10^{-4}}{5*10^{-3}}](https://tex.z-dn.net/?f=%20Volume%20%3D%20%5Cfrac%7B2.00%2A10%5E%7B-4%7D%7D%7B5%2A10%5E%7B-3%7D%7D%20%20%20%20)
![Volume = 0.04L](https://tex.z-dn.net/?f=%20Volume%20%3D%200.04L%20)
Volume of HNO₃ in mL = ![0.04 L * \frac{1000 mL}{1L} = 40 mL](https://tex.z-dn.net/?f=%20%200.04%20L%20%2A%20%5Cfrac%7B1000%20mL%7D%7B1L%7D%20%3D%2040%20mL%20%20)
Volume of HNO₃ needed for titration is 40 mL
Answer:
The wavelength of light is 68 nm.
Explanation:
Given data:
Binding energy of electron = 176 × 10³ Kj/ mol or (1.76× 10⁶ j/mol)
Wavelength of light require to remove the electron = ?
The number of grams present in 0.025 mole of Demerol hydrochloride is 7.094 grams.
<h3>What is Demerol hydrochloride?</h3>
It is a drug used in treating severe or moderate pain.
The moles are given = 0.025
The formula is C₁₅H₂₂CINO₂
The elements that present are C, H, Cl, N, O.
The molecular weight are given:
Carbon: 12.0107
Hydrogen: 1.00784
Chlorine: 35.453
Nitrogen: 14.007
Oxygen: 15.999
Multiply the molecular weight with moles and add them
(15 x 12.0107) + (22 x 1.00784) + (1 x 35.453) + (1 x 14.007) + (2 x 15.999) = 283.79098 grams per mole.
Now, multiply the number of moles given with the total molecular weight
283.79098 x 0.025 = 7.094
Thus, the number of grams are 7.094.
Learn more about Demerol hydrochloride
brainly.com/question/15499376
#SPJ1