Answer : The initial volume was, 71.2 mL
Explanation :
To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.
The equation follows:
where,
are the initial pressure, volume and temperature of the gas
are the final pressure, volume and temperature of the gas
We are given:
![P_1=0.85atm\\V_1=?\\T_1=66^oC=[66+273]K=339K\\P_2=0.60atm\\V_2=94mL\\T_2=43^oC=[43+273]K=316K](https://tex.z-dn.net/?f=P_1%3D0.85atm%5C%5CV_1%3D%3F%5C%5CT_1%3D66%5EoC%3D%5B66%2B273%5DK%3D339K%5C%5CP_2%3D0.60atm%5C%5CV_2%3D94mL%5C%5CT_2%3D43%5EoC%3D%5B43%2B273%5DK%3D316K)
Now put all the given values in above equation, we get:
Therefore, the initial volume was, 71.2 mL
I believe it's C. Amino acid is a monomer not a polymer. Q2 is B. Q3 I believe is A. Q4 is D.
Answer:
34.3 g
Explanation:
Step 1: Write the balanced equation
2 CH₃CH₂OH ⇒ CH₃CH₂OCH₂CH₃ + H₂O
Step 2: Calculate the moles corresponding to 50.0 g of CH₃CH₂OH
The molar mass of CH₃CH₂OH is 46.07 g/mol.
50.0 g × 1 mol/46.07 g = 1.09 mol
Step 3: Calculate the theoretical moles of CH₃CH₂OCH₂CH₃ produced
The molar ratio of CH₃CH₂OH to CH₃CH₂OCH₂CH₃ is 2:1. The moles of CH₃CH₂OCH₂CH₃ theoretically produced are 1/2 × 1.09 mol = 0.545 mol.
Step 4: Calculate the real moles of CH₃CH₂OCH₂CH₃ produced
The percent yield of the reaction is 85%.
0.545 mol × 85% = 0.463 mol
Step 5: Calculate the mass corresponding to 0.463 moles of CH₃CH₂OCH₂CH₃
The molar mass of CH₃CH₂OCH₂CH₃ is 74.12 g/mol.
0.463 mol × 74.12 g/mol = 34.3 g
Helium bc when u vape it goes into it lungs but also bringing helium at the same time.
