Answer is: 1,92 mol/L·s.
Chemical reaction: 2D(g) + 3E(g) + F(g) → <span>2G(g) + H(g).
</span>H is increasing at 0,64 mol/L·<span>s.
From chemical reaction n(H) : n(E) = 1 : 3.
0,64 mol : n(E) = 1 : 3.
n(E) = 1,92 mol.
</span>E is decreasing at 1,92 mol/L·s.
Answer:
0.75 g/cm³
Explanation:
Given data:
Mass of wooden block = 180 g
Length of block = 10 cm
Width of block = 6 cm
Height or thickness = 4 cm
Density of block = ?
Solution:
Volume of block = height × length × width
Volume of block = 4 cm × 10 cm× 6 cm
Volume of block = 240 cm³
Density of block:
density = mass/ volume
d = 180 g/ 240 cm³
d = 0.75 g/cm³
The mass in grams of butane at standard room temperature is 53.21 grams.
<h3>How can we determine the mass of an organic substance at room temperature?</h3>
The gram of an organic substance at room temperature can be determined by using the ideal gas equation which can be expressed as:
PV = nRT
- Pressure = 1.00 atm
- Volume = 22.4 L
- Rate = 0.0821 atm*L/mol*K
- Temperature = 25° C = 298 k
1 × 22.4 L = n × (0.0821 atm*L/mol*K× 298 K)
n = 22.4/24.4658 moles
n = 0.91556 moles
Recall that:
- number of moles = mass(in grams)/molar mass
mass of butane = 0.91556 moles × 58.12 g/mole
mass of butane = 53.21 grams
Learn more about calculating the mass of an organic substance here:
brainly.com/question/14686462
#SPJ12
Answer:
a) 88.48%
b) 0.05625 mol
Explanation:
2CH₃CH₂OH(l) → CH₃CH₂OCH₂CH₃(l) + H₂O(g) Reaction 1
CH₃CH₂OH(l) → CH₂═CH₂(g) + H₂O(g) Reaction 2
a) CH₃CH₂OH = 46.0684 g/mol
CH₃CH₂OCH₂CH₃ = 74.12 g/mol
1 mol CH₃CH₂OH ______ 46.0684 g
x ______ 50.0 g
x = 1.085 mol CH₃CH₂OH
1 mol CH₃CH₂OCH₂CH₃ ______ 74.12 g g
y ______ 35.9 g
y = 0.48 mol CH₃CH₂OCH₂CH₃
100% yield _____ 0.5425 mol CH₃CH₂OCH₂CH₃
w _____ 0.48 mol CH₃CH₂OCH₂CH₃
w = 88.48%
b) Only 0.96 mol of ethanol reacted to form diethyl ether. This means that 0.125 mol of ethanol did not react. 45% of 0.125 mol reacted to form ethylene. Therefore, 0.05625 mol of ethanol reacted by the side reaction (reaction 2). Since 1 mol of ethanol leads to 1 mol of ethylene, 0.05625 mol of ethanol produces 0.05625 mol of ethylene.
Answer:
38.9 grams of 
Explanation:
0.187 mol BaCl2 x 
0.187 m x 208 g/m
0.187 x 208 g
38.896 g --> 38.9 g BaCl2
