About 1 gram. 1 part of 50 is 2%
Answer:
109.7178g of H2O
Explanation:
First let us generate a balanced equation for the reaction. This is illustrated below:
2C3H8O + 9O2 —> 6CO2 + 8H2O
Next we will calculate the molar mass and masses of C3H8O and H20. This is illustrated below:
Molar Mass of C3H8O = (3x12.011) + (8x1.00794) + 15.9994 = 36.033 + 8.06352 + 15.9994 = 60.09592g/mol.
Mass of C3H8O from the balanced equation = 2 x 60.09592 = 120.19184g
Molar Mass of H2O = (2x1.00794) + 15.9994 = 2.01588 + 15.9994 = 18.01528g/mol
Mass of H2O from the balanced equation = 8 x 18.01528 = 144.12224g
From the equation,
120.19184g of C3H8O produced 144.12224g of H20.
Therefore, 91.5g of C3H8O will produce = (91.5 x 144.12224) /120.19184 = 109.7178g of H2O
It would cause a drop <span>but I am not sure double check other answers </span>
substitute: <span><span>t<span>1/2</span></span>=<span><span>ln(2)</span>k</span>→k=<span><span>ln(2)</span><span>t<span>1/2</span></span></span></span>
Into the appropriate equation: <span>[A<span>]t</span>=[A<span>]0</span>∗<span>e<span>−kt</span></span></span>
<span>[A<span>]t</span>=[A<span>]0</span>∗<span>e<span>−<span><span>ln(2)</span><span>t<span>1/2</span></span></span>t</span></span></span>
<span>[A<span>]t</span>=(250.0 g)∗<span>e<span>−<span><span>ln(2)</span><span>3.823 days</span></span>(7.22 days)</span></span>=67.52 g</span>
