To determine the mass of sucrose from a given volume of solution, we need to convert the volume into mass by using the density of the solution. We calculate as follows:
mass solution = 3.50 ( 1118 ) = 3913 g
mass of sucrose = 3913 g solution ( .485 g sucrose / g solution ) = 1897.805 g sucrose is present in the solution.
<span>The structural formula of 2-methylbutan-2-ol is in Word document below.
</span>2-methyl-2-butanol is organic compound and belongs to alcohols. Hydroxyl <span>functional group is on second saturated carbon atom of butane and also methyl group (-CH</span>₃) is on second saturated carbon atom of main chain (butane).<span>
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Answer:48kg of SiO2, 0.5kg of Al2O3, and 1.5kg of B2O3
Will be the final product
Explanation:
I) 96wt% of SiO2 will amount to 96/100*50 = 0.96*50=48kg of SiO2
ii) 1wt% of Al2O3 will amount to 1/100*50 = 0.01*50=0.5kg of Al2O3
III) 3wt% of B2O3 will amount to 3/100*50 = 0.03*50=1.5kg of B2O3..
The overall product form 48+ 0.5+1.5= 50kg
Explanation:
A clastic sedimentary rock is a rock that is formed from pre-existing rock materials and minerals. This materials have been transported to their new positions by the agents of denudation.
- Clastic sedimentary rocks are made up of rock fragments predominantly.
- The matrix is usually remnants of other rocks.
- A cementing material can form between clasts and holds them in place. Cements are usually iron oxides, calcite e.t.c.
- These rock types are usually formed from mechanincal weathering of rocks
- Mechanical weathering breaks down rocks into chunks.
- Examples are breccia, conglomerates, e.t.c.
Learn more:
Sedimentary rocks brainly.com/question/2740663
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We cannot solve this problem without using empirical data. These reactions have already been experimented by scientists. The standard Gibb's free energy, ΔG°, (occurring in standard temperature of 298 Kelvin) are already reported in various literature. These are the known ΔG° for the appropriate reactions.
<span>glucose-1-phosphate⟶glucose-6-phosphate ΔG∘=−7.28 kJ/mol
fructose-6-phosphate⟶glucose-6-phosphate ΔG∘=−1.67 kJ/mol
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Therefore, the reaction is a two-step process wherein glucose-6-phosphate is the intermediate product.
glucose-1-phosphate⟶glucose-6-phosphate⟶fructose-6-phosphate
In this case, you simply add the ΔG°. However, since we need the reverse of the second reaction to end up with the terminal product, fructose-6-phosphate, you'll have to take the opposite sign of ΔG°.
ΔG°,total = −7.28 kJ/mol + 1.67 kJ/mol = -5.61 kJ/mol
Then, the equation to relate ΔG° to the equilibrium constant K is
ΔG° = -RTlnK, where R is the gas constant equal to 0.008317 kJ/mol-K.
-5.61 kJ./mol = -(0.008317 kJ/mol-K)(298 K)(lnK)
lnK = 2.2635
K = e^2.2635
K = 9.62
