Answer:
1.21 g of Tris
Explanation:
Our solution if made of a solute named Tris
Molecular weight of Tris is 121 g/mol
[Tris] = 100 mM
This is the concentration of solution:
(100 mmoles of Tris in 1 mL of solution) . 1000
Notice that mM = M . 1000 We convert from mM to M
100 mM . 1 M / 1000 mM = 0.1 M
M = molarity (moles of solute in 1 L of solution, or mmoles of solute in 1 mL of solution). Let's determine the mmoles of Tris
0.1 M = mmoles of Tris / 100 mL
mmoles of Tris = 100 mL . 0.1 M → 10 mmoles
We convert mmoles to moles → 10 mmol . 1mol / 1000mmoles = 0.010 mol
And now we determine the mass of solute, by molecular weight
0.010 mol . 121 g /mol = 1.21 g
D! It’s a pure substance and a compound
When waters salinity increases it’s freezing point will decrease. This is one of the reasons why people salt the roads after a snow storm.
32.8 g of Butane is required and 99.3 g of CO₂ is produced
<u>Explanation:</u>
The above mentioned reaction can be written as,
C₄H₁₀(g) + 13 O₂(g) → 4CO₂(g) + 5 H₂O(g) where ΔH (rxn)= -2658 kJ
It is given that 1.5 × 10³ kJ of energy is produced, the original reaction says that 2658 kJ of heat is produced, which means that less than one mole of butane is used in the reaction.
That is
of butane reacted
Now this moles is converted into mass by multiplying it with its molar mass = 0.564 mol × 58.122 g / mol
= 32.8 g of butane.
Mass of CO₂ produced = 0.564 ×44.01 g /mol × 4 mol
= 99.3 g of CO₂
Thus 32.8 g of Butane is required and 99.3 g of CO₂ is produced
A logarithmic scale is a nonlinear scale used when there is a large range of quantities
