The amount of CO that would be required to generate 635 g of CO2 will be 404.14 g
<h3>Stoichiometric problem</h3>
First, let us get the equation of the reaction:
From the equation, we can see that the mole ratio of CO to that of CO2 is 1:1.
635 g of CO2 is to be generated.
Mole of 635 g CO2 = mass/molar mass = 635/44.01 = 14.43 moles
Thus, the equivalent mole of CO required will also be 14.43 moles.
Mass of 14.43 moles CO = moles x molar mass = 14.43 x 28.01 = 404.14 g
Hence, 404.14 g of CO will be required to produce 635 g of CO2
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Answer:
1.68 × 10²³ Molecules
Explanation:
As we know that 1 mole of any substance contains exactly 6.022 × 10²³ particles which is also called as Avogadro's Number. So in order to calculate the number of particles (molecules) contained by 0.280 moles of Br₂, we will use following relation,
Moles = Number of Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹
Solving for Number of Molecules,
Number of Molecules = Moles × 6.022 × 10²³ Molecules.mol⁻¹
Putting values,
Number of Molecules = 0.280 mol × 6.022 × 10²³ Molecules.mol⁻¹
Number of Molecules = 1.68 × 10²³ Molecules
Hence,
There are 1.68 × 10²³ Molecules present in 0.280 moles of Br₂.
700 mL ....................
Answer:
[∝] = +472
Explanation:
Specific rotation in a solution is defined as:
[∝] = ∝ / c×l
Where:
[∝] is specific rotation, ∝ is observed rotation (In degrees), c is concentration in g/mL and l is path length (In dm).
∝: +47.2°
c: 2.0g / 50mL = 0.04g/mL
l: 25cm × (1dm /10cm) = 2.5dm
Replacing:
[∝] = +47.2° / 0.04g/mL×2.5dm = <em>+472</em>
I hope it helps!
Answer:
All three vehicles have the same kinetic energy because they have equal speed.
