Explanation:
The answer for this question depends on the type of meniscus in the cylinder. If it is an upright meniscus like in water, the reading should be taken at the bottom of the meniscus. However if it is an inverted meniscus like in mercury, the reading should be taken at the top of the meniscus.
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The limiting reagent when 5 g of NaOH and 4.4 g CO₂ allowed to react will be NaOH
<h3>What is Limiting reagent ?</h3>
The limiting reactant (or limiting reagent) is the reactant that gets consumed first in a chemical reaction and therefore limits how much product can be formed.
Given chemical equation in balanced form ;
2NaOH(s) + CO₂(g) → Na₂CO₃(s) + H₂O(l).
According to the Chemical equation ;
- The limiting reagent when 5 g of NaOH and 4.4 g CO₂ allowed to react will be NaOH
If 44 g CO₂ requires 80 g of NaOH, therefore, 4.4 g CO₂ will require atleast 8 g of NaOH.
But the available quantity is 5 g NaOH. thus, NaOH is the Limiting reagent.
- 6.625 g of Na₂CO₃ are expected to be produced 5.0 g of NaOH and 4.4 g of CO₂ are allowed to react
As 80 g NaOH produces 106 g of Na₂CO₃.
Therefore 5 g NaoH will produce ;
106 / 80 x 5 = 6.625 g
Learn more about limiting reagent here ;
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Answer:
B. Ca2+ import into the ER because it has the steeper concentration gradient
Explanation:
ΔGt = RT㏑(C₂/C₁)
where ΔGt is the free energy change for transport; R = 8.315 J/mol; T = 298 K; C₂/C₁ is ratio of concentrations inside and outside each organelle.
For Ca²⁺ import
ΔGt = 8.315 J/mol * 298 K * ㏑(10⁻³/10⁻⁷)
ΔGt= 3.42 kJ/mol
For H⁺ import
ΔGt = 8.315 J/mol * 298 K * ㏑ (10⁻⁴/10⁻⁷)
ΔGt = 2.73 kJ/mol
From the above values, ΔGt is greater for Ca²⁺ import because it has a steeper concentration gradient
Answer:
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Explanation:
Answer:
The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.
Explanation:
Using Boyle's law
Given ,
V₁ = 3.6 L
V₂ = ?
P₁ = 1.0 atm
P₂ = 13.3 atm
Using above equation as:
<u>The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.</u>
