Answer:
0.544 M
Explanation:
First find the moles in the final solution
0.8 mols/L *1.7L
1.36 mols
so there is 1.36 mols in 2.5L
concentration will be 1.36/2.5
0.544 M
Answer:
The balanced equations for those dissociations are:
Ba(OH)₂(aq) → Ba²⁺(aq) + 2OH⁻ (aq)
H₂SO₄ (aq) → 2H⁺(aq) + SO₄⁻²(aq)
Explanation:
As a strong base, the barium hidroxide gives OH⁻ to the solution
As a strong acid, the sulfuric acid gives H⁺ to the solution
Ba(OH)₂, is a strong base so the dissociation is complete.
H₂SO₄ is considerd a strong acid, but only the first deprotonation is strong.
The second proton that is released, has a weak dissociation.
H₂SO₄ (aq) → H⁺(aq) + HSO₄⁻(aq)
HSO₄⁻(aq) ⇄ H⁺ (aq) + SO₄⁻² (aq) Ka
Answer:
This question is incomplete, the complete part of the question is as follows:
This best demonstrates which type of an interaction between the plants?
A. cooperation
B. parasitism
C. commensalism
D. competition
The answer is D
Explanation:
Organisms in their natural environment interact with one another in so many ways. The ways by which this interaction occurs are; competition, predation, commensalism etc.
Competition is the interaction between two organisms in which one or both organisms are harmed due to limited resources. Competition occurs when the organisms involved occupy the same niche or utilize the same limited resources.
In this question involving corn plants and milkweed plants. They are said to grow in the same area. Over several years, the milkweed plants have taken over the field and the corn plants no longer have space to grow. In this case, there is a limited space for growth, hence, the corn plant and milkweed COMPETE.
Answer:
0.0192 mL per min.
Explanation:
IV rate = 36 mg per 30 min.
IV concentration = 125 mg per 2.0 mL
36 mg per 30 min. IV rate = 36/30 = 1.2 mg per min
If 125 mg methylprednisolone is present in 2.0 mL of the IV nag, how many mL would contain 1.2 mg?
= 2x1.2/125
= 0.0192 mL
<em>Therefore, the flow rate of the IV must be </em><em>0.0192 mL per min</em><em>. in order to be able to deliver 36 mg per 30 min. </em>
T eg(1) : 97.5°C, T eg(2): 98.5°C, T eg(3): 99.2°C
∆T water(1): -2.5°C, ∆T water(2): -1.5°C, ∆T water(3): -0.8°C
∆T metal(1): 77.5°C, ∆T metal(2): 80.5°C, ∆T metal(3): 80.2°C.
ft= (m1 cp1 t1 + m2 cp2 t2 + .... + mn cpn tn) / (m1 cp1 + m2 cp2 + .... + mn cpn) (1)
where,
1000g = 1kg
ft(t eg)= final mixed temperature (°C)
m = mass of substance (kg)
cp = specific heat of substance (J/kg°C)
t = temperature of substance (°C)
