You have a stock solution of epinephrine at a concentration of 1 mg/mL. Knowing that the pipette you will use delivers 20 drops/mL, a. calculate the number of drops of the stock solution that must be added to a smooth muscle bath containing 25 mL of Locke’s solution so that the final concentration of epinephrine in the muscle bath will be 100 µg/mL, and
Answers
Answer : The number of drops pf the stock solution mist be added are, 50 drops.
Explanation :
As we are given that the concentration of stock solution 1 mg/mL and pipette delivers 20 drops/mL. That means,
1 mg of epinephrine = 1 mL = 20 drops
The final volume of Locke's solution = 25 mL
and the final concentration needed =
As, 1 mL of solution contains =
So, 25 mL of solution contains =
Conversion used :
Now we have to determine the number of drops needed.
As, 1 mg of epinephrine contains 20 drops.
So, 2.5 mg of epinephrine contains 2.5 × 20 = 50 drops.
Therefore, the number of drops pf the stock solution mist be added are, 50 drops.
Final answer:
If you want to achieve a final concentration of 100 µg/mL of epinephrine in a 25 mL solution, when using a stock solution of 1 mg/mL and a pipette that delivers 20 drops/mL, you need to add 50 drops of your stock solution.
Explanation:
Since we are asked to find the number of drops of stock solution required to achieve a final concentration of 100 µg/mL in a 25 mL solution, the first step is to convert the concentration of the stock solution to the same units, µg/mL. Hence, 1 mg/mL is equal to 1000 µg/mL. Further, we know that 1 mL of the stock solution contains 1000 µg of epinephrine, and our pipette delivers 20 drops/mL, so 1 drop of stock solution contains 1000 µg / 20 drops = 50 µg. Thus, if we need a 100 µg/mL concentration in 25 mL, we need a total of 100 µg/mL * 25 mL = 2500 µg of epinephrine. Therefore, to achieve this, we must add 2500 µg / 50 µg/drop = 50 drops of our stock solution. Hence,
50 drops
of the stock solution should be added to achieve the desired concentration.
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Calcium and bromine have formed a bond. Leading up to this, calcium gave up electrons. It was a(n)
Answers
Answer:
23.84g CH30H
32.81g H2O
Explanation:
We will be using the definition of mole fraction to determine the relationship between the number of moles of methanol,
CH3OH , and the number of moles of water.
But mole fraction gives the ratio between the number of moles of a component i of a solution to the total number of moles present in that solution.
CHECK THE ATTACHMENT FOR DETAILED EXPLANATION
Final answer:
In a 66.0g aqueous solution of methanol with a mole fraction of 0.290, the mass of the methanol is approximately 19.14g and the mass of the water is approximately 46.86g.
Explanation:
In this aqueous solution of methanol (CH4O), we know that its mass is 66.0g and the mole fraction of methanol is 0.290. The mole fraction is defined as the ratio of the number of moles of a component to the total number of moles of all components in the solution.
In order to find the mass of each component, namely the methanol and the water, we first need to establish that if the mole fraction of methanol is 0.290, the mole fraction of water must be 0.710 (because the total of all mole fractions in a solution is always equal to 1).
We then can set up the following proportion: mass of methanol/mass of water = mole fraction of methanol/mole fraction of water. After solving this equation, the mass of methanol will be approximately 19.14g and the mass of the water will be approximately 46.86g.
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b. Is colorless in the presence of water
c. Does not exist as a hydrate
d. Forms different hydrates which have different colors
Answers
Answer:Forms different hydrates which have different colors
Explanation:
CoCl2 in its anhydrous form is blue in colour. This anhydrous compound could absorb moisture, first forming the purple dihydrate and absorbs more water molecules to form the hexahydrate. Hence various hydrates of cobalt II chloride have different colours as stated above. Equations of reaction for the formation of the two hydrates are attached.
Calculate the molar solubility of NiS in 3.1 M NH3. g
Answers
Answer:
The molar solubility of NiS is 7.7 * 10⁻⁷ M
Explanation:
To answer this question, we need to keep in mind two equilibriums.
First, we have the solubilization of NiS:
NiS ⇄ Ni²⁺ + S²⁻ ksp= 3.0 * 10⁻²¹ (we know this from standard tables)
Second, we have the formation of the complex:
Ni²⁺ + 6NH₃ ⇄ [Ni(NH₃)₆]²⁺ kf=2.0 * 10⁻⁸
Combine the two equilibriums and we have
NiS + 6NH₃ ⇄ [Ni(NH₃)₆]²⁺ + S²⁻ K= ksp * kf =6.0* 10⁻¹³=
The molar solubility s is equal to both [Ni(NH₃)₆²⁺] and [S²]
At equilibrium, [NH₃]= 3,1 M - 6s
Thus, if we replace those terms in the formula for K, we're left with:
Using an approximation we can ignore the denominator and we have
- s²=6.0 * 10⁻¹³
- s=7.7 * 10⁻⁷
Answers
Answer:
1. The empirical formula is C₄H₅N₂O
2. The molecular formula is C₈H₁₀N₄O₂
Explanation:
The following data were obtained from the question:
Mass of compound = 200 g
Carbon (C) = 98.061 g
Hydrogen (H) = 10.381 g
Oxygen (O) = 32.956 g
Empirical formula =?
Molecular formula =?
Next, we shall determine the mass of nitrogen in the compound. This can be obtained as follow:
Nitrogen (N) = 200 – (98.061 + 10.381 + 32.956)
Nitrogen (N) = 200 – 141.398
Nitrogen (N) = 58.602 g
1. Determination of the empirical formula of the compound.
C = 98.061 g
H = 10.381 g
O = 32.956 g
N = 58.602 g
Divide by their molar masses
C = 98.061 /12 = 8.172
H = 10.381 /1 = 10.381
O = 32.956 /16 = 2.060
N = 58.602 /14 = 4.186
Divide by the smallest
C = 8.172 /2.060 = 4
H = 10.381 / 2.060 = 5
O = 2.060 / 2.060 = 1
N = 4.186 / 2.060 = 2
Thus, the empirical formula of the compound is C₄H₅N₂O
2. Determination of the molecular formula of the compound.
Empirical formula of the compound => C₄H₅N₂O
Molar mass of compound = 194.101 g/mol
Molecular formula =.?
[C₄H₅N₂O]n = 194.101
[(12×4) + (1×5) + (14×2) + 16]n = 194.101
[48 + 5 + 28 + 16]n = 194.101
97n = 194.101
Divide both side by 97
n = 194.101 /97
n = 2
Molecular formula => [C₄H₅N₂O]n
=> [C₄H₅N₂O]2
=> C₈H₁₀N₄O₂
Answers
Answer:
Explanation:
When an electron moves from a lower energy level to a higher energy level, energy is absorbed by the atom. When an electron moves from a higher to a lower energy level, energy is released and photon is emitted.
this emitted photon is depicted as a small wave-packet being expelled by the atom in a well-defined direction.
Answers
Answer:
ΔG°′ = 1.737 KJ/mol
Explanation:
The reaction involves the transfer of two electrons in the form of hydride ions from reduced coenzyme Q, CoQH₂ to fumarae to form succinate and oxidized coenzyme Q, CoQ.
The overall equation of reaction is as follows:
fumarate²⁻ + CoQH₂ ↽⇀ succinate²⁻ + CoQ ; ΔE∘′=−0.009 V
Using the equation for standard free energy change; ΔG°′ = −nFΔE°′
where n = 2; F = 96.5 KJ.V⁻¹.mol⁻¹; ΔE°′ = 0.009 V
ΔG°′ = - 2 * 96.5 KJ.V⁻¹.mol⁻¹ * 0.009 V
ΔG°′ = 1.737 KJ/mol